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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–1. The floor of a heavy storage warehouse building is made
of 6-in.-thick stone concrete. If the floor is a slab having a
length of 15 ft and width of 10 ft, determine the resultant force
caused by the dead load and the live load.
Solution
From Table 1–3,
DL = [12 lb>ft2 # in.(6 in.)](15 ft)(10 ft) = 10,800 lb
From Table 1–4,
LL = (250 lb>ft2)(15 ft)(10 ft) = 37, 500 lb
Total load:
F = 48,300 lb = 48.3 kAns.
Ans.
F = 48.3 k
1
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–2. The wall is 15 ft high and consists of 2 * 4 in. studs,
plastered on one side. On the other side there is 4-in. clay brick.
Determine the average load in lb>ft of length of wall that the
wall exerts on the floor.
Solution
Using the data tabulated in Table 1–3,
4@in. clay brick: (39 lb>ft2)(15 ft) = 585 lb>ft
2 * 4@in. studs plastered
on one side: (12 lb>ft2) (15 ft) = 180 lb>ft
wD = 765 lb>ft
Ans.
Ans.
wD = 765 lb>ft
2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–3. A building wall consists of 12-in. clay brick and 12 -in.
fiberboard on one side. If the wall is 10 ft high, determine the
load in pounds per foot that it exerts on the floor.
Solution
From Table 1–3,
12@in. clay brick:
(115 lb>ft2)(10 ft) = 1150 lb>ft
1>2@in. fiberboard:
(0.75 lb>ft2)(10 ft) = 7.5 lb>ft
Total: 1157.5 lb>ft = 1.16 k>ft
Ans.
Ans.
w = 1.16k>ft
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–4. The “New Jersey” barrier is commonly used during
highway construction. Determine its weight per foot of length
if it is made from plain stone concrete.
4 in.
758
12 in. 558
6 in.
24 in.
Solution
1
1
Cross@sectional area = 6(24) + a b(24 + 7.1950)(12) + a b(4 + 7.1950)(5.9620)
2
2
Use Table 1–2.
= 364.54 in2
w = 144 lb>ft3(364.54 in2) a
l ft2
b = 365 lb>ft
144 in2
Ans.
Ans.
w = 365 lb>ft
4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–5. The precast floor beam is made from concrete having a
specific weight of 23.6 kN>m3. If it is to be used for a floor of an
office building, calculate its dead and live loadings per foot
length of beam.
1.5 m
0.15 m
0.3 m
0.15 m
Solution
The dead load is caused by the self-weight of the beam.
wD = [(1.5 m)(0.15 m) + (0.15 m)(0.3 m)](23.6 kN>m3)
= 6.372 kN>m = 6.37 kN>m
Ans.
For the office, the recommended line load for design in
Table 1–4 is 2.4 kN>m2. Thus,
wL = (2.40 kN>m2)(1.5 m) = 3.60 kN>m
Ans.
Ans.
wD = 6.37 kN>m
wL = 3.60 kN>m
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–6. The floor of a light storage warehouse is made of
150-mm-thick lightweight plain concrete. If the floor is a slab
having a length of 7 m and width of 3 m, determine the resultant
force caused by the dead load and the live load.
Solution
From Table 1–3,
DL = [0.015 kN>m2 # mm (150 mm)](7 m)(3 m) = 47.25 kN
From Table 1–4,
LL = (6.00 kN>m2)(7 m)(3 m) = 126 kN
Total Load:
F = 126 kN + 47.25 kN = 173 kN
Ans.
Ans.
F = 173 kN
6
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–7. The precast inverted T-beam has the cross section
shown. Determine its weight per foot of length if it is made
from reinforced stone concrete and twelve 34 -in.-diameter
cold-formed steel reinforcing rods.
9 in.
36 in.
12 in.
48 in.
Solution
From Table 1–2, the specific weight of reinforced stone concrete and the cold-formed steel are
gC = 150 lb>ft3 and gH = 492 lb>ft3, respectively.
Reinforced stone concrete:
Cold@formed steel:
c 12a
wD = (950.09 lb>ft) a
ca
2
48
12
9
36
p 0.75
ft b a ft b + a ft b a ft b - 12a b a
ft b d (150 lb>ft)
12
12
12
12
4
12
= 931.98 lb>ft
2
18.11 lb>ft
p
0.75
ba
ft b d (492 lb>ft3) =
4
12
950.09 lb>ft
1k
b = 0.950 k>ft
1000 lb
Ans.
Ans.
wD = 0.950 k>ft
7
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–8. The hollow core panel is made from plain stone
concrete. Determine the dead weight of the panel. The holes
each have a diameter of 4 in.
7 in.
12 in.
12 in.
12 in.
12 in.
12 ft
12 in.
12 in.
Solution
From Table 1–2,
W = (144 lb>ft3)[(12 ft)(6 ft) a
2
7
2
ft b - 5(12 ft)(p) a
ft b ] = 5.29 k
12
12
Ans.
Ans.
W = 5.29 k
8
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–9. The floor of a light storage warehouse is made of
6-in.-thick cinder concrete. If the floor is a slab having a length
of 10 ft and width of 8 ft, determine the resultant force caused
by the dead load and that caused by the live load.
Solution
From Table 1–3,
DL = (6 in.)(9 lb>ft2 # in.)(8 ft)(10 ft) = 4.32 k
Ans.
From Table 1–4,
LL = (125 lb>ft2)(8 ft)(10 ft) = 10.0 k
Ans.
Ans.
DL = 4.32 k
LL = 10.0 k
9
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–10. The interior wall of a building is made from 2 * 4
wood studs, plastered on two sides. If the wall is 12 ft high,
determine the load in lb>ft of length of wall that it exerts on
the floor.
Solution
From Table 1–3,
w = (20 lb>ft2)(12 ft) = 240 lb>ft
Ans.
Ans.
w = 240 lb>ft
10
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–11. The second floor of a light manufacturing building is
constructed from a 5-in.-thick stone concrete slab with an
added 4-in. cinder concrete fill as shown. If the suspended
ceiling of the first floor consists of metal lath and gypsum
plaster, determine the dead load for design in pounds per
square foot of floor area.
4 in. cinder fill
5 in. concrete slab
ceiling
Solution
From Table 1–3,
5@in. concrete slab
= (12)(5)
= 60.0
4@in. cinder fill
= (9)(4)
= 36.0
metal lath & plaster
= 10.0
Total dead load
= 106.0 lb>ft2
Ans.
Ans.
DL = 106 lb>ft2
11
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–12. A two-story hotel has interior columns for the rooms
that are spaced 6 m apart in two perpendicular directions.
Determine the reduced live load supported by a typical interior
column on the first floor under the public rooms.
Solution
Table 1–4:
Lo = 4.79 kN>m2
AT = (6 m)(6 m) = 36 m2
KLL = 4
KLLAT = 4(36) = 144 m2 7 37.2 m2
From Eq. 1-1,
LL = Lo a0.25 +
4.57
b
1KLLAT
LL = 4.79a0.25 +
LL = 3.02 kN>m2
2
4.57
b
14(36)
Ans.
2
3.02 kN>m 7 0.4Lo = 1.916 kN>m
OK
Ans.
LL = 3.02 kN>m2
12
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–13. A four-story office building has interior columns
spaced 30 ft apart in two perpendicular directions. If the
flat-roof live loading is estimated to be 30 lb>ft2, determine the
reduced live load supported by a typical interior column
located at ground level.
Solution
From Table 1–4,
Lo = 50 psf
AT = (30)(30) = 900 ft2
KLLAT = 4(900) = 3600 ft2 7 400 ft2
From Eq. 1-1,
L = La a0.25 L = 50a0.25 % reduction =
15
b
1KLLAT
15
b = 25 psf
14(900)
25
= 50% 7 40% (OK)
50
F = 3[(25 psf)(30 ft)(30 ft)] + 30 psf(30 ft)(30 ft) = 94.5 k
Ans.
Ans.
LL = 94.5 k
13
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–14. The office building has interior columns spaced 5 m
apart in perpendicular directions. Determine the reduced live
load supported by a typical interior column located on the first
floor under the offices.
Solution
From Table 1–4,
Lo = 2.40 kN>m2
AT = (5 m)(5 m) = 25 m2
KLL = 4
L = Lo a0.25 +
4.57
b
1KLLAT
L = 2.40 a0.25 +
L = 1.70 kN>m2
2
4.57
b
14(25)
Ans.
2
1.70 kN>m 7 0.4 Lo = 0.96 kN>m
OK
Ans.
LL = 1.70 kN>m2
14
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–15. A hospital located in Chicago, Illinois, has a flat roof,
where the ground snow load is 25 lb>ft2. Determine the design
snow load on the roof of the hospital.
Solution
Ce = 1.2
Ct = 1.0
I = 1.2
pf = 0.7 CeCt Ipg
pf = 0.7(1.2)(1.0)(1.2)(25) = 25.2 lb>ft2
Ans.
Ans.
pf = 25.2 lb>ft2
15
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–16. Wind blows on the side of a fully enclosed 30-ft-high
hospital located on open flat terrain in Arizona. Determine the
design wind pressure acting over the windward wall of the building
at the heights 0–15 ft, 20 ft, and 30 ft. The roof is flat. Take Ke = 1.0.
Solution
V = 120 mi>h
Kzt = 1.0
Kd = 1.0
Ke = 1.0
qz = 0.00256 KzKztKdKeV 2
= 0.00256Kz (1.0)(1.0)(1.0)(120)2
= 36.86 Kz
From Table 1–5,
z
Kz
qz
0–15
0.85
31.33
20
0.90
33.18
25
0.94
34.65
30
0.98
36.13
Thus,
p = qGCp - qh(GCpi)
= q(0.85)(0.8) - 36.13( {0.18)
= 0.68q|6.503
p0 - 15 = 0.68(31.33) | 6.503 = 14.8 psf or 27.8 psf
Ans.
p20 = 0.68(33.18) | 6.503 = 16.1 psf or 29.1 psf
Ans.
p25 = 0.68(34.65) | 6.503 = 17.1 psf or 30.1 psf
Ans.
p30 = 0.68(36.13) | 6.503 = 18.1 psf or 31.1 psf
Ans.
Ans.
p0 - 15 = 14.8 psf or 27.8 psf
p20 = 16.1 psf or 29.1 psf
p25 = 17.1 psf or 30.1 psf
p30 = 18.1 psf or 31.1 psf
16
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–17. Wind blows on the side of the fully enclosed hospital
located on open flat terrain in Arizona. Determine the external
pressure acting on the leeward wall, if the length and width of
the building are 200 ft and the height is 30 ft.
Solution
V = 120 mi>h
Kzt = 1.0
Kd = 1.0
Ke = 1.0
qh = 0.00256KzKztKdKeV 2
= 0.00256Kz(1.0)(1.0)(1.0)(120)2
= 36.864Kz
From Table 1–5, for z = h = 30 ft, Kz = 0.98
qh = 36.864(0.98) = 36.13
From the text,
L
200
=
= 1 so that Cp = - 0.5
B
200
p = qGCp - qh(GCpi)
p = 36.13(0.85)( - 0.5) - 36.13(|0.18)
p = -21.9 psf or - 8.85 psf
Ans.
Ans.
p = -21.9 psf or - 8.85 psf
17
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–18. The light metal storage building is on open flat terrain in
central Oklahoma. If the side wall of the building is 14 ft high,
what are the two values of the design wind pressure acting on
this wall when the wind blows on the back of the building? The
roof is essentially flat and the building is fully enclosed.
Solution
V = 105 mi>h
Kzt = 1.0
Kd = 1.0
Ke = 1.0
qz = 0.00256KzKztKdKeV 2
= 0.00256Kz(1.0)(1.0)(1.0)(105)2
= 28.22Kz
From Table 1–5,
For 0 … z … 15 ft, Kz = 0.85
Thus,
qz = 28.22(0.85) = 23.99
p = qGCp - qh(GCpi)
p = 23.99(0.85)(0.7) - (23.99)( { 0.18)
p = - 9.96 psf or p = - 18.6 psf
Ans.
Ans.
p = -9.96 psf or - 18.6 psf
18
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–19. Determine the resultant force acting on the face of the
sign if qh = 25.5 lb>ft2. The sign has a width of 32 ft and a
height of 8 ft as indicated.
32 ft
8 ft
8 ft
Solution
Here, G = 0.85 since the structure that supports the sign can
be considered rigid. Since B>s = 32 ft>8 ft = 4, Table 1–6
can be used to obtain Cf . Here, s>h = 8 ft>(8 ft + 8 ft) = 0.5.
Then, Cf = 1.70.
F = qhGCfAs
= (25.5 lb>ft2)(0.85)(1.70)[(32 ft)(8 ft)]
= 9.433(103) lb
= 9.43 k
Ans.
Ans.
F = 9.43 k
19
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–20. The barn has a roof with a slope of 40 mm>m. It is located
in an open field where the ground snow load is
1.50 kN>m2. Determine the snow load that is required to design
the roof of the stall.
Solution
40 mm
b * 100%
1000 mm
= 4% 6 5%. Then the roof can be considered flat. Since
Here, the slope of the roof = a
the barn is located in an open terrain, is unheated and is an
­agricultural building, Ce = 0.8, Ct = 1.2, and Is = 0.80,
respectively. Here, pg = 1.50 kN>m2.
pf = 0.7 CeCt Is pg
= 0.7(0.8)(1.2)(0.8)(1.50 kN>m2)
= 0.8064 kN>m2 = 0.806 kN>m2
Ans.
Ans.
pf = 0.806 kN>m2
20
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–21. The stall has a flat roof with a slope of 40 mm>m.
It is located in an open field where the ground snow load is
0.84 kN>m2. Determine the snow load that is required to design
the roof of the stall.
Solution
40 mm
b * 10%
1000 mm
= 4% 6 5%. Then the roof can be considered flat. Since the
barn is located in an open terrain, is unheated and is an agricultural building, Ce = 0.8, Ct = 1.2 and Is = 0.8, respectively.
Here, pg = 0.84 kN>m2.
Here, the slope of the roof = a
pf = 0.7CeCt Is pg
= 0.7(0.8)(1.2)(0.8)(0.84 kN>m2)
= 0.4516 kN>m2 = 0.452 kN>m2
Ans.
Ans.
pf = 0.452 kN>m2
21
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–22. An urban hospital located in central Illinois has a flat
roof. Determine the snow load in kN>m2 that is required to
design the roof.
Solution
In central Illinois, pg = 0.96 kN>m2. Because the hospital is in
an urban area, Ce = 1.2.
pf = 0.7CeCt Is pg
pf = 0.7(1.2)(1.0)(1.20)(0.96)
= 0.968 kN>m2
Ans.
Ans.
pf = 0.968 kN>m2
22
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–23. The school building has a flat roof. It is located in an
open area where the ground snow load is 0.68 kN>m2.
Determine the snow load that is required to design the roof.
Solution
pf = 0.7CeCtIspg
pf = 0.7(0.8)(1.0)(1.20)(0.68)
= 0.457 kN>m2
Ans.
Ans.
pf = 0.457 kN>m2
23
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–24. Wind blows on the side of the fully enclosed
agriculture building located on open flat terrain in Oklahoma.
Determine the external pressure acting over the windward
wall, the leeward wall, and the side walls. Also, what is the
internal pressure in the building which acts on the walls? Use
linear interpolation to determine qh.
B
A
108
100 ft
wind
C
D
15 ft
50 ft
Solution
qz = 0.00256Kz Kzt Kd Ke V 2
qz = 0.00256Kz (1)(1)(1)(105)2
q15 = 0.00256(0.85)(1)(1)(1)(105)2 = 23.9904 psf
q20 = 0.00256(0.90)(1)(1)(1)(105)2 = 25.4016 psf
h = 15 +
1
(25 tan 10°) = 17.204 ft
2
qh - 23.9904
25.4016 - 23.9904
=
17.204 - 15
20 - 15
qh = 24.612 psf
External pressure on windward wall:
pmax = qz GCp = 23.990410.85210.82 = 16.3 psf
External pressure on leeward wall:
Ans.
L
50
=
= 0.5
B
100
p = qh GCp = 24.612(0.85)( -0.5) = - 10.5 psf
Ans.
External pressure on side walls:
p = qh GCp = 24.61210.85)1 -0.72 = -14.6 psf
Ans.
Internal pressure:
p = - qh(GCpi) = - 24.612(0.18) = { 4.43 psf
Ans.
Ans.
External pressure on windward wall
pmax = 16.3 psf
External pressure on leeward wall
p = -10.5 psf
External pressure on side walls
p = -14.6 psf
Internal pressure
p = {4.43 psf
24
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–25. Wind blows on the side of the fully enclosed agriculture
building located on open flat terrain in Oklahoma. Determine
the external pressure acting on the roof. Also, what is the
internal pressure in the building which acts on the roof? Use
linear interpolation to determine qh and Cp in Fig. 1–13.
B
A
108
100 ft
wind
C
D
15 ft
50 ft
Solution
qz = 0.00256Kz Kzt Kd Ke V 2
= 0.00256Kz (1)(1)(1)(105)2
q15 = 0.00256(0.85)(1)(1)(1)(105)2 = 23.9904 psf
q20 = 0.00256(0.90)(1)(1)(1)(105)2 = 25.4016 psf
1
(25 tan 10°) = 17.204 ft
2
qh - 23.9904
25.4016 - 23.9904
=
17.204 - 15
20 - 15
h = 15 +
qh = 24.612 psf
External pressure on windward side of roof:
p = qh GCp
h
17.204
=
= 0.3441
L
50
( -0.9 - Cp)
[ -0.9 - ( -0.7)]
=
(0.5 - 0.25)
(0.5 - 0.3441)
Cp = -0.7753
p = 24.612(0.85)( - 0.7753) = -16.2 psf
Ans.
External pressure on leeward side of roof:
[ -0.5 - ( -0.3)]
(0.5 - 0.25)
=
( -0.5 - Cp)
(0.5 - 0.3441)
Cp = -0.3753
p = qh GCp
= 24.612(0.85)( -0.3753) = -7.85 psf
Internal pressure:
p = - qh(GCpi) = - 24.612({ 0.18) = { 4.43 psf
Ans.
Ans.
Ans.
External pressure on windward side of roof
p = -16.2 psf
External pressure on leeward side of roof
p = -7.85 psf
Internal pressure
p = {4.43 psf
25
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–1. The steel framework is used to support the
reinforced stone concrete slab that is used for an office.
The slab is 200 mm thick. Sketch the loading that acts
along members BE and FED. Take a = 2 m, b = 5 m.
Hint: See Tables 1.2 and 1.4.
C
B
D
A
a
E
b
F
a
Solution
b
5m
=
= 2.5, the concrete slab will behave as a one-way slab.
a
2m
Thus, the tributary area for this beam is rectangular, as shown in Fig. a, and the intensity of the uniform distributed load is
Beam BE.
Since
200 mm thick reinforced stone concrete slab:
(23.6 kN>m3)(0.2 m)(2 m) = 9.44 kN>m
Live load for office: (2.40 kN>m2)(2 m) = 480 kN>m
Ans.
14.24 kN>m
Due to symmetry, the vertical reactions at B and E are
By = Ey = (14.24 kN>m)(5)>2 = 35.6 kN
The loading diagram for beam BE is shown in Fig. b.
Beam FED. The only load this beam supports is the vertical reaction of beam BE
at E, which is Ey = 35.6 kN. The loading diagram for this beam is shown in Fig. c.
Ans.
Live load for office: 14.24 kN>m
26
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2–2.
Solve Prob. 2–1 with a = 3 m, b = 4 m.
C
B
D
A
a
E
b
Solution
Beam BE.
Since
F
a
b
4
= 6 2, the concrete slab will behave as a two-way slab.
a
3
Thus, the tributary area for this beam is the hexagonal area shown in Fig. a, and the
­maximum intensity of the distributed load is:
200-mm-thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(3 m)
= 14.16 kN>m
Live load for office: [(2.40 kN>m2 )(3 m)] = 7.20 kN>m
21.36 kN>m
Ans.
Due to symmetry, the vertical reactions at B and E are
1
2c (21.36 kN>m)(1.5 m) d + (21.36 kN>m)(1 m)
2
By = Ey =
2
= 26.70 kN
The loading diagram for beam BE is shown in Fig. b.
Beam FED. The loadings that are supported by this beam are the vertical reaction of
beam BE at E, which is Ey = 26.70 kN and the triangular distributed load of which its
tributary area is the triangular area shown in Fig. a. Its maximum intensity is
200-mm-thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(1.5 m)
= 7.08 kN>m
Live load for office: (2.40 kN>m2)(1.5 m)
=
3.60 kN>m
10.68 kN>m
Ans.
The loading diagram for beam FED is shown in Fig. c.
Ans.
Live load for office:
21.36 kN>m
Live load for office:
3.60 kN>m
10.68 kN>m
27
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–3. The floor system used in a school classroom consists of a
4-in. reinforced stone concrete slab. Sketch the loading that
acts along the joist BF and side girder ABCDE. Set
a = 10 ft, b = 30 ft. Hint: See Tables 1.2 and 1.4.
E
a
a
a
a
b
D
C
B
F
A
Solution
b
30 ft
=
= 3, the concrete slab will behave as a one-way slab.
a
10 ft
Thus, the tributary area for this joist is the rectangular area shown in Fig. a, and the
intensity of the uniform distributed load is
Joist BF. Since
4@in.@thick reinforced stone concrete slab: (0.15 k>ft3) a
4
ft b (10 ft) = 0.5 k>ft
12
Live load for classroom: (0.04 k>ft2)(10 ft) = 0.4 k>ft
0.9 k>ft
Ans.
Due to symmetry, the vertical reactions at B and F are
Ans.
By = Fy = (0.9 k>ft)(30 ft)>2 = 13.5 k
The loading diagram for joist BF is shown in Fig. b.
Girder ABCDE. The loads that act on this girder are the vertical reactions of the
joists at B, C, and D, which are By = Cy = Dy = 13.5 k, and 6.75-k end loads from
the joists at A and E. The loading diagram for this girder is shown in Fig. c.
Ans.
Live load for classroom: 0.9 k>ft
By = 13.5 k
28
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–4.
Solve Prob. 2–3 with a = 10 ft, b = 15 ft.
E
a
a
a
a
b
D
C
B
A
F
Solution
b
15 ft
=
= 1.5 6 2, the concrete slab will behave as a two-way
a
10 ft
slab. Thus, the tributary area for the joist is the hexagonal area, as shown in
Fig. a, and the maximum intensity of the distributed load is:
Joist BF.
Since
4@ in.@thick reinforced stone concrete slab: (0.15 k>ft3) a
4
ft b (10 ft 2 = 0.5 k>ft
12
Live load for classroom: (0.04 k>ft2)(10 ft) = 0.4 k>ft
0.9 k>ft Ans.
Due to symmetry, the vertical reactions at B and G are
1
2c (0.9 k>ft)(5 ft) d + (0.9 k>ft)(5 ft)
2
B y = Fy =
= 4.50 k
2
Ans.
The loading diagram for beam BF is shown in Fig. b.
Girder ABCDE. The loadings that are supported by this girder are the vertical reactions of the joist at B, C and D, which are By = Cy = Dy = 4.50 k, the 2.25-k end loads
from the joists at A and E, and the triangular distributed load shown in Fig. a. Its maximum intensity is
4-in.-thick reinforced stone concrete slab:
(0.15 k>ft3) a
4
ft b (5 ft) = 0.25 k>ft
12
Live load for classroom: (0.04 k>ft2)(5 ft)
=
0.20 k>ft
0.45 k>ft
Ans.
The loading diagram for the girder ABCDE is shown in Fig. c.
Ans.
Live load for classroom: 0.9 k>ft
By = 4.50 k
Live load for classroom:
29
0.20 k>ft
0.45 k>ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–5.
Solve Prob. 2–3 with a = 7.5 ft, b = 20 ft.
E
a
a
a
a
b
D
C
B
F
A
Solution
b
20 ft
=
= 2.7 7 2, the concrete slab will behave as a one-way
a
7.5 ft
slab. Thus, the tributary area for this beam is a rectangle, as shown in Fig. a, and the
intensity of the distributed load is:
Beam BF.
Since
4@in.@thick reinforced stone concrete slab: (0.15 k>ft3) a
4
ft b (7.5 ft) = 0.375 k>ft
12
Live load from classroom: (0.04k>ft2)(7.5 ft) = 0.300 k>ft
0.675 k>ft
Ans.
Due to symmetry, the vertical reactions at B and F are
B y = Fy =
(0.675 k>ft)(20 ft)
2
Ans.
= 6.75 k
The loading diagram for beam BF is shown in Fig. b.
Beam ABCD. The loading diagram for this beam is shown in Fig. c.
Ans.
Live load from classroom: 0.675 k>ft
By = 6.75 k
30
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–6. The frame is used to support the wood deck in a
residential dwelling where the live load is 40 lb>ft2. Sketch the
loading that acts along members BG and ABCD. Set
b = 10 ft, a = 5 ft.
D
C
B
E
A
F
a
G
b
a
H
a
Solution
From Table 1–4,
LL = 40 psf
L2
b
10
=
=
= 2 … 2
L1
a
5
Two-way slab:
Reaction at B: 750 lb c
Reaction at A: 1.5 k c
Ans.
Ans.
Ans.
Reaction at B: 750 lb c
Reaction at A: 1.5 k c
31
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–7.
Solve Prob. 2–6 if b = 8 ft, a = 8 ft.
D
C
B
E
A
F
a
G
b
a
H
a
Solution
From Table 1–4,
LL = 40 psf
L2
b
8
=
= = 1 … 2
L1
a
8
Two-way slab:
Reaction at B: 640 lb c Ans.
Reaction at A: 1920 lb c Ans.
Ans.
Reaction at B: 640 lb c
Reaction at A: 1920 lb c
32
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*2–8.
Solve Prob. 2–6 if b = 15 ft, a = 10 ft.
D
C
B
E
A
F
a
G
b
a
H
a
Solution
From Table 1.4,
LL = 40 psf
b>a = 15>10 = 1.5 … 2
Two-way slab:
Reaction at B: 2 k c
Reaction at A: 4.5 k c
Ans.
Ans.
Ans.
Reaction at B: 2 k c
Reaction at A: 4.5 k c
33
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
D
2–9. The steel framework is used to support the 4-in.
reinforced stone concrete slab that carries a uniform live
loading of 400 lb>ft2. Sketch the loading that acts along
members BE and FED. Set a = 9 ft, b = 12 ft. Hint: See
Table 1.2.
E
C
F
a
B
a
b
A
Solution
b
12 ft
4
=
= 6 2, the concrete slab will behave as a two-way
a
9 ft
3
slab. Thus, the tributary area for this beam is the shaded octagonal area shown in
Fig. a, and the maximum intensity of the trapezoidal distributed load is:
Beam BE. Since
4@in.@thick reinforced stone concrete slab: (0.15 k>ft3) a
4
ft b (9 ft) = 0.45 k>ft
12
Floor live load: (0.4 k>ft2)(9 ft) =
3.60 k>ft
4.05 k>ft
Ans.
Due to symmetry, the vertical reactions at B and E are
1
(4.05 k>ft)(3 ft + 12 ft)
2
By = Ey =
= 15.2 k
2
Ans.
The loading diagram for beam BE is shown in Fig. a.
Beam FED. The loadings that are supported by this beam are the vertical reactions
of beam BE at E, which is Ey = 15.19 k and the triangular distributed load contributed
by dotted triangular tributary area shown in Fig. a. Its maximum intensity is
4@in.@thick concrete slab: (0.15 k>ft3) a
4
ft b (4.5 ft) = 0.225 k>ft
12
Floor live load: (0.4 k>ft2)(4.5 ft) =
1.800 k>ft
2.025 k>ft
Ans.
The loading diagram for beam FED is shown in Fig. c.
Ans.
Floor live load:
3.60 k>ft
4.05 k>ft
By = Ey = 15.2 k
Beam BE.
Beam FED.
w max = 4.05 k>ft
Floor live load:
15.2 k at E, w max = 2.025 k>ft
34
1.800 k>ft
2.025 k>ft
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2–10.
Solve Prob. 2–9, with a = 6 ft, b = 18 ft.
D
E
C
F
a
B
a
b
A
Solution
b
18 ft
=
= 3 7 2, the concrete slab will behave as a one-way
a
6 ft
slab. Thus, the tributary area for this beam is the shaded rectangular area shown in
Fig. a, and the intensity of the uniform distributed load is:
Beam BE.
Since
4@in.@thick reinforced stone concrete slab: (0.15 k>ft3) a
4
ft b(6 ft) = 0.30 k>ft
12
Floor live load: (0.4 k>ft2)(6 ft) =
2.40 k>ft
2.70 k>ft
Ans.
Due to symmetry, the vertical reactions at B and E are
By = Ey =
(2.70 k>ft)(18 ft)
2
= 24.3 k
Ans.
The loading diagram of this beam BE is shown in Fig. b.
Beam FED. The only load this beam supports is the vertical
reaction of beam BE at E, which is Ey = 24.3 k.
The loading diagram of beam FED is shown in Fig. c.
Ans.
Floor live load:
2.40 k>ft
2.70 k>ft
By = Ey = 24.3 k
35
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–11. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify the
degree of indeterminacy.
Solution
(a) r = 2
3n = 3(1) = 3
r 6 3n
Unstable
(b) r = 4
3n = 3(1) = 3
r - 3n = 4 - 1 = 1
Stable and statically indeterminate to the first degree
(c) r = 9
3n = 3(3) = 9
r = 3n
Stable and statically determinate
(d) r = 8
3n = 3(2) = 6
r - 3n = 8 - 6 = 2
Stable and statically indeterminate to the second degree
(e) r = 7
3n = 3(2) = 6
r - 3n = 7 - 6 = 1
Stable and statically indeterminate to the first degree
(a)
Ans.
Ans.
Ans.
(b)
Ans.
Ans.
(c)
(d)
(e)
Ans.
(a) Unstable
(b) Stable and statically indeterminate to the first degree
(c) Stable and statically determinate
(d) Stable and statically indeterminate to the second degree
(e) Stable and statically indeterminate to the first degree
36
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*2–12. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify the
degree of indeterminacy.
Solution
(b)
(a)
(a) r 7 3n
4 7 3(1)
Statically indeterminate to the first degree.
Ans.
(b) Parallel reactions
Ans.
Unstable.
(c) r 7 3n
6 7 3(1)
Statically indeterminate to the third degree.
Ans.
(c)
(d)
(d) Parallel reactions
Ans.
Unstable.
(e)
(f)
(g)
(h)
Ans.
(a) Statically indeterminate to the first degree
(b) Unstable
(c) Statically indeterminate to the first degree
(d) Unstable
37
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2–13. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify the
degree of indeterminacy. The supports or connections are to be
assumed as stated.
pin
fixed
pin
(a)
(a) r = 7
3n = 3(2) = 6
3n = 3(2) = 6
(c) r = 4
3n = 3(1) = 3
Ans.
r = 3n
Stable and statically determinate
roller
roller
pin
Ans.
(c)
r - 3n = 4 - 3 = 1
Stable and statically indeterminate to first degree
roller
(b)
r - 3n = 7 - 6 = 1
Stable and statically indeterminate to first degree.
(b) r = 6
pin
fixed
Solution
Ans.
Ans.
(a) Stable and statically indeterminate to first degree
(b) Stable and statically determinate
(c) Stable and statically indeterminate to first degree
38
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–14. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify
the degree of indeterminacy. The supports or connections are
to be assumed as stated.
pin
Solution
(a) r = 5
fixed
3n = 3(2) = 6
r 6 3n
(a)
Ans.
Unstable
(b) r = 9
pin
rocker
3n = 3(3) = 9
r = 3n
Stable and statically determinate
(c) r = 8
Ans.
3n = 3(2) = 6
r - 3n = 8 - 6 = 2
Stable and statically indeterminate to the second degree
Ans.
roller
pin
roller
fixed
pin
(b)
fixed
pin
fixed
fixed
(c)
Ans.
(a) Unstable
(b) Stable and statically determinate
(c) Stable and statically indeterminate to the second degree
39
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2–15. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify
the degree of indeterminacy.
(a)
Solution
Ans.
(a) Since the lines of action of the reactive forces are concurrent, the structure is unstable.
(b) r = 6
3n = 3(2) = 6
r = 3n
Ans.
Stable and statically determinate
(c) r = 3
(b)
3n = 3(1) = 3
r = 3n
Ans.
Stable and statically determinate
(c)
Ans.
(a) Unstable
(b) Stable and statically determinate
(c) Stable and statically determinate
40
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–16. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify
the degree of indeterminacy.
(a)
Solution
(a) r = 6
3n = 3(1) = 3
r - 3n = 6 - 3 = 3
Stable and statically indeterminate to the third degree
(b) r = 5
Ans.
(b)
3n = 3(1) = 3
r - 3n = 5 - 3 = 2
Stable and statically indeterminate to the second degree
(c) r = 5
Ans.
3n = 3(1) = 3
r - 3n = 5 - 3 = 2
Stable and statically indeterminate to the second degree
(d) r = 6
Ans.
3n = 3(2) = 6
(c)
r = 3n
Stable and statically determinate.
Ans.
(d)
Ans.
(a) Stable and statically indeterminate to the third degree
(b) Stable and statically indeterminate to the second degree
(c) Stable and statically indeterminate to the second degree
(d) Stable and statically determinate
41
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2–17. Classify each of the structures as statically determinate,
statically indeterminate, or unstable. If indeterminate, specify
the degree of indeterminacy.
Solution
(a)
(a) r = 2
3n = 3(1) = 3
r 6 3n
Unstable.
(b) r = 12
3n = 3(2) = 6
r 7 3n
r - 3n = 12 - 6 = 6
Statically indeterminate to the sixth degree.
(c) r = 6
3n = 3(2) = 6
r = 3n
Stable and statically determinate.
(d) Unstable since the lines of action of the reactive force
components are concurrent.
Ans.
Ans.
Ans.
Ans.
(b)
(c)
(d)
Ans.
(a) Unstable
(b) Statically indeterminate to the sixth degree
(c) Stable and statically determinate
(d) Unstable since the lines of action of the reactive force
components are concurrent
42
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–18.
Determine the reactions on the beam.
12 k
6 k>ft
3 k>ft
A
B
12 ft
12 ft
Solution
Equations of Equilibrium. Referring to the FBD of the beam
shown in Fig. a, NA and By can be determined directly by writing the moment equations of equilibrium about points B and
A, respectively.
a + ΣMB = 0;
1
(3)(12)(4) + 3(12)(6) + 12(12) - NA(24) = 0
2
Ans.
NA = 18.0 k
a + ΣMA = 0;
By(24) - 12(12) - 3(12)(18) -
1
(3)(12)(20) = 0
2
By = 48.0 k
Ans.
Write the force equation of equilibrium along the x-axis.
+
S ΣFx = 0,
bx = 0
Ans.
Ans.
NA = 18.0 k
By = 48.0 k
Bx = 0
43
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–19.
Determine the reactions at the supports.
3 k>ft
3 k>ft
A
B
9 ft
9 ft
9 ft
Solution
Equations of Equilibrium. Referring to the FBD of the beam
shown in Fig. a, NB and Ay can be determined directly by writing the moment equations of equilibrium about points A and
B, respectively.
a + ΣMA = 0;
NB(18) -
1
1
(3)(18)(9) - (3)(9)(24) = 0
2
2
Ans.
NB = 31.5 k
a + ΣMB = 0;
1
1
(3)(18)(9) - (3)(9)(6) - Ay(18) = 0
2
2
Ay = 9.00 k
Ans.
Write the force equation of equilibrium along the x axis.
+
S ΣFx = 0;
Ax = 0
Ans.
Ans.
NB = 31.5 k
Ay = 9.00 k
Ax = 0
44
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–20.
Determine the reactions on the beam.
10 k
12 k
1 k>ft
A
3 k>ft
5 ft
B
4 ft
4 ft
4 ft
Solution
Equations of Equilibrium. Referring to the FBD of the beam
shown in Fig. a, NA can be obtained directly by writing the moment
equation of equilibrium about point B.
a + ΣMB = 0;
10(8.667) + B(6.50) + 25(4.333) - NA(13) = 0
Ans.
NA = 21.5 k
Using this result to write the force equation of equilibrium along
the x and y axis,
+
S ΣFx = 0;
Bx + 21.5a
Bx = 10.19 k = 10.2 k
b ΣFy = 0;
By + 21.5a
By = 24.46 k = 24.5 k
5
5
b - (10 + 13 + 25) a b = 0
13
13
Ans.
12
12
b - (10 + 13 + 25) a b = 0
13
13
Ans.
Ans.
NA = 21.5 k
Bx = 10.2 k
By = 24.5 k
45
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–21. Determine the reactions at the supports A and B of the
compound beam. There is a pin at C.
4 kN>m
A
C
6m
B
4.5 m
Solution
Member AC:
a + ΣMC = 0; -Ay(6) + 12(2) = 0
Ay = 4.00 kN
Ans.
+ c ΣFy = 0; Cy + 4.00 - 12 = 0
Cy = 8.00 kN
+
S ΣFx = 0; Cx = 0
Member CB:
a + ΣMB = 0; - MB + 8.00(4.5) + 9(3) = 0
MB = 63.0 kN # m
Ans.
+ c ΣFy = 0; By - 8 - 9 = 0
By = 17.0 kN
Ans.
+
S ΣFx = 0; Bx = 0
Ans.
Ans.
Ay = 4.00 kN
MB = 63.0 kN # m
By = 17.0 kN
Bx = 0
46
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–22.
Determine the reactions at the supports.
200 lb>ft
100 lb>ft
A
B
5 ft
4 ft
Solution
+
S ΣFx = 0;
a + ΣMB = 0;
Ax = 0
900(4.5) + 200(1.333) - Ay(9) = 0
Ay = 480 lb
+ c ΣFy = 0;
Ans.
Ans.
480 - 1100 + By = 0
By = 620 lb
Ans.
Ans.
Ax = 0
Ay = 480 lb
By = 620 lb
47
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–23. Determine the reactions at the supports A and C of
the compound beam. Assume C is fixed, B is a pin, and A is
a roller.
20 kN
15 kN>m
A
B
3m
C
1m
1m
Solution
Equations of Equilibrium. First consider the equilibrium of the FBD of segment AB
in Fig. a. NA and By can be determined directly by writing the moment equations of
equilibrium about points B and A, respectively.
a + ΣMB = 0;
1
(15)(3)(2) - NA(3) = 0 NA = 15.0 kN
2
a + ΣMA = 0; By(3) -
Ans.
1
(15)(3)(1) = 0 By = 7.50 kN
2
Write the force equation of equilibrium along the x axis.
+
S ΣFx = 0; Bx = 0.
Then consider the equilibrium of the FBD of segment BC using the results of Bx
and By.
+
S ΣFx = 0; Cx = 0
+ c ΣFy = 0; Cy - 20 - 7.50 = 0 Cy = 27.5 kN
Ans.
a + ΣMC = 0; 7.50(2) + 20(1) - MC = 0 MC = 35.0 kN # m
Ans.
Ans.
NA = 15.0 kN
Cx = 0
Cy = 27.5 kN
MC = 35.0 kN # m
48
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–24. Determine the reactions on the beam. The support at
B can be assumed to be a roller.
2 k>ft
A
B
12 ft
12 ft
Solution
Equations of Equilibrium:
a + ΣMA = 0;
a + ΣMB = 0;
+
S ΣFx = 0;
1
(2)(12)(16) = 0
2
NB = 14.0 k
1
(2)(12)(8) + 2(12)(18) - Ay(24) = 0
2
Ay = 22.0 k
NB(24) - 2(12)(6) -
Ax = 0
Ans.
NB = 14.0 k
Ay = 22.0 k
Ax = 0
49
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2–25. Determine the horizontal and vertical components of
reaction at the pins A and C.
A
12 kN>m
608
60 kN ?m
C
B
6m
Solution
Equations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, FAB
and Cy can be determined directly by writing the moment equations of equilibrium
about points C and B, respectively,
a + ΣMC = 0;
a + ΣMB = 0;
1
(12)(6)(2) + 60 - (FAB sin 60°)(6) = 0
2
FAB = 25.40 kN
Cy(6) + 60 -
1
(12)(6)(4) = 0
2
Cy = 14.0 kN
Ans.
Using the result of FAB to write the force equation of equilibrium along the x axis,
+
S Σ Fx = 0; 25.40 cos 60° - Cx = 0 Cx = 12.70 kN = 12.7 kN
Ans.
Referring to the FBD of pin A, Fig. b, the force equations of equilibrium written along
the x and y axis give
+
S ΣFx = 0; Ax - 25.40 cos 60° = 0 Ax = 12.70 kN = 12.7 kN
Ans.
Ans.
+ c ΣFy = 0; Ay - 25.40 sin 60° = 0 Ay = 22.0 kN
Ans.
Cy = 14.0 kN
Cx = 12.7 kN
Ax = 12.7 kN
Ay = 22.0 kN
50
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–26. Determine the reactions at the truss supports A
and B. The distributed loading is caused by wind.
600 lb>ft
400 lb>ft
20 ft
A
B
48 ft
48 ft
Solution
a + ΣMA = 0; By(96) + a
-a
12
5
b20.8(72) - a b20.8(10)
13
13
12
5
b31.2(24) - a b31.2(10) = 0
13
13
By = 5.117 k = 5.12 k
+ c ΣFy = 0; Ay - 5.117 + a
Ay = 14.7 k
+
S ΣFx = 0;
- Bx + a
Ans.
12
12
b20.8 - a b31.2 = 0
13
13
Ans.
5
5
b31.2 + a b20.8 = 0
13
13
Ans.
Bx = 20.0 k
Ans.
By = 5.12 k
Ay = 14.7 k
Bx = 20.0 k
51
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6 kN
2–27. The compound beam is fixed at E and supported by
rockers at A and B. There are hinges (pins) at C and D.
Determine the reactions at the supports. The 4-kN load is
applied just to the right of the pin at D.
4 kN
6 kN>m
B
A
2m
2m
D
C
2m
3m
E
3m
Solution
Equations of Equilibrium.
Fig. b, first
a + ΣMC = 0;
a + ΣMD = 0;
+
S ΣFx = 0;
Consider the equilibrium of the FBD of segment CD,
1
(6)(3)(2) = 0
2
Dy = 6.00 kN
1
(6)(3)(1) - Cy(3) = 0
2
Cy = 3.00 kN
Dy(3) -
Dx - Cx = 0
(1)
Then, using the result of Cy, the equilibrium of the FBD of segment ABC, Fig. a, gives
a + ΣMA = 0;
a + ΣMB = 0;
+
S ΣFx = 0;
Then from Eq (1),
NB(4) - 6(2) - 3.00(6) = 0
NB = 7.50 kN
Ans.
6(2) - 3.00(2) - NA(4) = 0
NA = 150 kN
Ans.
Cx = 0
Dx = 0
Finally, using the results of Dx and Dy, the equilibrium of the FBD of segment DE gives
+
S ΣFx = 0;
Ex = 0
+ c ΣFy = 0;
a + ΣME = 0;
Ey - 4 - 6.00 = 0
Ey = 10.0 kN
ME = 30.0 kN # m
4(3) + 6.00(3) - ME = 0
Ans.
Ans.
Ans.
NB = 7.50 kN
NA = 150 kN
Ex = 0
Ey = 10.0 kN
ME = 30.0 kN # m
52
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–28.
Determine the reactions on the beam.
20 kN
20 kN
20 kN
20 kN
A
608
B
3m
3m
3m
3m
3m
8 kN
Solution
a + ΣMA = 0;
- 20 kN(3 m) - 20 kN(6 m) - 20 kN(9 m) - 20 kN(12 m)
- 8 kN(sin 60°)(15 m) + By(9 m) = 0
+
S ΣFx = 0;
Ans.
By = 78.2 kN
- Ax + 8 kN(cos 60°) = 0
Ans.
Ax = 4 kN
+ c ΣFy = 0;
- 20 kN - 20 kN - 20 kN - 20 kN - 8 kN(sin 60°)
Ay = 8.71 kN
+ 78.2 kN + Ay = 0
Ans.
Ans.
By = 78.2 kN
Ax = 4 kN
Ay = 8.71 kN
53
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–29. The construction features of a cantilever truss bridge are
shown in the figure. Here it can be seen that the center
truss CD is suspended by the cantilever arms ABC and DEF.
C and D are pins. Determine the vertical reactions at the supports
A, B, E, and F if a 15-k load is applied to the center truss.
A
C
B
200 ft
100 ft 100 ft
15 k
200 ft
D
E
150 ft
F
150 ft
Solution
Truss CD:
a + ΣMD = 0;
+ c ΣFy = 0;
15(200) - Cy(300) = 0
Cy = 10.0 k
- 15 + 10 + Dy = 0
Dy = 5.0 k
Truss ABC:
a + ΣMA = 0;
+ c ΣFy = 0;
Truss DEF:
By(200) - 10(300) = 0
By = 15.0 k c Ans.
15 - 10 - Ay = 0
Ay = 5.0 kT Ans.
a + ΣMF = 0;
+ c ΣFy = 0;
5(300) - Ey(150) = 0
Ey = 10.0 k c Ans.
-5 + 10 - Fy = 0
Fy = 5.0 kT Ans.
Ans.
By = 15.0 k c
Ay = 5.00 kT
Ey = 10.0 k c
Fy = 5.00 kT
54
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–30. Determine the reactions at the supports A and C of the
compound beam. Assume C is a roller, B is a pin, and A is
fixed.
12 kN>m
C
B
A
3m
6m
Solution
Equations of Equilibrium. Consider the equilibrium of the FBD of segment BC, Fig. b, first.
a + ΣMB = 0;
a + ΣMC = 0;
+
S ΣFx = 0;
1
(12)(6)(2) = 0
2
NC = 12.0 kN
1
(12)(6)(4) - By(6) = 0
2
By = 24.0 kN
Nc(6) -
Ans.
Bx = 0
Using the results of Bx and By, the equilibrium of the FBD of segment AB, Fig. a, gives
+
S
ΣFx = 0;
Ax = 0
+ c ΣFy = 0;
Ay - 12(3) - 24.0 = 0
a + ΣMA = 0;
MA - 12(3)(1.5) - 24.0(3) = 0
Ans.
Ay = 60.0 kN
Ans.
MA = 126 kN # m
Ans.
Ans.
NC = 12.0 kN
Ax = 0
Ay = 60.0 kN
MA = 126 kN # m
55
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
2–31. The beam is subjected to the two concentrated loads as
shown. Assuming that the foundation exerts a linearly varying
load distribution on its bottom, determine the load intensities
w1 and w2 for equilibrium (a) in terms of the parameters
shown; (b) set P = 500 lb, L = 12 ft.
L
__
3
2P
L
__
3
L
__
3
w1
w2
Solution
Equations of Equilibrium: The load intensity w1 can be determined directly by s­ umming
moments about point A.
a + ΣMA = 0;
Pa
L
L
b - w1L a b = 0
3
6
w1 =
If P = 500 lb and L = 12 ft,
w2 =
Ans.
1
2P
2P
aw2 bL +
(L) - 3P = 0
2
L
L
+ c ΣFy = 0;
w1 =
2P
L
2(500)
12
4(500)
12
w2 = a
4P
b
L
Ans.
= 83.3 lb>ftAns.
Ans.
= 167 lb>ft
Ans.
2P
L
4P
w2 =
L
w1 = 83.3 lb>ft
w2 = 167 lb>ft
w1 =
56
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*2–32. Determine the horizontal and vertical components of
reaction at the supports A and C. Assume the members are pin
connected at A, B, and C.
6 kN
3 kN>m
A
0.3 m
B
3m
C
4m
2m
Solution
Equations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, FBC can be
obtained directly by writing the moment equation of equilibrium about point A.
4
3
a + ΣMA = 0; FBC a b(0.3) + FBC a b(4) - 3(4)(2) - 6(6) = 0
5
5
FBC = 22.73 kN
Using the result of FBC to write the force equation of equilibrium along the x and y axis,
+
S ΣFx = 0;
4
22.73a b - Ax = 0 Ax = 18.18 kN = 18.2 kN
5
3
+ c ΣFy = 0; Ay + 22.73a b - 3(4) - 6 = 0 Ay = 4.364 kN = 4.36 kN
5
Ans.
Ans.
Using the result of FBC, write the force equation of equilibrium along the x and y axis by
referring to the FBD of pin C, Fig. b,
4
+
S ΣFx = 0; Cx - 22.73a b = 0 Cx = 18.18 kN = 18.2 kN
5
3
+ c ΣFy = 0; Cy - 22.73a b = 0 Cy = 13.64 kN = 13.6 kN
5
Ans.
Ans.
Ans.
Ax = 18.2 kN
Ay = 4.36 kN
Cx = 18.2 kN
Cy = 13.6 kN
57
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2–33. Determine the horizontal and vertical components of
reaction at the supports A and C.
6 kN >m
B
4m
A
C
3m
Solution
Equations of Equilibrium. Member BC is a two-force member, which is reflected
in the FBD diagram of member AB, Fig. a. FBC Ax can be determined directly by
writing the moment equations of equilibrium about A and B, respectively.
a + ΣMA = 0;
a + ΣMB = 0;
3
FBC a b(4) - 24(2) = 0
5
24(2) - Ax(4) = 0
FBC = 20.0 kN
Ax = 12.0 kN
Ans.
Write the force equation of equilibrium along the y axis using the result of FBC.
c + ΣFy = 0;
4
20.0a b - Ay = 0
5
Ay = 16.0 kN
Ans.
Cx = 12.0 kN
Ans.
Cy = 16.0 kN
Ans.
Then consider the FBD of pin at C, Fig. b.
+
S ΣFx = 0;
+ c ΣFy = 0;
3
20.0a b - Cx = 0
5
4
Cy - 20.0a b = 0
5
Ans.
Ax = 12.0 kN
Ay = 16.0 kN
Cx = 12.0 kN
Cy = 16.0 kN
58
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2–34. Determine the horizontal and vertical components of
force at the connections A, B, and C. Assume each of these
connections is a pin.
4k
4 ft
400 lb>ft
C
B
3k
6 ft
3 ft
C
12 ft
A
Solution
Member AB:
a + ΣMA = 0;
+
d ΣFx = 0;
Bx(12) - 4.8(6) = 0
Bx = 2.40 k
Ax + 2.4 - 4.8 = 0
Ax = 2.40 k d + c ΣFy = 0;
Ans.
Ans.
Ay - By = 0(1)
Member BC:
a + ΣMC = 0;
- By(13) + 4(9) + 3(3) = 0
By = 3.46 k
+ c ΣFy = 0;
+
d ΣFx = 0;
Ans.
Cy + 3.46 - 4 - 3 = 0
Cy = 3.54 k c Ans.
Cx - 2.40 = 0
Cx = 2.40 k d Ans.
Ay = 3.46 k c Ans.
From Eq. (1),
Ans.
Bx = 2.40 k
Ax = 2.40 k d
By = 3.46 k
Cy = 3.54 k c
Cx = 2.40 k d
Ay = 3.46 k c
59
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5k
2–35. Determine the reactions at the supports A and B of the
frame. Assume that the support at A is a roller.
7k
8 ft
10 k
2k
6 ft
6 ft
A
8 ft
0.5 k
6 ft
B
Solution
b + ΣMB = 0;
- (0.5)(6) + (2)(6) - (7)(6) - (5)(14) + Ay (14) = 0
Ay = 7.36 k c + c ΣFy = 0;
Ans.
7.36 - 5 - 7 - 10 - 2 + By = 0
By = 16.6 k c +
S ΣFx = 0;
Ans.
- 0.5 + Bx = 0
Bx = 0.500 k S Ans.
Ans.
Ay = 7.36 k c
By = 16.6 k c
Bx = 0.500 k S
60
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–36. Determine the resultant forces at pins B and C on
member ABC of the four-member frame.
5 ft
2 ft
150 lb>ft
A
B
C
4 ft
F
E
2 ft
D
5 ft
Solution
a + a MA = 0;
- 150(7)(3.5) +
a + a MF = 0;
FCD(7) -
4
F (5) - FCD(7) = 0(1)
5 BE
4
F (2) = 0(2)
5 BE
Solving Eqs. (1) and (2) simultaneously,
FBE = 1531 lb = 1.53 k (C)
Ans.
FCD = 350 lb (T)
Ans.
Ans.
FBE = 1.53 k (C)
FCD = 350 lb (T)
61
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–37. Determine the horizontal and vertical reactions at
A and C of the two-member frame.
2 kN >m
A
B
3m
C
4m
Solution
Equations of Equilibrium.
segment AB, Fig. a.
a + ΣMA = 0;
12 kN >m
First, consider the equilibrium of the FBD of
a + ΣMB = 0;
By(4) - 2(4)(2) = 0
By = 4.00 kN
2(4)(2) - Ay(4) = 0
Ay = 4.00 kN
+
S ΣFx = 0;
Ax - Bx = 0(1)
Ans.
Then, using the result of By, the equilibrium of the FBD of Segment BC,
Fig. b, gives
a + ΣMC = 0;
+
S ΣFx = 0;
+ c ΣFy = 0;
1
(12)(3)(1) - 4.00(4) - Bx(3) = 0
Bx = 0.6667 kN
2
1
Cx + 0.6667 - (12)(3) = 0
Cx = 17.33 kN = 17.3 kN
2
Cy - 4.00 = 0
Ans.
Ans.
Cy = 4.00 kN
From Eq. (1),
Ans.
Ax = 0.6667 kN = 0.667 kN
Ans.
Ay = 4.00 kN
Cx = 17.3 kN
Cy = 4.00 kN
Ax = 0.667 kN
62
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–38. The frame supports a load of 600 lb. Determine the
horizontal and vertical components of reaction at the pins A
and D. Also, what is the force in the cable?
D
6 ft
1.5 ft
C
B
Solution
6 ft
Equations of Equilibrium. First, consider the equilibrium of
pulley E, Fig. a.
6 ft
A
E
+ c ΣFy = 0; 2T - 600 = 0 T = 300 lb
Ans.
Then, consider the equilibrium of member ABC, Fig. b.
a + ΣMA = 0; 300(12) + 300(13.5) - 300(1.5) - By(6) = 0
600 lb
By = 1200 lb
a + ΣMB = 0; 300(6) + 300(7.5) - 300(1.5) - Ay(6) = 0
Ay = 600 lb
+
S ΣFx = 0; 300 + Bx - Ax = 0
Ans.
(1)
Finally, consider the equilibrium of member BD, Fig. c.
a + ΣND = 0; 1200(6) - 300(4.5) - Bx(6) = 0 Bx = 975 lb
+
S ΣFx = 0; Dx - 300 - 975 = 0 Dx = 1275 lb
Ans.
+ c ΣFy = 0; Dy - 1200 = 0 Dy = 1200 lb
Ans.
Substitute the result of Bx into Eq (1).
Ans.
300 + 975 - Ax = 0 Ax = 1275 lb
Ans.
T = 300 lb
Ay = 600 lb
Dx = 1275 lb
Dy = 1200 lb
Ax = 1275 lb
63
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–39. Determine the horizontal and vertical force
components that the pin supports at C and D exert on members
AC and BD, respectively.
6 kN>m
3 kN>m
A
C
2.25 m
D
B
4 kN>m
3m
Solution
Equations of Equilibrium. The equilibrium of the FBD of
member shown in Fig. a will be considered first.
a + ΣMC = 0; 3(3)(1.5) +
a + ΣMA = 0; Cy(3) +
S ΣFx = 0; Cx = 0
1
(3)(3)(2) - FAB(3) = 0 FAB = 7.50 kN
2
1
(3)(3)(1) - 3(3)(1.5) = 0 Cy = 6.00 kN
2
Ans.
Ans.
Then using the result of FAB to consider the equilibrium of the
FBD of member BD shown in Fig. b,
3
a + ΣMD = 0; 7.50(3) - 4(3)(1.5) - FBC a b(3) = 0 FBC = 2.5 kN
5
a + ΣMB = 0;
- Dy(3) + 4(3)(1.5) = 0 Dy = 6.00 kN
4
+
S ΣFx = 0; 2.5a b - Dx = 0 Dx = 2.00 kN
5
Ans.
Ans.
Ans.
Cy = 6.00 kN
Cx = 0
Dy = 6.00 kN
Dx = 2.00 kN
64
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*2–40. Determine the reactions at the supports A and D.
Assume A is fixed and B, C, and D are pins.
w
w
B
C
L
A
D
1.5L
Solution
Member BC:
a + a MB = 0; Cy(1.5L) - (1.5wL) a
Cy = 0.75 wL
1.5L
b = 0
2
+ c a Fy = 0; By - 1.5wL + 0.75wL = 0
By = 0.75wL
Member CD:
a + a MD = 0; Cx = 0
+
S ΣFx = 0; Dx = 0
+ c a Fy = 0; Dy - 0.75wL = 0
Dy = 0.75wL c Ans.
Ans.
Member BC:
+
S ΣFx = 0; Bx - 0 = 0; Bx = 0
0.75 wL
Member AB:
+
S ΣFx = 0; wL - Ax = 0
Ax = wL d Ans.
Ay = 0.75wL c Ans.
wL2
b B
2
Ans.
+ c a Fy = 0; Ay - 0.75wL = 0
1
a + a MA = 0; MA - wL a b = 0
2
MA = a
Ans.
Dx = 0
Dy = 0.75wL c
Ax = wL d
Ay = 0.75wL c
65
MA = a
wL2
bB
2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–41. Determine the components of reaction at the pinned
supports A and C of the two-member frame. Neglect the
thickness of the members. Assume B is a pin.
6 kN>m
6 kN>m
B
2 kN>m
2.5 m
4m
C
A
6m
Solution
Equations of Equilibrium. Referring to the FBD of members AB and BC
shown in Fig. a and b, respectively, we notice that Bx and By can be determined
by solving simultaneously the moment equations of equilibrium written about
A and C, respectively.
a + ΣMA = 0;
Bx(6.5) + By(6) - (6)( 142.25) a
- (6)( 142.25) a
a + ΣMC = 0;
2.5
b(5.25)
142.25
6
b(3) - (2)(4)(2) = 0
142.25
6.5Bx + 6By = 202.75(1)
(6)( 142.25) a
2.5
6
b(5.25) + (6)( 142.25) a
b(3)
142.25
142.25
+ By(6) - Bx(6.5) = 0
6.5Bx - 6By = 186.75(2)
66
6m
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2–41.
(Continued)
Solving Eq (1) and (2) yields
Bx = 29.96 k By = 1.333 k
Using these results and writing the force equation of equilibrium by referring to the FBD of member AB, Fig a,
+
S ΣFx = 0;
2(4) + (6)( 142.25) a
2.5
b - 29.96 + Ax = 0
142.25
Ax = 6.962 k = 6.96 k
+ c ΣFy = 0;
Ans.
6
b = 0
142.25
Ans.
2.5
b - Cx = 0
142.25
Ans.
6
b = 0
142.25
Ans.
Ay + 1.333 - 6( 142.25) a
Ay = 34.67 k = 34.7 k
Referring to the FBD of member BC, Fig b,
+
S ΣFx = 0;
29.96 - (6)( 142.25) a
Cx = 14.96 k = 15.0 k
+ c ΣFy = 0;
Cy - 1.333 - (6)( 142.25) a
Cy = 37.33 k = 37.3 k
Ans.
Ax = 6.96 k
Ay = 34.7 k
Cx = 15.0 k
Cy = 37.3 k
67
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2–42. Determine the horizontal and vertical components of
reaction at A, B, and D. Assume the frame is pin connected at
A, B, and D, and there is a fixed-connected joint at C.
2 k>ft
6k
B
8 ft
6 ft
6 ft
C
12 ft
12 k
D
12 ft
Solution
Equations of Equilibrium. We will consider the equilibrium of
the FBD of member AB shown in Fig. a first.
a + ΣMA = 0; Bx(20) - 6(20) - 12(12) = 0 Bx = 13.2 k
A
D
Ans.
a + ΣMB = 0; 12(8) - Ax(20) = 0 Ax = 4.80 k
Ans.
+ c ΣFy = 0; By - Ay = 0(1)
Using the result of Bx to consider the equilibrium of the FBD
of member BCD shown in Fig. b,
a + ΣMD = 0;
1
(2)(6)(4) + 2(6)(9) - 13.2(12) + By(12) = 0
2
By = 2.20 k
Ans.
+
S ΣFx = 0; 13.2 - Dx = 0 Dx = 13.2 k
Ans.
+ c ΣFy = 0; Dy - 2(6) -
1
(2)(6) - 2.20 = 0 Dy = 20.2 k Ans.
2
From Eq. (1),
2.20 - Ay = 0 Ay = 2.20 k
Ans.
Ans.
Bx = 13.2 k
Ax = 4.80 k
By = 2.20 k
Dx = 13.2 k
Dy = 20.2 k
Ay = 2.20 k
68
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2–43. The bridge frame consists of three segments which can
be considered pinned at A, D, and E, rocker supported at C
and F, and roller supported at B. Determine the horizontal and
vertical components of reaction at all these supports due to the
loading shown.
4 k>ft
30 k
10 k
A
E
10 ft
B
8 ft
10 ft
D
5 ft
5 ft
12 ft
C
4 ft
F
4 ft
Solution
Equations of Equilibrium. We will consider the equilibrium
of the FBD of member BD shown in Fig. b first.
a + ΣMD = 0; 30(10) - NB(18) = 0 NB = 16.67 k = 16.7 k
Ans.
a + ΣMB = 0; Dy(18) - 30(8) = 0 Dy = 13.33 k = 13.3 k
+
S ΣFx = 0; Dx = 0
Ans.
Ans.
Next, using the result of NB to consider the equilibrium of the
FBD of member ABC, shown in Fig. a,
a + ΣMA = 0; NC(4) - 4(10)(5) - 16.67(10) = 0 NC = 91.67 k = 91.7 k
+
S
ΣFx = 0; Ax = 0
Ans.
+ c ΣFy = 0; 91.67 - 4(10) - 16.67 - Ay = 0 Ay = 35.0 k
Ans.
Ans.
Finally, using the result of Dx and Dy, the equilibrium of the
FBD of member DEF, Fig. c, gives
a + ΣME = 0; 10(5) + 13.33(10) - NF(4) = 0 NF = 45.83 k = 45.8 k
+
S
ΣFx = 0; Ex = 0
Ans.
+ c ΣFy = 0; 45.83 - 13.33 - 10 - Ey = 0 Ey = 22.5 k
Ans.
Ans.
Ans.
NB = 16.7 k
Dy = 13.3 k
Dx = 0
NC = 91.7 k
Ax = 0
Ay = 35.0 k
NF = 45.8 k
Ex = 0
Ey = 22.5 k
69
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*2–44. Determine the horizontal and vertical reactions at the
connections A and C of the gable frame. Assume that A, B, and
C are pin connections. The purlin loads such as D and E are
applied perpendicular to the center line of each girder.
400 lb 400 lb
600 lb
800 lb
E
D
600 lb
F
800 lb
G
B
5 ft
120 lb>ft
10 ft
A
C
6 ft
6 ft
6 ft
6 ft
Ans.
Solution
Member AB:
a + ΣMA = 0;
Bx(15) + By(12) - (1200)(5) - 600a
- 400a
12
5
b(6) - 600a b(12.5)
13
13
12
5
b(12) - 400a b(15) = 0
13
13
Bx(15) + By(12) = 18,946.154(1)
Member BC:
a + ΣMC = 0;
- Bx(15) + By(12) + 600a
400a
12
5
b(6) + 600a b(12.5)
13
13
12
5
b(12) + 400a b(15) = 0
13
13
Bx(15) - By(12) = 12,446.15(2)
Solving Eqs. (1) and (2),
Bx = 1063.08 lb
By = 250.0 lb
Member AB:
+
S ΣFx = 0;
-Ax + 1200 + 1000a
Ax = 522 lb
+ c ΣFy = 0;
Ay - 800 - 1000a
Ay = 1473 lb
5
b - 1063.08 = 0
13
12
b + 250 = 0
13
Ans.
Ans.
Member BC:
+
S ΣFx = 0;
-Cx - 1000a
Cx = 678 lb
+ c ΣFy = 0;
5
b + 1063.08 = 0
13
Cy - 800 - 1000a
Cy = 1973 lb
12
b - 250.0 = 0
13
Ans.
Ans.
70
Ans.
Ax = 522 lb
Ay = 1473 lb
Cx = 678 lb
Cy = 1973 lb
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P
2–1P. The railroad trestle bridge shown in the photo is
supported by reinforced concrete bents. Assume the two simply
supported side girders, track bed, and two rails have a weight of
0.5 k>ft and the load imposed by a train is 7.2 k>ft. Each girder
is 20 ft long. Apply the load over the entire bridge and
determine the compressive force in the columns of each bent.
For the analysis assume all joints are pin connected and neglect
the weight of the bent. Are these realistic assumptions?
P
8 ft
B
C
18 ft
A
758
758
D
Solution
Maximum reactions occur when the live load is over the entire span.
Load = 7.2 + 0.5 = 7.7 k>ft
R = 7.7(10) = 77 k
2(77)
= 77 k
2
All members are two-force members.
Then P =
a + ΣMB = 0;
- 77(8) + F sin 75°(8) = 0
Ans.
F = 79.7 k
It is not reasonable to assume the members are pin connected, since
Ans.
such a framework is unstable.
Ans.
F = 79.7 k
It is not reasonable to assume the members are pin
connected, since such a framework is unstable
71
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3–1. Classify each of the following trusses as statically
determinate, statically indeterminate, or unstable. If
indeterminate, state its degree.
(a)
Solution
(a) By inspection, the truss is internally and externally stable.
Here, b = 9, r = 3 and j = 6. Since b + r = 2j, the truss
Ans.
is statically determinate.
(b) By inspection, the truss is internally and externally
stable. Here, b = 10, r = 4 and j = 6. Since b + r 7 2j
and (b + r) - 2j = (10 + 4) - 2(b) = 2, the truss is
­statically indeterminate to the second degree.
Ans.
(c) By inspection, the truss is internally and externally
stable. Here, b = 14, r = 3 and j = 8. Since b + r 7 2j
and (b + r) - 2j = (14 + 3) - 2(8) = 1, the truss is
­statically indeterminate to the first degree.
Ans.
(b)
(c)
Ans.
(a) Statically determinate.
(b) Statically indeterminate to the second degree.
(c) Statically indeterminate to the first degree.
72
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3–2. Classify each of the following trusses as statically
determinate, statically indeterminate, or unstable. If
indeterminate, state its degree.
(a)
Solution
(a) Here, b = 10, r = 3 and j = 7. Since b + r 6 2j, the
truss is unstable.
Ans.
(b) By inspection, the truss is internally and externally
stable. Here, b = 14, r = 4 and j = 8. Since b + r 7 2j
and (b + r) - 2j = (14 + 4) - 2(8) = 2, the truss is
­statically indeterminate to the second degree.
Ans.
(c) By inspection, the truss is internally and externally
stable. Here, b = 12, r = 3 and j = 7. Since b + r 7 2j
and (b + r) - 2j = (12 + 3) - 2(7) = 1, the truss is
­statically indeterminate to the first degree.
Ans.
(b)
(c)
Ans.
(a) Unstable.
(b) Statically indeterminate to the second degree.
(c) Statically indeterminate to the first degree.
73
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3–3. Classify each of the following trusses as statically
determinate, statically indeterminate, or unstable. If indeterminate
state its degree.
(a)
Solution
(a) By inspection, the truss is internally and externally stable.
Here, b = 9, r = 3 and j = 6. Since b + r = 2j, the truss
Ans.
is statically determinate.
(b) By inspection, the truss is internally and externally stable.
Here, b = 9, r = 4 and j = 6. Since b + r = 2j, and
(b + r) - 2j = 1, the truss is statically indeterminate to
Ans.
the first degree.
(c) By inspection, the truss is internally unstable: the left and
right ends are not rigidly joined across the middle. Note
also that b = 17, r = 3, and j = 11, so that b + r 6 2j.
(b)
Ans.
(c)
Ans.
(a) Internally and externally stable. Statically determinate.
(b) Internally and externally stable. Statically indeterminate
to the first degree.
(c) Internally unstable.
74
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*3–4. Classify each of the following trusses as statically
determinate, statically indeterminate, or unstable. If
indeterminate state its degree.
(a)
Solution
(a) By inspection, the truss is internally and externally stable.
Here, b = 16, r = 3 and j = 8. Since b + r 7 2j and
(b + r) - 2j = (16 + 3) - 2(8) = 3, the truss is statically
indeterminate to the third degree.
Ans.
(b) By inspection, the truss is internally and externally stable.
Here, b = 9, r = 3 and j = 6. Since b + r = 2j, the truss
is statically determinate.
Ans.
(c) Here, b = 15, r = 3 and j = 10. Since b + r 6 2j, the
truss is unstable.
(b)
Ans.
(c)
Ans.
(a) Statically indeterminate to third degree.
(b) Statically determinate.
(c) Unstable.
75
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3–5. Determine the force in each member of the truss. State if
the members are in tension or compression. Assume all
members are pin connected.
G
F
E
3m
A
Solution
B
2m
C
2m
5 kN
D
2m
5 kN
5 kN
Joint D:
3
+ c ΣFy = 0; FED a b - 5 = 0; FED = 8.33 kN (T)
5
Ans.
4
+
S
ΣFx = 0; FCD - (8.33) = 0;
5
FCD = 6.67 kN (C)
Joint C:
+
S
Ans.
ΣFx = 0; FBC - 6.67 = 0;
Ans.
FBC = 6.67 kN (C)
+ c ΣFy = 0; FCE - 5 = 0; FCE = 5 kN (T)
Ans.
ΣFx = 0; FGF - 20 = 0; FGF = 20 kN (T) Ans.
+ c ΣFy = 0; 15 - FGA = 0; FGA = 15 kN (T)
Ans.
Joint G:
+
S
Joint A:
+ c ΣFy = 0; 15 - FAF (sin 56.3°) = 0;
+
S
ΣFx = 0;
Ans.
FAF = 18.0 kN (C)
- FAB - 18.0(cos 56.3°) + 20 = 0;
FAB = 10.0 kN (C)
Ans.
4
- FBE a b + 10.0 - 6.67 = 0;
5
FBE = 4.17 kN (C)
Ans.
Joint B:
+
S
ΣFx = 0;
3
+ c ΣFy = 0; FFB - 5 - 4.17a b = 0;
5
Ans.
FFB = 7.50 kN (T)
Joint F:
3
+ c ΣFy = 0; 18(sin 56.3°) - 7.5 - FFE a b = 0;
5
Ans.
FFE = 12.5 kN (T)
76
Ans.
FED = 8.33 kN (T); FCD = 6.67 kN (C);
FBC = 6.67 kN (C); FCE = 5 kN (T);
FGF = 20 kN (T); FGA = 15 kN (T);
FAF = 18.0 kN (C); FAB = 10.0 kN (C);
FBE = 4.17 kN (C); FFB = 7.50 kN (T);
FFE = 12.5 kN (T)
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–6. Determine the force in each member of the truss. State if
the members are in tension or compression.
F
A
608
608
B
15 ft
Solution
Support Reactions. Referring to the FBD of the entire truss, Fig. a,
a + ΣMA = 0; ND(45) - 30(15) - 15(30) = 0 ND = 20.0 k
Method of Joints. The joint’s equilibrium analysis will be performed in the ­
sequence D, C, E, F and B.
Joint D. Fig. b:
77
C
15 ft
30 k
+ c ΣFy = 0; 20.0 - FDE sin 60° = 0 FDE = 23.09 k (C) = 23.1 k (C)
+
S
ΣFx = 0; 23.09 cos 60° - FCD = 0 FCD = 11.547 k (T) = 11.5 k (T) E
Ans.
Ans.
15 ft
15 k
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–6.
(Continued)
Joint C. Fig. c:
+
S
ΣFx = 0; 11.547 - FBC = 0 FBC = 11.547 k (T) = 11.5 k (T)
+ c ΣFy = 0; FCE - 15 = 0 FCD = 15.0 k (T)
Ans.
Ans.
Joint E. Fig. d:
+ c ΣFy = 0;
+
S
ΣFx = 0;
Joint F. Fig. e:
+
S
ΣFx = 0;
+ c ΣFy = 0;
23.09 sin 60° - 15 - FBE sin 60° = 0 FBE = 5.774 k (T) = 5.77 k (T)
Ans.
FEF = 14.43 k = 14.4 k (C)
Ans.
FEF - 5.774 cos 60° - 23.09 cos 60° = 0
FAF cos 60° - 14.43 = 0 FAF = 28.87 k (C) = 28.9 k (C)
28.87 sin 60° - FBF = 0 FBF = 25.0 k (T)
Joint B. Fig. f:
+
S
ΣFx = 0; 11.547 + 5.774 cos 60° - FAB = 0 FAB = 14.43 k (T) = 14.4 k (T)
+ c ΣFy = 0; 5.774 sin 60° + 25.0 - 30 = 0 (Check!!)
Ans.
Ans.
Ans.
Ans.
FDE = 23.1 k (C)
FCD = 11.5 k (T)
FBC = 11.5 k (T)
FCE = 15.0 k (T)
FBE = 5.77 k (T)
FEF = 14.4 k (C)
FAF = 28.9 k (C)
FBF = 25.0 k (T)
FAB = 14.4 k (T)
78
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3–7. Determine the force in each member of the truss. State
whether the members are in tension or compression. Set
P = 8 kN.
4m
B
A
608
C
608
E
4m
D
4m
P
Solution
Method of Joints: In this case, the support reactions are not
required for determining the member forces.
Joint D:
+ c ΣFy = 0; FDC sin 60° - 8 = 0
Ans.
FDC = 9.238 kN (T) = 9.24 kN (T)
+
S
ΣFx = 0; FDE - 9.238 cos 60° = 0
FDE = 4.619 kN (C) = 4.62 kN (C)
Ans.
Joint C:
+ c ΣFy = 0 ; FCE sin 60° - 9.328 sin 60° = 0
FCE = 9.238 kN (C) = 9.24 kN (C)
+
S
ΣFx = 0; 2(9.238 cos 60°) - FCB = 0
FCB = 9.238 kN (T) = 9.24 kN (T)
Ans.
Ans.
Joint B:
+ c ΣFy = 0; FBE sin 60° - FBA sin 60° = 0
FBE = FBA = F
+
S ΣFx = 0; 9.238 - 2F cos 60° = 0
F = 9.238 kN
Thus,
FBE = 9.24 kN (C) FBA = 9.24 kN (T)
Ans.
Joint E:
+ c ΣFy = 0; Ey - 2(9.238 sin 60°) = 0 Ey = 16.0 kN
+
S
ΣFx = 0; FBA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0
Ans.
FEA = 4.62 kN (C)
Note: The support reactions Ax and Ay can be determined by
analyzing Joint A using the results obtained above.
Ans.
FDC = 9.24 kN (T); FDE = 4.62 kN (C);
FCE = 9.24 kN (C); FCB = 9.24 kN (T);
FBE = 9.24 kN (C); FBA = 9.24 kN (T);
FEA = 4.62 kN (C)
79
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–8. If the maximum force that any member can support is
8 kN in tension and 6 kN in compression, determine the
maximum force P that can be supported at joint D.
4m
B
A
608
C
608
E
4m
D
4m
P
Solution
Method of Joints: In this case, the support reactions are not required for
determining the member forces.
Joint D:
+ c ΣFy = 0;
+
S
ΣFx = 0;
FDC sin 60° - P = 0
FDC = 1.1547P (T)
FDE - 1.1547P cos 60° = 0
FDE = 0.57735P (C)
Joint C:
+ c ΣFy = 0;
+
S
ΣFx = 0;
FCE sin 60° - 1.1547P sin 60° = 0
FCE = 1.1547P (C)
2(1.1547P cos 60°) - FCB = 0
FCB = 1.1547P (T)
FBE sin 60° - FBE sin 60° = 0
FBE = FBA = F
Joint B:
+ c ΣFy = 0;
+
S
ΣFx = 0;
Thus,
Joint E:
+
S
ΣFx = 0;
1.1547P - 2F cos 60° = 0
F = 1.1547P
FBE = 1.1547P (C)
FBA = 1.1547P (T)
FEA + 1.1547P cos 60° - 1.1547P cos 60° - 0.57735P = 0
FEA = 0.57735P (C)
From the above analysis, the maximum compression and tension in the truss
members is 1.1547P. For this case, compression controls which requires
1.1547P = 6
P = 5.20 kN
Ans.
Ans.
P = 5.20 kN
80
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–9. Determine the force in each member of the truss. State if
the members are in tension or compression.
E 12 kN
F
8 kN 3 m
A
B
458
4m
24 kN
4m
Solution
Note that the line of action of support reaction NA does not
coincide with member AF.
Support Reactions. Referring to the FBD of the entire truss,
Fig. a,
a + ΣMC = 0; 24(4) + 12(3) - 8(4) - NA cos 45°(8) = 0 NA = 17.68 kN
Method of Joints. The joints’ equilibrium analysis will be performed in the sequence A, F, B, D and E.
Joint A. Fig. b:
3
+ c ΣFy = 0; 17.68 cos 45° - FAF a b = 0 FAF = 20.83 kN (C) = 20.8 kN (C)
5
4
+
S
ΣFx = 0; FAB + 17.68 sin 45° - 20.83 a b = 0
5
FAB = 4.167 kN (T) = 4.17 kN (T)
Ans.
Ans.
81
D
C
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–9.
(Continued)
Joint F. Fig. c:
4
+
S
ΣFx = 0; 20.83a b - FEF = 0 FEF = 16.67 kN (C) = 16.7 kN (C) 5
3
+ c ΣFy = 0; 20.83a b - FBF = 0 FBF = 12.5 kN (T)
5
Ans.
Ans.
Joint B. Fig. d:
3
+ c ΣFy = 0; FBE a b + 12.5 - 24 = 0 FBE = 19.17 kN (T) = 19.2 kN (T)
5
4
+
S
ΣFx = 0; 19.17a b - 4.167 - FBC = 0 FBC = 11.17 kN (C) = 11.2 kN (C) 5
Joint D. Fig. e:
3
+ c ΣFy = 0; FDE a b - 8 = 0 FDE = 13.33 kN (T) = 13.3 kN (T)
5
4
+
S
ΣFx = 0; FCD - 13.33a b = 10 FCD = 10.67 kN (C) = 10.7 kN (C) 5
Joint E. Fig. f:
3
3
+ c ΣFy = 0; FCE - 19.17a b - 13.33a b = 0 FCE = 19.5 kN (C)
5
5
Ans.
Ans.
Ans.
Ans.
Ans.
4
4
+
S
ΣFx = 0; 13.33a b + 16.67 - 19.17a b - 12 = 0 (Check!!)
5
5
Ans.
FAF = 20.8 kN (C)
FAB = 4.17 kN (T)
FEF = 16.7 kN (C)
FBF = 12.5 kN (T)
FBE = 19.2 kN (T)
FBC = 11.2 kN (C)
FDE = 13.3 kN (T)
FCD = 10.7 kN (C)
FCE = 19.5 kN (C)
82
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–10. Determine the force in each member of the truss. State
if the members are in tension or compression.
20 kN
F
15 kN
10 kN
E
G
4m
A
Solution
3m
83
D
C
B
2m
2m
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–10.
(Continued)
4
Support Reactions. From the geometry of the truss, u = tan-1 a b = 38.66° and
5
h = 3 tan 38.66° = 2.4 m. Referring to the FBD of the entire truss, Fig. a,
a + ΣMA = 0; ND(10) + (10 sin 38.66°)(2.4) - (10 cos 39.66°)(7)
- 20(5) - (15 sin 38.66°)(2.4) - (15 cos 38.66°)(3) = 0
ND = 19.73 kN
Method of Joints. The joints’ equilibrium analysis will be performed in the sequence D, E, C, F,
G and B.
Joint D. Fig. b:
+ c ΣFy = 0; 19.73 - FDE sin 38.66° = 0 FDE = 31.58 kN (C) = 31.6 kN (C)
+
S
ΣFx = 0; 31.58 cos 38.66° - FCD = 0 FCD = 24.66 kN (T) = 24.7 kN (T) Ans.
Ans.
Joint E. Fig. c:
+ QΣFy = 0; FCB cos 38.66° - 10 = 0 FCE = 12.81 kN (C) = 12.8 kN (C)
Ans.
+ RΣFx = 0; FEF - 12.81 sin 38.66° - 31.58 = 0 FEF = 39.58 kN (C) = 39.6 kN (C)
Ans.
4
Joint C. Here f = tan - 1 a b = 63.43; Fig. d:
2
+ c ΣFy = 0; FCF sin 63.43 - 12.81 = 0 FCF = 14.32 kN (T) = 14.3 kN (T)
+
S
ΣFx = 0; 24.66 - 14.33 cos 63.43° - FBC = 0 FBC = 18.26 kN (T) = 18.3 kN (T) Ans.
Ans.
Joint F. Fig. e:
+
S
ΣFx = 0; FFG cos 38.66° - FBF cos 63.43° + 14.32 cos 63.43° - 39.58 cos 38.66° = 0
0.7809FFG - 0.4472FBF = 24.5059 (1)
+ c ΣFy = 0; FFG sin 58.66 - FBF sin 63.43° + 39.58 sin 38.66° - 14.32 sin 63.43° - 20 = 0
0.6247FFG - 0.8944FBF = 8.0791 (2)
Solving Eqs. (1) and (2),
FFG = 43.68 kN (C) = 43.7 kN (C)
Ans.
FBF = 21.48 kN (T) = 21.5 kN (T)
Ans.
Joint G. Fig. f:
+ aΣFx = 0; FBG cos 38.66° - 15 = 0 FBG = 19.21 kN (C) = 19.2 kN (C)
Ans.
+ QΣFy = 0; FAG + 19.21 sin 38.66° - 43.68 = 0 FAG = 31.68 kN (C) = 31.7 kN (C)
Ans.
Joint B. Fig. g:
+
S
ΣFx = 0; 18.26 + 21.48 cos 63.43° - FAB = 0 FAC = 27.86 kN (T) = 27.9 kN (T) + c ΣFy = 0; 21.48 sin 63.93° - 19.21 = 0 (Check!!)
Ans.
Ans.
FDE = 31.6 kN (C)
FCD = 24.7 kN (T)
FCE = 12.8 kN (C)
FEF = 39.6 kN (C)
FCF = 14.3 kN (T)
FBC = 18.3 kN (T)
FFG = 43.7 kN (C)
FBF = 21.5 kN (T)
FBG = 19.2 kN (C)
FAG = 31.7 kN (C)
FAB = 27.9 kN (T)
84
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3–11. Determine the force in each member of the truss. State
if the members are in tension or compression.
F
E
D
Solution
Fx =
B
30
45
30
45
A
C
30
30
2m
2m
2 kN
Joint F
2.31 kN d S FFE
Joint F:
+
S
ΣFx = 0;
- 2.31 + FFE = 0 FFE = 2.31 kN (T)
Joint E:
+ c ΣFy = 0; FEA = FEC
+
S
ΣFx = 0; 2.31 - 2 FEA sin 30° = 0
FEA = 2.31 kN (C)
Ans.
FEC = 2.31 kN (T)
Ans.
Joint A:
+
S
ΣFx = 0; 2.31 - 2.31 sin 30° - FAB cos 30° + FAD cos 45° = 0 (1)
+ c ΣFy = 0; 2 - 2.31 cos 30° + FAD sin 45° - FAB sin 30° = 0 (2)
Solving Eqs. (1) and (2) simultaneously,
Joint B:
+
S
FAD = 2.24 kN (T)
Ans.
FAB = 3.16 kN (C)
Ans.
ΣFx = 0; FBC = 3.16 kN (C) Ans.
+ c ΣFy = 0; 2(3.16) sin 30° - FBD = 0
Ans.
FBD = 3.16 kN (C)
Joint D:
Ans.
FDC = 2.24 kN (T)
85
Ans.
FFE = 2.31 kN (T)
FEA = 2.31 kN (C)
FEC = 2.31 kN (T)
FAD = 2.24 kN (T)
FAB = 3.16 kN (C)
FBC = 3.16 kN (C)
FBD = 3.16 kN (C)
FDC = 2.24 kN (T)
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–12. Determine the force in each member of the truss. All
interior angles are 60°. State if the members are in tension or
compression. Assume all members are pin connected.
600 lb
300 lb
400 lb
E
F
D
Solution
5 ft
60
A
C
B
5 ft
5 ft
Joint F:
+ c ΣFy = 0; FFA (sin 60°) - 300 = 0; FFA = 346.4 lb = 346 lb (C)
+
S
ΣFx = 0; FFE - 346.4(cos 60°) = 0; FFE = 173.2 lb = 173 lb (T) Ans.
Ans.
Joint A:
+ c ΣFy = 0;
- 346.4 (sin 60°) + 925 - FAE (sin 60°) = 0
FAE = 721.7 lb = 722 lb (C)
+
S ΣFx = 0; FAB - 721.7 (cos 60°) + 346.4 (cos 60°) = 0
FAB = 187.6 lb = 188 lb (T)
Ans.
Ans.
Joint E:
+ c ΣFy = 0; 721.7 (sin 60°) - FEB (sin 60°) - 600 = 0
+
S
ΣFx = 0;
FEB = 28.89 lb = 28.9 lb (T)
Ans.
- 173.2 + 721.7 (cos 60°) + 28.89 (cos 60°) - FED = 0
FED = 202.1 lb = 202 lb (C)
Ans.
Joint B:
+ c ΣFy = 0; 28.89 (sin 60°) - FBD (sin 60°) = 0
+
S
ΣFx = 0;
FBD = 28.89 lb = 28.9 lb (C)
Ans.
- 187.6 - 28.89 (cos 60°) - 28.89 (cos 60°) + FBC = 0
FBC = 216.5 lb = 217 lb (T)
Ans.
Joint C:
+ c ΣFx = 0; 375 - FDC (sin 60°) = 0
Ans.
FDC = 433 lb (C)
Ans.
FFA = 346 lb (C)
FFE = 173 lb (T)
FAE = 722 lb (C)
FAB = 188 lb (T)
FEB = 28.9 lb (T)
FED = 202 lb (C)
FBD = 28.9 lb (C)
FBC = 217 lb (T)
FDC = 433 lb (C)
86
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–13. The truss shown is used to support the floor deck. The
uniform load on the deck is 3 k>ft. This load is transferred from
the deck to the floor beams, which rest on the top joints of the
truss. Determine the force in each member of the truss, and
state if the members are in tension or compression. Assume all
members are pin connected.
3 k/ft
J
H
I
G
F
16 ft
A
B
12 ft
Solution
Support Reactions. The concentrated load acting on joints J, F
is 3(6) = 18 k and joints G, H, I is 3(12) = 36 k. Referring to
the FBD of the entire truss shown in Fig. a,
a + ΣMA = 0; NE(48) - 18(48) # 36(36) - 36(24) - 36(12) = 0 NE = 72.0 k
Method of Joints. The joint’s equilibrium analysis will be
performed in the sequence E, F, G, D and C
Joint E, Fig b.
+
S
ΣFx = 0; FDE = 0
+ c ΣFy = 0; 72.0 - FEF = 0 FEF = 72.0 k (C)
87
12 ft
E
D
C
12 ft
12 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–13.
(Continued)
Joint F, Fig. c.
4
+ c ΣFy = 0; 72.0 - 18 - FDF a b = 0 FDF = 67.5 k (T)
5
3
+
S
ΣFx = 0; FFG - 67.5a b = 0 FFG = 40.5 k (C)
5
Joint G, Fig. d.
+
S
ΣFx = 0; FGH - 40.5 = 0 FGH = 40.5 k (C)
+
S
ΣFy = 0; FDG - 36 = 0 FDG = 36.0 k (C)
Joint D, Fig. e.
4
4
+ c ΣFy = 0; 67.5 a b - 36.0 - FDH a b = 0 FDH = 22.5 k (C)
5
5
3
3
+
S
ΣFx = 0; 67.5 a b + 22.5a b - FCD = 0 FCD = 54.0 k (T)
5
5
Joint C, Fig. f.
+ c ΣFy = 0; FCH = 0
Due to symmetry,
Ans.
FAJ = FEF = 72.0 k (C)
FIJ = FFG = 40.5 k (C)
Ans.
FBJ = FDF = 67.5 k (T)
Ans.
FHI = FGH = 40.5 k (C)
Ans.
FBI = FDG = 36.0 k (C)
Ans.
FBH = FDH = 22.5 k (C)
Ans.
FBC = FCD = 54.0 k (T)
Ans.
FAB = FDE = 0
Ans.
Ans.
FCH = 0
FAJ = FEF = 72.0 k (C)
FIJ = FEG = 40.5 k (C)
FBJ = FDF = 67.5 k (T)
FHI = FGH = 40.5 k (C)
FBI = FDG = 36.0 k (C)
FBH = FDH = 22.5 k (C)
FBC = FCD = 54.0 k (T)
FAB = FDE = 0
88
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–14. Determine the force in each member of the truss.
Indicate if the members are in tension or compression. Assume
all members are pin connected.
E
6 kN
F
6 kN
4.5 m
G
3 kN
B
A
Solution
2m
Support Reactions. Not required.
Method of Joints. The joints’ equilibrium analysis will be
performed in the sequence A, G, B, F and C.
Joint A, Fig a.
3
4
+ c ΣFy = 0; FAG a b - 3a b = 0 FAG = 4.00 kN (T)
5
5
4
3
+
S
ΣFx = 0; 4.00 a b + 3a b - FAB = 0 FAB = 5.00 kN (C) 5
5
89
Ans.
Ans.
C
2m
D
2m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–14.
(Continued)
Joint G, Fig. b.
+
S
ΣFx = 0;
4
4
3
- FFG a b - 4.00a b + 6 a b = 0 FFG = 0.500 kN (C) 5
5
5
4
4
3
+ c ΣFy = 0; FBG - 6a b - 4.00a b - 0.500a b = 0 FBG = 7.50 kN (C)
5
5
5
Ans.
Ans.
Joint B, Fig. c.
+ c ΣFy = 0; FBF a
3
b - 7.50 = 0 FBF = 2.5113 kN (T) = 9.01 kN (T)
113
2
+
S
ΣFx = 0; 2.5113 a
b + 5 - FBC = 0 FBC = 10.0 kN (C) 113
Ans.
Ans.
Joint F, Fig. d.
4
3
4
2
+
S
ΣFx = 0; FEF a b + 6a b + 0.500a b - 2.5113 a
b = 0
5
5
5
113
FEF = 1.25 kN (T)
Ans.
3
3
4
3
+ c ΣFy = 0; FCF + 1.25a b + 0.500a b - 6a b - 2.5113a
b = 0
5
5
5
113
Ans.
FCF = 11.25 kN (C)
Joint C, Fig. e.
+ c ΣFy = 0; FCE a
4.5
b - 11.25 = 0 FCE = 12.31 kN (T) = 12.3 kN (T)
124.25
+
S
ΣFx = 0; 12.31a
2
b + 10.0 - FCD = 0 FCD = 15.0 kN (C) 124.25
Ans.
Ans.
Ans.
FAG = 4.00 kN (T)
FAB = 5.00 kN (C)
FFG = 0.500 kN (C)
FBG = 7.50 kN (C)
FBF = 9.01 kN (T)
FBC = 10.0 kN (C)
FEF = 1.25 kN (T)
FCF = 11.25 kN (C)
FCE = 12.3 kN (T)
FCD = 15.0 kN (C)
90
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–15. Determine the force in each member of the truss. State
if the members are in tension or compression. Assume all
members are pin connected.
3k
2k
6 ft
G
3k
2k
6 ft
F
E
4 ft
B
C
4 ft
Solution
A
91
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–15.
(Continued)
Equations of Equilibrium. Referring to the FBD of the entire
truss shown in Fig. a,
a + ΣMA = 0; ND 1122 - 21122 - 3162 - 3182 = 0 ND = 5.50 k
Method of Joints. By inspecting Joints C, B and E, we notice that
members CE, BG and EF are zero force members. Thus,
Ans.
FCE = FEG = FEF = 0
The joints’ equilibrium analysis will be performed in the
­sequence E, G, D, C, F, and B.
Joint E, Fig. b.
+ c ΣFy = 0; FDE - 2 = 0 FDE = 2.00 k 1C2
Ans.
Joint G, Fig. c.
+
S
ΣFx = 0; 3 - FEG = 0 FEG = 3.00 k 1C2 Ans.
+ c ΣFy = 0; FAG - 2 = 0 FAG = 2.00 k 1C2
Ans.
Joint D, Fig. d.
+ c ΣFy = 0; 5.50 - 2.00 - FCD a
4
b = 0 FCD = 4.375 k 1C2
5
3
+
S
ΣFx = 0; 4.375a b - FAD = 0 FAD = 2.625 k 1T2 5
Ans.
Ans.
Joint C, Fig. e.
Ans.
+ RΣFx = 0; FCF - 4.375 = 0 FCF = 4.375 k 1C2
Joint F, Fig. f.
3
3
+
S
ΣFx = 0; 3.00 - 4.375a b - FBF a b = 0 FBF = 0.625 k 1T2 5
5
+ c ΣFy = 0; 4.375a
Joint B, Fig. g.
Ans.
4
4
b - 3 - 0.625a b = 0 (Check!!)
5
5
Ans.
+ QΣFx = 0; 0.625 - FAB = 0 FAB = 0.625 k 1T2
92
Ans.
FCE = FEG = FEF = 0
FDE = 2.00 k 1C2
FFG = 3.00 k 1C2
FAG = 2.00 k 1C2
FCD = 4.375 k 1C2
FAD = 2.625 k 1T2
FCF = 4.375 k 1C2
FBF = 0.625 k 1T2
FAB = 0.625 k 1T2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–16. The members of the truss have a mass of 5 kg>m.
Lifting is done using a cable connected to joints E and G.
Determine the largest member force and specify if it is in
tension or compression. Assume half the weight of each
member can be applied as a force acting at each joint.
P
H
308 308
F 308
E
G
608
A
2m
Solution
Ans.
The weights for the truss members are as follows:
B
1m
608
C
1m
308
2m
D
Ans.
WAG = WGF = WFE = WED = 1.73215219.812 = 84.957 N
WAB = WBF = WBC = WCF = WCD = 215219.812 = 98.1 N
WGB = WCI = 115219.812 = 49.50 N
The total weight of the truss is
Wtotal = P = 4184.9572 + 519.812 + 2149.502 = 928.428 N
The force acting on each joint for the truss:
FD = F A =
FG = F E =
FB = F C =
FF =
184.957 + 98.12
2
= 91.53 N
[84.957122 + 49.05]
2
[98.1132 + 49.05]
2
[84.957122 + 98.1122]
2
= 109.48 N
= 171.68 N
= 183.06 N
Joint H:
+ c ΣFy = 0; 928.428 - 2T cos 30° = 0
Ans.
T = 536.028 = 536 N
Joint A:
+ c ΣFy = 0; FAG sin 30° - 91.529 = 0
FAG = 183.057 = 183 N 1T2
+
S
ΣFx = 0; - FAB + 183.057 cos 30° = 0
FAB = 158.532 = 159 N 1C2
Ans.
Ans.
Joint G:
+ QΣFx = 0;
- FGF + 536.028 cos 30° - 183.057 - 109.482 sin 30° = 0
Ans.
FGF = 226.416 = 226 N 1C2
+ aΣFy = 0;
- FGB - 109.482 cos 30° + 536.028 sin 30° = 0
FGB = 173.200 = 173 N 1T2
Ans.
Joint B:
+ c ΣFy = 0; 173.200 sin 60° + FBF sin 60° - 171.675 = 0
FBF = 25.033 = 25.0 N 1T2
+
S
ΣFy = 0; 25.033 cos 60° - FBC - 173.200 cos 60° + 158.532 = 0
FBC = 84.4 N 1C2
Thus, by comparison,
FGF = FEF = 226 N 1T2
93
Ans.
Ans.
Ans.
Ans.
T = 536 N
FAG = 183 N 1T2
FAB = 159 N 1C2
FGF = 226 N 1C2
FGB = 173 N 1T2
FBF = 25.0 N 1T2
FBC = 84.4 N 1C2
FGF = 226 N 1T2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–17. Determine the force in each member of the truss in
terms of the load P, and indicate whether the members are in
tension or compression.
B
P
d
C
A
D
F
d
E
Solution
d
Support Reactions:
4
3
a + ΣMB = 0; P12d2 - Ay a db = 0 Ay = P
2
3
+ c ΣFy = 0;
4
4
P - Ey = 0 Ey = P
3
3
+
S
ΣFx = 0 Ex - P = 0 Ex = P
Method of Joints: By inspection of joint C, members CB and
CD are zero-force members. Hence,
Ans.
FCB = FCD = 0
Joint A:
+ c ΣFy = 0; FAB a
1
4
b- P = 0
3
13.25
Ans.
FAB = 2.404P 1C2 = 2.40P 1C2
1.5
+
S
ΣFx = 0; FAF - 2.404P a
b = 0
13.25
FAF = 2.00P 1T2
Ans.
Joint B:
1.5
+
S
ΣFx = 0; 2.404P a
b - P
13.25
- FBF a
0.5
0.5
b - FBD a
b = 0
11.25
11.25
1.00P - 0.4472FBF - 0.4472FBD = 0 (1)
+ c ΣFy = 0; 2.404P a
1
1
1
b + FBD a
b - FBF a
b = 0
13.25
11.25
11.25
1.333P + 0.8944FBD - 0.8944FBF = 0 (2)
Solving Eqs. [1] and [2] yields
Ans.
FBF = 1.863P 1T2 = 1.86P 1T2
Joint F:
Ans.
FBD = 0.3727P 1C2 = 0.373P 1C2
+ c ΣFy = 0; 1.863P a
1
1
b - FFE a
b = 0
11.25
11.25
Ans.
FFE = 1.863P 1T2 = 1.86P 1T2
0.5
+
S
ΣFx = 0; FFD + 2c 1.863P a
b d - 2.00P = 0
11.25
FFD = 0.3333P 1T2 = 0.333P 1T2
Ans.
94
d/ 2
d/ 2
d
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–17.
(Continued)
Joint D:
+ c ΣFy = 0; FDX a
1
1
b - 0.3727P a
b = 0
11.25
11.25
FDE = 0.3727P 1C2 = 0.373P 1C2
0.5
+
S
ΣFx = 0; 2c 0.3727P a
b d - 0.3333P = 0
11.25
Ans.
(Check!!)
95
Ans.
FCB = FCD = 0
FAB = 2.40P 1C2
FAF = 2.00P 1T2
FBF = 1.86P 1T2
FBD = 0.373P 1C2
FFE = 1.86P 1T2
FFD = 0.333P 1T2
FDE = 0.373P 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–18. If the maximum force that any member can support is
4 kN in tension and 3 kN in compression, determine the
maximum force P that can be supported at point B. Take
d = 1 m.
B
P
d
C
A
D
F
d
E
d
Solution
Support Reactions:
4
3
a + ΣME = 0; P12d2 - Ay a db = 0 Ay = P
2
3
+ c ΣFy = 0;
4
4
P - Ey = 0 Ey = P
3
3
+
S
ΣFx = 0 Ex - P = 0 Ex = P
Method of Joints: By inspection of joint C, members CB and
CD are zero-force members. Hence,
FCB = FCD = 0
Joint A:
+ c ΣFy = 0; FAB a
1
4
b - P = 0 FAB = 2.404P 1C2
3
13.25
1.5
+
S
ΣFx = 0; FAF - 2.404P a
b = 0 FAF = 2.00P 1T2
13.25
Joint B:
+
S
ΣFx = 0; 2.404P a
15
b - P
13.25
- FBF a
0.5
0.5
b - FBD a
b = 0
11.25
11.25
1.00P - 0.4472FBF - 0.4472FBD = 0
+ c ΣFy = 0; 2.404P a
(1)
1
1
1
b + FBD a
b - FBF a
b = 0
13.25
11.25
11.25
1.333P + 0.8944FBD - 0.8944FBF = 0
(2)
Solving Eqs. [1] and [2] yields
Joint F:
FBF = 1.863P 1T2
+ c ΣFy = 0; 1.863P a
FBD = 0.3727P 1C2
1
1
b - FFE a
b = 0
11.25
11.25
FFE = 1.863P 1T2
0.5
+
S
ΣFx = 0; FFD + 2c 1.863P a
b d - 2.00P = 0
11.25
FFD = 0.3333P 1T2
96
d/ 2
d/ 2
d
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–18.
(Continued)
Joint D:
+ c ΣFy = 0; FDE a
1
1
b - 0.3727P a
b = 0
11.25
11.25
FDE = 0.3727P 1C2
0.5
+
S
ΣFx = 0; 2c 0.3727P a
b d - 0.3333P = 0 (Check!!)
11.25
From the above analysis, the maximum compression and tension in
the truss members are 2.404P and 2.00P, respectively. For this case,
compression controls, which requires
2.404P = 3
P = 1.25 kN
Ans.
Ans.
Ans.
P = 1.25 kN
97
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8k
3–19. Determine the force in members AE, BE, and BC of the
truss and indicate if the members are in tension or compression.
6k
D
E
6 ft
C
A
B
8 ft
8 ft
12 k
Solution
Support Reactions. Referring to the FBD of the entire truss shown
in Fig. a,
a + ΣMC = 0; 121162 + 8182 - 6162 - NB 182 = 0 NB = 27.5 k
Method of Sections. Referring to the FBD of the left segment of
the truss sectioned through a–a, Fig. b,
a + ΣMA = 0; 27.5182 - FBE 182 = 0 FBE = 27.5 k 1C2
Ans.
a + ΣME = 0; 12182 - FBC 162 = 0 FBC = 16.0 k 1C2
Ans.
a + ΣMB = 0; 12182 - FAE a
3
b 182 = 0 FAE = 20.0 k 1T2
5
Ans.
98
Ans.
FBE = 27.5 k 1C2
FAE = 20.0 k 1T2
FBC = 16.0 k 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–20. Determine the force in members JK, JN, and CD.
State if the members are in tension or compression. Identify all
the zero-force members.
J
K
M
A
L
B
I
N
D
C
20 ft
H
O
30 ft
2k
20 ft
G
F
E
20 ft
2k
Solution
Reactions:
Ax = 0,
Ay = 2.0 k,
a + ΣMJ = 0;
Fy = 2.0 k
FCD 1202 + 21152 - 21352 = 0
FCD = 2.00 k 1T2
+ c ΣFy = 0;
Jy = 0
+
S
ΣFx = 0;
-Jx + 2.00 = 0;
Ans.
Jx = 2.00 k
Joint J:
a + ΣFy = 0;
- FJN sin 23.39° + 2 sin 29.74° = 0
+ QΣFx = 0;
FJN = 2.50 k 1T2
Ans.
FJK cos 29.74° - 2.50 cos 23.39°
FJK = 4.03 k 1C2
Members KN, NL, MB, BL, CL, IO, OH, GE, EH, HD
are ­zero-force members.
Ans.
Ans.
Ans.
FCD = 2.00 k 1T2; FJN = 2.50 k 1T2; FJK = 4.03 k 1C2;
Members KN, NL, MB, BL, CL, IO, OH, GE, EH, HD are
zero-force members.
99
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3–21. Determine the force in members FC, BC, and FE. State
if the members are in tension or compression. Assume all
members are pin connected.
1.5 k
2k
F
E
D
6 ft
C
6 ft
B
6 ft
A
8 ft
8 ft
8 ft
Solution
a + ΣMD = 0;
2182 -
3
1F )1162 = 0
5 FC
FFC = 1.67 k 1T2
a + ΣMF = 0;
1.8331162 - 2182 -
a + ΣMC = 0;
FBC = 1.39 k 1T2
3
1F 21162 = 0
5 BC
811.8332 - 61FEF 2 = 0
FEF = 2.44 k 1C2
100
Ans.
FFC = 1.67 k 1T2
FBC = 1.39 k 1T2
FEF = 2.44 k 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–22. Determine the force in members KJ, NJ, ND, and CD
of the K-truss. Indicate if the members are in tension or
compression.
L
K
J
I
H
15 ft
15 ft
A
M
N
B
C
O
P
G
D
E
F
20 ft
20 ft
1200 lb 1500 lb 1800 lb
20 ft
20 ft
20 ft
20 ft
Solution
Support Reactions at A:
a+ ΣMG = 0; 1.2011002 + 1.501802 + 1.801602 - Ay 11202 = 0
Ay = 2.90 k
+
S ΣFx = 0; Ax = 0
Method of Sections: From section a–a, FKJ and FCD can be obtained
directly by summing moments about points C and K, respectively.
a + ΣMC = 0; FKJ 1302 + 1.201202 - 2.901402 = 0
Ans.
FKJ = 3.067 k 1C2 = 3.07 k 1C2
a + ΣMK = 0; FCD 1302 + 1.201202 - 2.901402 = 0
Ans.
FCD = 3.067 k 1T2 = 3.07 k 1T2
From sec. b–b, summing forces along the x and y axes yields
4
4
+
S
ΣFx = 0; FND a b - FNJ a b + 3.067 - 3.067 = 0
5
5
FND = FNJ + c ΣFy = 0; 2.90 - 1.20 - 1.50 - FND a
(1)
3
3
b - FNJ a b = 0
5
5
FND + FNJ = 0.3333(2)
Solving Eqs. [1] and [2] yields
FND = 0.167 k 1T2
FNJ = 0.167 k 1C2
Ans.
101
Ans.
FKJ = 3.07 k
FCD = 3.07 k
FND = 0.167 k 1T2
FNJ = 0.167 k 1C2
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3 kN
3–23. Determine the force in members HG, HC, HB, and AB
of the truss. State if the members are in tension or compression.
Assume all members are pin connected.
3 kN
I
3 kN
H
3 kN
G
4m
F
A
C
B
2m
2m
E
D
2m
2m
Solution
Support Reactions. Referring to the FBD of the entire truss,
Fig. a,
a + ΣME = 0;
3122 + 3142 + 3162 + 3182 - NA 182 = 0
NA = 7.50 kN
Method of Joints. By inspecting joint B, we notice that member HB is a zero-force member. Thus,
FHB = 0
Ans.
Method of Sections. Consider the FBD of the left segment of
the truss sectioned through a–a, Fig. b.
a + ΣMH = 0;
a + ΣME = 0;
a + ΣMC = 0;
FAB 132 + 3122 - 7.50122 = 0
Ans.
3
b 142 = 0
113
FHC = 5.408 kN 1T2 = 5.41 kN 1T2
Ans.
1
b 142 + 3122 + 3142 - 7.50142 = 0
15
FHG = 6.708 kN 1C2 = 6.71 kN 1C2
Ans.
3162 + 3182 - 7.50182 + FHC a
FHG a
FAB = 3.00 kN 1T2
102
Ans.
FHB = 0; FAB = 3.00 kN 1T2;
FHC = 5.41 kN 1T2;
FHG = 6.71 kN 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–24. Determine the force in members GF, GC, HC, and BC
of the truss. State if the members are in tension or compression.
Assume all members are pin connected.
3 kN
3 kN
I
3 kN
H
3 kN
G
4m
F
A
C
B
2m
2m
E
D
2m
2m
Solution
Support Reactions. Referring to the FBD of the entire truss,
Fig. a,
+
S
ΣFx = 0;
Ex = 0
a + ΣMA = 0;
Ey 182 - 3122 - 3142 - 3162 = 0
Ey = 4.50 kN
Method of Joints.
Joint G, Fig. b.
+ QΣFy = 0;
FGC = 3.00 kN 1C2
FGC cos u - 3 cos u = 0
Ans.
Method of Sections. Consider the FBD of the right segment
of the truss sectioned trough b–b, Fig. c.
a + ΣMH = 0;
a + ΣMC = 0;
a + ΣME = 0;
4.50162 - 3142 - 3.00122 - FBC 132 = 0
FBC = 3.00 kN 1T2
4.50142 - 3122 - FGF a
1
b 142 = 0
15
3122 + 3.00142 - FHC a
3
b 142 = 0
113
Ans.
FGF = 6.708 kN 1C2 = 6.71 kN 1C2 Ans.
FHC = 5.408 kN 1T2 = 5.41 kN 1T2 Ans.
103
Ans.
FGC = 3.00 kN 1C2; FBC = 3.00 kN 1T2;
FGF = 6.71 kN 1C2; FHC = 5.41 kN 1T2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–25. Determine the force in members EF, EP, and LK of the
Baltimore bridge truss and state if the members are in tension
or compression. Also, indicate all zero-force members.
M
L
N
O
K
2m
J
P
A
I
B C
2 kN
D
E
F
5 kN 3 kN
G
2m
H
2 kN
8 @ 2 m = 16 m
Solution
Support Reaction at I:
a + ΣMA = 0; Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0
Iy = 6.375 kN
Method of Joints: By inspection, BN, NC, DO, OC, HJ, JG and
Ans.
LE are zero-force members.
Method of Sections:
a + ΣMK = 0;
a + ΣME = 0;
+ c ΣFy = 0;
3122 + 6.375142 - FEF 142 = 0
FEF = 7.875 kN 1T2
Ans.
6.375182 - 2142 - 3122 - FLK 142 = 0
FLK = 9.25 kN 1C2
Ans.
FEP = 1.94 kN 1T2
Ans.
6.375 - 3 - 2 - FEP sin 45° = 0
104
Ans.
By inspection, BN, NC, DO, OC, HJ,
JG and LE are zero-force members.
FEF = 7.875 kN 1T2
FLK = 9.25 kN 1C2
FEP = 1.94 kN 1T2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–26. Determine the forces in members JI, JD, and DE of the
truss. State if the members are in tension or compression.
30 kN
5 kN
15 kN 15 kN
L
A
B
K
20 kN
10 kN
J
I
E
C
5 kN
H
G
F
3@1m53m
D
6 @ 3 m 5 18 m
Solution
Entire truss:
+
S
ΣFx = 0;
Ax = 0
Gy = 49.17 kN
- 15132 - 15162 - 30192 - 201122 - 101152 - 51182 + Gy 1182 = 0
a + ΣMA = 0;
+ c ΣFy = 0; Ay - 5 - 15 - 15 - 30 - 20 - 10 - 5 + 49.17 = 0
Ay = 50.833 kN
Section:
a + ΣME = 0;
- FJI 122 - 10132 - 5162 + 49.17162 = 0
FJI = 117.5 kN 1C2
a + ΣMJ = 0;
Ans.
- 20132 - 10162 - 5192 + 49.17192
- FDE cos 18.43°122 - FDE sin 18.43°132 = 0
Joint D:
+
S
ΣFx = 0;
Ans.
FDE = 97.5 kN 1T2
97.5 cos 18.43° - FCD cos 18.43° = 0
FCD = 97.5 kN 1T2
+ c ΣFy = 0;
2197.5 sin 18.43°2 - FJD = 0
FJD = 61.7 kN 1C2
Ans.
105
Ans.
FJI = 117.5 kN 1C2; FDE = 97.5 kN 1T2;
FJD = 61.7 kN 1C2
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12 kN/m
3–27. Determine the force in members BC, CI, DI, and HI of
the truss and state if the members are in tension or compression.
B
A
C
D
E
F
G
3m
H
I
J
K
5 @ 2 m = 10 m
Solution
Support Reactions. Not required. The concentrated load acting on joints
A and F is (12 kN/m)(1m) = 12 kN and joints B, C, D and E is
(12 kN/m)(2m) - 24 kN.
Method of Joints. The force in member CI can be determined by analyzing
the equilibrium of joint C, Fig a.
+ c ΣFy = 0; FCI - 24 = 0 FCI = 24.0 kN 1C2
Ans.
Method of Sections. Referring to the FBD of the right segment of the truss
sectioned through a- a, Fig b,
a + ΣMI = 0; FBC 11.82 - 24122 - 24142 - 12162 = 0
FBC = 120 kN 1T2
a + ΣMF = 0; 24122 + 24142 + 24162 - 24.0162 - FDI a
18
b 142 = 0
1724
Ans.
3
b 142 - 12142 - 24122 + 24122 - 24.0122 = 0
1109
Ans.
FDI = 53.81 kN 1C2 = 53.8 kN 1C2
a + ΣMD = 0; FHI a
Ans.
FHI = 83.52 kN 1C2 = 83.5 kN 1C2
106
Ans.
FCI = 24.0 kN 1C2
FBC = 120 kN 1T2
FDI = 53.8 kN 1C2
FHI = 83.5 kN 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–28. Determine the force in members GF, FB, and BC of
the Fink truss and state if the members are in tension or
compression.
600 lb
800 lb
800 lb
F
E
G
A
608
308
10 ft
B
608
10 ft
308
C
D
10 ft
Solution
Support Reactions: Due to symmetry, Dy = Ay
+ c ΣFy = 0;
+
S
ΣFx = 0;
2Ay - 800 - 600 - 800 = 0
Ay = 1100 lb
Ax = 0
Method of Sections:
a + ΣMB = 0;
FGF sin 30°(10) + 800(10 - 10 cos 2 30°) - 1100(10) = 0
a + ΣMA = 0;
FGF = 1800 lb (C) = 1.80 k (C)
a + ΣMF = 0;
FFB = 692.82 lb (T) = 693 lb (T)
Ans.
FFB sin 60°(10) - 800(10 cos 2 30°) = 0
Ans.
2
FBC (15 tan 30°) + 800(15 - 10 cos 30°) - 1100(15) = 0
Ans.
FBC = 1212.43 lb (T) = 1.21 k (T)
Ans.
FGF = 1.80 k (C); FFB = 693 lb (T);
FBC = 1.21 k (T)
107
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6k
3–29. Specify the type of compound truss and determine the
forces in members FH, BC, and FG. State if the members are in
tension or compression. Assume all members are pin
connected.
F
G
H
4k
E
I
6 ft
D
B
A
15 ft
15 ft
C
10 ft
10 ft
15 ft
Solution
Support Reactions. Referring to the FBD of the entire truss
shown in Fig. a,
a + ΣMD = 0; 4(15) + 6(25) - Ay(50) = 0 Ay = 4.20 k
+
S
ΣFx = 0;
Ax - 4 = 0
Ax = 4.00 k
Method of Sections. Referring to the FBD of the left segment
of the truss sectioned through a- a,
a + ΣMF = 0; FBC(15) + 4.00(15) - 4.20(25) = 0
a + ΣMA = 0; FFH a
a + ΣMB = 0; FFG a
Ans.
FBC = 3.00 k (T)
3
b = 0 FFH = 0
113
Ans.
3
b(15) - 4.20(15) = 0
134
Ans.
FFG = 8.163 k (C) = 8.16 k (C)
Ans.
FBC = 3.00 k (T)
FFH = 0
FFG = 8.16 k (C)
108
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3–30. Determine the force in members HI, CH, and CD of
the truss. State if the members are in tension or compression.
Assume all members are pin connected.
K
J
I
H
15 ft
G
F
A
B
C
D
E
6k
6k
6k
6k
3k
5 @ 9 ft = 45 ft
Solution
Support Reactions. Not required.
Method of Sections. Referring to the FBD of the right segment of the truss sectioned through a- a, Fig. a,
a + ΣMC = 0; FHI a
1
b(27) - 3(27) - 6(18) - 6(9) = 0
110
FHI = 9110 k (T) = 28.5 k (T)
Ans.
a + ΣMF = 0; 6(9) + 6(18) - FCH a
2
b(27) = 0
113
FCH = 3113 k (C) = 10.8 k (C)
a + ΣMH = 0; FCD(6) - 6(9) - 3(18) = 0
FCD = 18.0 k (C)
Ans.
Ans.
Ans.
FHI = 28.5 k (T)
FCH = 10.8 k (C)
FCD = 18.0 k (C)
109
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–31. Determine the force in members IJ, BI, and BC of the
truss. State if the members are in tension or compression.
Assume all members are pin connected.
K
J
I
H
15 ft
G
F
A
B
C
D
E
6k
6k
6k
6k
3k
5 @ 9 ft = 45 ft
Solution
Support Reactions. Not required.
Method of Sections. Referring to the FBD of the right segment of the
truss sectioned through a- a, Fig. a,
a + ΣMB = 0; FIJ a
1
b(36) - 3(36) - 6(27) - 6(18) - 6(9) = 0
110
FIJ = 12110 k (T) = 37.9 k (T)
a + ΣMF = 0; 6(9) + 6(18) + 6(27) - FBI a
1
b(36) = 0
12
Ans.
FBI = 912 k (C) = 12.7 k (C)
a + ΣMI = 0; FBC(9) - 6(9) - 6(18) - 3(27) = 0
Ans.
FBC = 27.0 k (C)
Ans.
Ans.
FIJ = 37.9 k (T)
FBI = 12.7 k (C)
FBC = 27.0 k (C)
110
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*3–32. The wooden headframe is subjected to the loading
shown. Determine the forces in members JI, JD, and ID. State
if the members are in tension or compression.
30 k
25 k
4 ft
F
G
4 ft
E
H
4 ft
D
I
4 ft
J
C
A
B
4 ft
8 ft
Solution
a + ΣMD = 0;
- FJI cos 7.125°(6) + 25(5) + 30(1) = 0
a + ΣMO = 0;
FJI = 26.0 k (C)
a + ΣMJ = 0;
FJD = 0.4543 k = 0.454 k (C)
Ans.
FJD cos 31.61°(24) + FJD sin 31.61°(3) - 30(2) + 25(2) = 0
Ans.
FCD cos 7.125°(7) - 25(1.5) - 30(5.5) = 0
Ans.
FCD = 29.15 k (C)
Joint D:
+ QΣFx = 0; 0.4543 cos 24.48° - FID cos 7.125° = 0
Ans.
FID = 0.417 k (T)
u = tan-1
2
= 7.125°
16
7
4 sec u
=
cos (u - f)
sin f
f = 31.61°
Ans.
FJI = 26.0 k (C)
FJD = 0.454 k (C)
FCD = 29.15 k (C)
FID = 0.417 k (T)
111
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–33. The wooden headframe is subjected to the loading
shown. Determine the forces in members HI, ED, and EI. State
if the members are in tension or compression.
30 k
25 k
4 ft
F
G
4 ft
E
H
4 ft
D
I
4 ft
J
C
A
B
4 ft
8 ft
Solution
a + ΣMO = 0; 25(2) - 30(2) + FEI cos 36.03°(20) + FEI sin 36.03°(2.5) = 0
FEI = 0.567 k (C)
a + ΣME = 0; 30(0.5) + 25(4.5) - FHI cos 7.125°(5) = 0
Ans.
FHI = 25.7 k (C)
Ans.
a + ΣMI = 0;
- 25(1) - 30(5) + FED cos 7.125°(6) = 0
Ans.
FED = 29.4 k (C)
u = tan -1
2
= 7.125°
16
6
4 sec u
=
cos (u - f)
sin f
f = 36.03°
Ans.
FEI = 0.567 k (C)
FHI = 25.7 k (C)
FED = 29.4 k (C)
112
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3–34. Determine the forces in all the members of the complex
truss. State if the members are in tension or compression. Hint:
Substitute member AD with one placed between E and C.
E
F
308
600 lb
308
D
12 ft
C
A
458
Solution
6 ft
458
B
6 ft
Si = S′i + x(Si 2
FEC = S′EC + (x) SEC = 0
747.9 + x(0.526) = 0
x = 1421.86
Thus,
FAF = SAF + (x) SAF
= 1373.21 + (1421.86)( - 1.41)
= -646.3 lb
FAF = 646 lb (C)
In a similar manner,
FAB = 580 lb (C)
FEB = 820 lb (T)
FBC = 580 lb (C)
FEF = 473 lb (C)
FCF = 580 lb (T)
FCD = 1593 lb (C)
FED = 1166 lb (C)
FDA = 1428 lb (T)
Ans.
FAF = 646 lb (C)
FAB = 580 lb (C)
FEB = 820 lb (T)
FBC = 580 lb (C)
FEF = 473 lb (C)
FCF = 580 lb (T)
FCD = 1593 lb (C)
FED = 1166 lb (C)
FDA = 1428 lb (T)
113
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3–35. Determine the forces in all the members of the complex
truss. State if the members are in tension or compression.
Assume all members are pin connected.
10 ft
20 ft
9k
10 ft
9k
F
E
D
G
7 ft
14 ft
C
A
B
20 ft
20 ft
Solution
Superposition required:
SGE = S′GE + xsGE = 0
19.8 + x(1.262) = 0
x = - 15.694
Member
Si ′
si
xsi
Si
CB
0
- 0.8192
12.85
12.9 (T)
CD
- 9.00
- 0.5735
9.00
0
DB
7.847
0.5
- 7.847
0
DE
- 7.847
- 0.5
7.847
0
BG
- 7.847
- 0.5
7.847
0
BA
12.857
0
0
12.9 (T)
AE
- 15.694
0
0
15.7 (C)
AG
0
0
0
0
FG
- 15.694
- 1.00
15.69
0
FE
- 12.857
- 1.638
25.71
12.9 (T)
CF
0
1.0
- 15.69
15.7 (C)
GE
19.8
1.262
- 19.8
0
Ans.
Ans.
FCB = 12.9 k (T)
FCD = 0
FDB = 0
FDE = 0
FBG = 0
FBA = 12.9 (T)
FAE = 15.7 (C)
FAG = 0
FFG = 0
FFE = 12.9 (T)
FCF = 15.7 (C)
FGE = 0
114
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*3–36. Determine the force in each member and state if the
members are in tension or compression.
4 kN
1m
4 kN
1m
E
1m
D
1m
F
C
2m
A
B
Solution
Reactions:
Ax = 0,
Ay = 4.00 kN,
Joint A:
+
S
ΣFx = 0;
+ c ΣFy = 0;
By = 4.00 kN
FAD = 0 4.00 - FAF = 0;
Ans.
FAF = 4.00 kN (C)
Ans.
Joint F:
a + ΣFy = 0;
4.00 sin 45° - FFD sin 18.43° = 0
Ans.
FFD = 8.944 kN = 8.94 kN (T)
+ QΣFx = 0;
4.00 cos 45° + 8.94 cos 18.43° - FFE = 0
Ans.
FFE = 11.313 kN = 11.3 kN (C)
Due to symmetrical loading and geometry,
FBC = 4.00 kN (C)
FBE = 0
Joint E:
+
S
Ans.
FCE = 8.94 kN (T)
Ans.
FCD = 11.3 kN (C)
ΣFx = 0;
- FED + 8.944 cos 26.56° + 11.31 cos 45° = 0
Ans.
FED = 16.0 kN (C)
+ c ΣFy = 0;
- 4 - 8.944 sin 26.56° + 11.31 sin 45° = 0 (Check!!)
Ans.
FAD = 0; FAF = 4.00 kN (C); FFD = 8.94 kN (T);
FFE = 11.3 kN (C); FBC = 4.00 kN (C);
FCE = 8.94 kN (T); FBE = 0; FCD = 11.3 kN (C);
FED = 16.0 kN (C)
115
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3 kN
3–37. Determine the forces in all the members of the complex
truss. State if the members are in tension or compression. Hint:
Substitute member AB with one placed between C and E.
2m
B
C
D
1m
G
H
F
E
2m
A
3m
1m
3m
Solution
Si = S′i + x(Si 2
FCE = S′CE + (x)SCE = 0
3.04 + x(0.779) = 0
x = -3.905
Thus:
FAB = 3.90 kN (C)
Ans.
FBC = 2.50 kN (C)
Ans.
FCD = 250 kN (C)
Ans.
FBG = FGF = FGH = FCH = FHE = 0
Ans.
Ans.
FAB = 3.90 kN (C); FBC = 2.50 kN (C);
FCD = 2.50 kN (C);
FBG = FGF = FGH = FCH = FHE = 0
116
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3–38. Determine the force in each member of the space truss.
State if the members are in tension or compression. The
supports at A and B are rollers and C is a ball-and-socket. Is
this truss stable?
z
D
4 ft
E
8k
C
B
3 ft
8 ft
x
Solution
3 ft
3 ft
A
y
Method of Joints: In this case, the support reactions are not
required for determining the member forces.
Joint E:
ΣFx = 0;
a
5
3
5
3
b a bFEB - a
b a bFEA = 0
189 5
189 5
FEB = FEA
ΣFz = 0;
2a
8
bFEB - 8 = 0
189
Ans.
FEB = FEA = 4.717 kN = 4.72 k (C)
ΣFy = 0;
- FED + 2a
5
4
b a b(4.717) = 0
189 5
Ans.
FED = 4.00 k (T)
Joint D:
ΣFy = 0;
4.00 - FDC sin 20.56° = 0
ΣFx = 0;
FDB sin 20.56° - FDA sin 20.56° = 0
ΣFz = 0;
2FDA cos 20.56° - 11.39 cos 20.56° = 0
Ans.
FDC = 11.39 k = 11.4 k (T)
FDB = FDA
Ans.
FDA = FDB = 5.696 k = 5.70 k (C)
Joint A:
ΣFy = 0;
FAC cos 45° - a
5
4
b a b(4.717) = 0
189 5
Ans.
FAC = 2.828 k = 2.83 k (C)
ΣFx = 0;
5.696 sin 20.56° + a
FAB = 5.50 k (T)
5
3
b a b(4.717) + 2.828 sin 45° - FAB = 0
189 5
Ans.
Joint B:
ΣFx = 0;
5.5 - 4.717
or from symmetry,
3
1
- 5.696 sin 20.56° FBC = 0
189
12
Ans.
FBC = 2.828 k = 2.83 k (C)
The truss is externally unstable since it can rotate about the
z axis.
Ans.
FEB = FEA = 4.72 k (C)
FED = 4.00 k (T)
FDC = 11.4 k (T)
FDA = 5.70 k (C)
FAC = 2.83 k (C)
FAB = 5.50 k (T)
FBC = 2.83 k (C)
The truss is externally
unstable, since it can
rotate about the z-axis.
117
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z
3–39. Determine the force in the members of the space truss,
and state whether they are in tension or compression.
2m
C
D
2m
2m
3 kN
B
A
x
5 kN
2m
y
Solution
Support Reactions. Not required.
Method of Joints.
Joint D, Fig. a.
1
1
b - FDB a
b = 0(1)
18
18
ΣFx = 0;
FDA a
ΣFy = 0;
- FDA a
ΣFz = 0;
- FDA a
2
2
b - FDB a
b - FDC + 3 = 0(2)
18
18
1.732
1.732
b - FDB a
b - 5 = 0(3)
18
18
Solving Eqs. (1), (2) and (3),
FDA = FDB = - 4.082 kN = 4.08 kN (C)
Ans.
FDC = 8.774 kN = 8.77 kN (T)
Ans.
Joint B, Fig. b.
ΣFz = 0;
FBC sin 60° - 4.082a
1.732
b = 0
18
Ans.
FBC = 2.887 kN = 2.89 kN (T)
ΣFx = 0;
FBA + 2.887 cos 60° - 4.082a
FBA = 0
ΣFy = 0;
- By - 4.082a
Joint A, Fig. c.
ΣFz = 0;
2
b = 0
18
FAC sin 60° - 4.082a
1
b = 0
18
Ans.
By = - 2.887 kN
1.732
b = 0
18
Ans.
FAC = 2.887 kN = 2.89 kN (T)
1
b - 2.887 cos 60° = 0
18
ΣFx = 0;
Ax + 4.082a
ΣFy = 0;
- Ay - 4.082a
2
b = 0
18
Ax = 0
Ay = - 2.887 kN
Ans.
FDA = 4.08 kN (C)
FDC = 8.77 kN (T)
FBC = 2.89 kN (T)
FBA = 0
FAC = 2.89 kN (T)
118
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*3–40. Determine the force in each member of the space truss
and state if the members are in tension or compression. The truss is
supported by ball-and socket joints at C, D, E, and G. Note:
Although this truss is indeterminate to the first degree, a solution is
possible due to symmetry of geometry and loading.
z
1m
G
1m
E
A
D
2m
x
F 5 3 kN
C
2m
B
1.5 m
y
Solution
Σ(MEG 2 x = 0;
2
2
4
FBC(2) +
FBD(2) - (3)(2) = 0
5
15
15
FBC + FBD = 2.683 kN
Ans.
Due to symmetry: FBC = FBD = 1.342 = 1.34 kN (C)
Joint A:
ΣFz = 0;
FAB -
4
(3) = 0
5
Ans.
FAB = 2.4 kN (C)
ΣFx = 0;
ΣFy = 0;
FAG = FAE
3
3
3
(3) FAE FAG = 0
5
15
15
Ans.
FAG = FAE = 1.01 kN (T)
Joint B:
ΣFx = 0;
ΣFy = 0;
ΣFz = 0;
1
1
1
1
(1.342) + FBE (1.342) - FBG = 0
3
3
15
15
2
2
2
2
(1.342) - FBE +
(1.342) - FBG = 0
3
3
15
15
2
2
F + FBG - 2.4 = 0
3 BE
3
FBG = 1.80 kN (T)
Ans.
FBE = 1.80 kN (T)
Ans.
Ans.
FBC = FBD = 1.34 kN (C); FAB = 2.4 kN (C);
FAG = FAE = 1.01 kN (T); FBG = 1.80 kN (T);
FBE = 1.80 kN (T)
119
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3–41. Determine the force in members FE and ED of the
space truss and state if the members are in tension or
compression. The truss is supported by a ball-and-socket joint
at C and short links at A and B.
z
G
D
4 kN
F
3m
E
C
A
3m
B
Solution
0.75 m
x 0.75 m
1m
1.5 m
y
Method of Joints.
Joint F. Here, FFG, FFD and FFC lie on the same plane shown
shaded in Fig a and the x¿ axis is the normal to this plane, thus,
ΣFx′ = 0; FFE cos u = 0 FFE = 0
Ans.
Joint E. Here, FEG, FEF, FEC and FEB lie on the same plane
shown shaded in Fig b and the x¿ axis is the normal to this
plane, thus,
ΣFx′ = 0; FED cos u = 0 FED = 0
Ans.
Ans.
FFE = 0; FED = 0
120
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z
3–42. Determine the force in members GD, GE, GF, and FD
of the space truss and state if the members are in tension or
compression.
G
D
4 kN
F
E
C
Solution
A
Support Reactions. Not required.
3m
B
Method of Joints.
0.75 m
x 0.75 m
Joint G. Fig. a.
ΣFx = 0; FGD a
3m
1
1
15
b - FGE a
b + FGF a
b = 0
137.5625
137.5625
138.8125
ΣFy = 0;
- FGD a
ΣFZ = 0;
- FGD a
0.75
0.75
0.75
b - FGE a
b - FGF a
b + 4 = 0
137.5625
137.5625
138.8125
6
6
6
b + FGE a
b + FGF a
b = 0
137.5625
137.5625
138.8125
1m
1.5 m
y
(1)
(2)
(3)
Solving Eqs. (1), (2) and (3),
FGF = 0
Ans.
FGD = 16.34 kN (T) = 16.3 kN (T)
Ans.
FGE = 16.34 kN (C) = 16.3 kN (C)
Ans.
Joint F. Referring to Fig. b, we notice that FFG, FFC and FFE lie in the same plane (shown shaded) and
x¿ axis is the normal to this plane. Thus,
ΣFx′ = 0;
FFD cos u = 0
Ans.
FFD = 0
Ans.
FGF = 0
FGD = 16.3 kN (T)
FGE = 16.3 kN (C)
FFD = 0
121
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3–43. Three identical trusses are pin connected to produce the
framework shown. If the framework rests on the smooth
supports at A, C, and E, determine the force in members CD,
DH, and CH. State if the members are in tension or compression.
2k
2k
F
2k
I
G
A
H
B
4 ft
4 ft
E
4 ft
4 ft
D
4 ft
C
4 ft
4 ft
Solution
h = 242 - 2.3093 = 3.2660 ft
Due to symmetrical system and loading,
Ay = Cy = Ey = 2.0 k
Joint C:
ΣFZ = 0;
2.0 -
3.2660
FCH = 0
4
FCH = 2.449 k = 2.45 k (C)
ΣFy = 0;
- FCB cos 30° +
Ans.
1.155
(2.449) = 0
4
FCB = 0.8165 k = 0.817 k (T)
ΣFx = 0;
- FCD - 0.8165 sin 30° +
2
(2.449) = 0
4
FCD = 0.8165 k = 0.817 k (T)
Ans.
Joint D:
Due to symmetrical geometry and loading,
FDE = FCD = 0.8165 k (T)
and FDH = FDF
ΣFy = 0;
2FDH cos 30° = 0
FDH = 0
Ans.
Ans.
FCH = 2.45 k (C)
FCD = 0.817 k (T)
FDH = 0
122
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z
*3–44. Determine the force in members AB, BD, and FE of
the space truss and state if the members are in tension or
compression.
D
F
3 ft
E
3 ft
C
A
x
3 ft
Solution
500 lb
Support Reactions. Not required.
B
4 ft
y
500 lb
Method of Joints.
Joint A, Fig. a.
ΣFz = 0;
FAF sin 60° - 500 = 0 FAF = 577.35 lb (T)
ΣFy = 0
577.35 cos 60° + FAB = 0 FAB = - 288.68 lb = 289 lb (C)
Ans.
Joint F, Fig. b. FAF , FFD and FFE lie on the same plane and FFB is out of this
plane. Then, FFB = 0.
ΣFZ = 0;
- 577.35 sin 60° - FFE a
2.598
b = 0
5
Ans.
FFE = - 962.25 lb = 962 lb (C)
Joint B, Fig. c.
ΣFZ = 0; FBD a
2.598
b - 500 = 0
5
Ans.
FBD = 962.25 lb = 962 lb (T)
Ans.
FAB = 287 lb (C); FFE = 962 lb (C);
FBD = 962 lb (T)
123
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3–45. Determine the force in members AF, AE, and FD of
the space truss and state if the members are in tension or
compression.
z
D
F
3 ft
E
3 ft
C
A
x
3 ft
500 lb
Solution
B
4 ft
y
500 lb
Support Reactions. Not required.
Method of Joints.
Joint A. (Fig. a)
ΣFx = 0;
FAE = 0
Ans.
ΣFZ = 0;
FAF sin 60° - 500 = 0 FAF = 577.35 lb = 577 lb (T)
Ans.
Joint F. (Fig. b) FAF FFD, and FFE lie on the same plane and FFB is out of this
plane. Then, FFB = 0.
ΣFZ = 0;
- 577.35 sin 60° - FFE a
2.598
b = 0
5
FFE = - 962.25 lb = 962.25 lb (C)
ΣFx = 0;
4
- FFD - ( - 962.25) a b = 0
5
Ans.
FFD = 769.80 lb = 770 lb (T)
Ans.
FAE = 0; FAF = 577 lb (T); FFD = 770 lb (T)
124
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3–1P. The Pratt roof trusses are uniformly spaced every
15 ft. The deck, roofing material, and the purlins have an
average weight of 5.6 lb>ft2. The building is located in
New York where the anticipated snow load is 20 lb>ft2 and the
anticipated ice load is 8 lb>ft2. These loadings occur over the
horizontal projected area of the roof. Determine the force in
each member due to dead load, snow, and ice loads. Neglect the
weight of the truss members and assume A is pinned and E is a
roller.
6 ft
G
H
A
B
8 ft
C
8 ft
F
6 ft
D
E
8 ft
8 ft
Solution
Loading:
At joints H, G, and F,
F = (20 + 8)(15)(8) + 5.6(15)(10) = 4.20 k
At joints A and E,
F = (20 + 8)(15)(4) + 5.6(15)(5) = 2.10 k
Due to symmetrical loading and geometry,
Ax = 0,
Ay = 8.40 k,
Ey = 8.40 k
Joint A:
+ c ΣFy = 0;
+
S
ΣFx = 0;
3
8.40 - 2.10 - FAH a b = 0
5
FAH = 10.5 k (C)
Ans.
4
FAB - 10.5a b = 0
5
FAB = 8.40 k (T)
Ans.
Joint B:
+
S
ΣFx = 0;
+ c ΣFy = 0;
FBC = 8.40 = 0
FBC = 8.40 k (T)
Ans.
FBH = 0
Ans.
Joint H:
+ aΣFy = 0;
- 4.20 cos 36.87° + FBC cos 16.26° = 0
Ans.
FHC = 3.50 k (C)
+ QΣFx = 0;
10.5 - FNG - 4.20 sin 36.87° - 3.50 sin 16.26° = 0
FHG = 7.00 k (C)
Ans.
4
4
(7.00) - FGF = 0
5
5
FGF = 7.00 k (C)
Ans.
Joint G:
+
S
ΣFx = 0;
+ c ΣFy = 0;
3
3
(7.00) + (7.00) - 4.20 - FGC = 0
5
5
Ans.
FGC = 4.20 k (T)
Due to symmetrical loading and geometry,
Ans.
Ans.
Ans.
Ans.
Ans.
FDE = FAB = 8.40 k (T)
FDC = FBC = 8.40 k (T)
FEF = FAH = 10.5 k (C)
FBH = FDF = 0
FHC = FFC = 3.50 k (C)
Ans.
FHG = 7.00 k (C); FGF = 7.00 k (C);
FGC = 4.20 k (T); FDE = FAB = 8.40 k (T);
FDC = FBC = 8.40 k (T);
FEF = FAH = 10.5 k (C);
FBH = FDF = 0; FHC = FFC = 3.50 k (C)
125
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4–1. Determine the internal normal force, shear force, and
bending moment in the beam at points C and D. Point D is
located just to the right of the concentrated force and
moment. Assume the support at B is a pin and A is a roller.
15 kN
15 kN>m
A
D
C
B
30 kN ? m
2m
2m
2m
Solution
Support Reactions. Referring to the FBD of the entire beam shown in
Fig. a,
a + ΣMA = 0;
a + ΣMB = 0;
+
S
ΣFx = 0;
By(6) - 15(2)(1) - 15(14) - 30 = 0
By = 20.0 kN
15(2) + 15(2)(5) - 30 - NA(6) = 0
NA = 25.0 kN
Bx = 0
Internal Loadings. Referring to the FBD of the left segment of the beam
sectioned through point C, Fig. b,
+
S
ΣFx = 0;
NC = 0
Ans.
+ c ΣFy = 0;
Ans.
a + ΣMC = 0;
25.0 - 15(2) - VC = 0
MC + 15(2)(1) - 25.0(2) = 0
VC = -5.00 kN
MC = 20.0 kN # m
Ans.
Referring to the right segment of the beam sectioned through point D,
Fig. c,
+
S
ΣFx = 0;
ND = 0
Ans.
+ c ΣFy = 0;
Ans.
a + ΣMD = 0;
20.0 + VD = 0
20.0(2) - MD = 0
VD = - 20.0 kN
MD = 40.0 kN # m
Ans.
Ans.
NC = 0
VC = -5.00 kN
MC = 20.0 kN # m
ND = 0
VD = -20.0 kN
MD = 40.0 kN # m
126
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–2. Determine the internal normal force, shear force, and
bending moment at point C. Point C is located just to the left of
the moment.
12 kN
12 kN>m
24 kN ? m
A
2m
2m
C
2m
B
2m
Solution
Support Reactions. Referring to the FBD of the entire beam shown
in Fig. a,
a + ΣMA = 0;
NB(4) + 24 - 12(6) = 0
NB = 12.0 kN
Internal Loadings. Referring to the FBD of the right segment of the
beam sectioned through point C, Fig. b,
+
S
ΣFx = 0;
NC = 0
Ans.
+ c ΣFy = 0;
a + ΣMC = 0;
VC + 12.0 - 12.0 = 0
Ans.
VC = 0
24 + 12.0(2) - 12(4) - MC = 0
MC = 0
Ans.
Ans.
NC = 0
VC = 0
MC = 0
127
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4–3. Determine the internal normal force, shear force, and
moment acting at point C and at point D, which is located just
to the right of the roller support at B.
300 lb>ft
200 lb>ft
200 lb>ft
D
F
A
4 ft
4 ft
C
B
4 ft
E
4 ft
Solution
Support Reaction.
a + ΣMA = 0;
From FBD (a),
By(8) + 800(2) - 2400(4) - 800(10) = 0
By = 2000 lb
Internal Forces. Applying the equations of equilibrium to
segment ED [FBD (b)], we have
+
S
ΣFx = 0;
ND = 0
Ans.
+ c ΣFy = 0;
Ans.
a + ΣMD = 0;
VD - 800 = 0
VD = 800 lb
- MD - 800(2) = 0
MD = - 1600 lb # ft = -1.60 k # ft
Ans.
Applying the equations of equilibrium to segment EC [FBD (c)],
we have
+
S
ΣFx = 0;
NC = 0
Ans.
+ c ΣFy = 0;
Ans.
a + ΣMC = 0;
VC + 2000 - 1200 - 800 = 0
VC = 0
2000(4) - 1200(2) - 800(6) - MC = 0
MC = 800 lb # ft
Ans.
Ans.
ND = 0
VD = 800 lb
MD = - 1.60 k # ft
NC = 0
VC = 0
MC = 800 lb # ft
128
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*4–4. Determine the internal normal force, shear force, and
bending moment at point D. Take w = 150 N>m.
w
B
A
D
4m
3m
4m
C
4m
Solution
a + ΣMA = 0;
- 150(8)(4) +
3
F (8) = 0
5 BC
FBC = 1000 N
+
S
ΣFx = 0;
Ax -
4
(1000) = 0
5
Ax = 800 N
+ c ΣFy = 0;
Ay - 150(8) +
+
S
ΣFx = 0;
ND = - 800 N
3
(1000) = 0
5
Ay = 600 N
+ c ΣFy = 0;
600 - 150(4) - VD = 0
a + ΣMD = 0;
VD = 0
Ans.
Ans.
- 600(4) + 150(4)(2) + MD = 0
MD = 1200 N # m = 1.20 kN # m
Ans.
Ans.
ND = - 800 N
VD = 0
MD = 1.20 kN # m
129
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4–5. The beam AB will fail if the maximum internal moment
at D reaches 800 N # m or the normal force in member BC
becomes 1500 N. Determine the largest load w it can support.
w
B
A
D
4m
3m
4m
C
4m
Solution
Assume maximum moment occurs at D.
a + ΣMD = 0;
MD -
8w
(4) + 4w (2) = 0
2
800 = 8w
a + ΣMA = 0;
w = 100 N>m
- 800(4) + TBC(0.6)(8) = 0
TBC = 666.7 N 6 1500 N
(OK!)
w = 100 N>m
Ans.
Ans.
w = 100 N>m
130
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4–6. Determine the internal normal force, shear force, and
bending moment in the beam at points C and D. Point C is
located just to the left of the roller support. Assume the support
at B is a roller and A is a pin.
18 kN>m
9 kN>m
A
B
D
3m
3m
C
3m
Solution
Support Reactions.
a + ΣMA = 0;
Referring to the FBD of the entire beam, shown in Fig. a,
NB(6) -
1
(9)(9)(3) - 9(9)(4.5) = 0
2
NB = 81.0 kN
Internal Loadings. Referring to the FBD of the right segment of the beam
sectioned through point C, Fig. b,
+
S ΣFx = 0;
NC = 0
Ans.
+ c ΣFy = 0;
VC + 81.0 -
a + ΣMC = 0;
- MC -
1
(3)(3) - 9(3) = 0
2
1
(3)(3)(1) - 9(3)(15) = 0
2
VC = -49.5 kN
Ans.
MC = -45.0 kN # m
Ans.
Referring to the FBD of the right segment of the beam sectioned through point D,
Fig. c,
+
S ΣFx = 0;
ND = 0
Ans.
1
(6)(6) - 9(6) = 0
2
+ c ΣFy = 0;
VD + 81.0 -
a + ΣMD = 0;
81.0(3) - 9(6)(3) -
VD = -9.00 kN
Ans.
1
(6)(6)(2) - MD = 0
2
MD = 45.0 kN # m
Ans.
Ans.
NC = 0
VC = - 49.5 kN
MC = - 45.0 kN # m
ND = 0
VD = - 9.00 kN
MD = 45.0 kN # m
131
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4–7. Determine the internal normal force, shear force, and
bending moment acting at point C, located just to the right of
the 12-kN force and 18 kN # m moment.
18 kN
12 kN
A
2m
C
B
18 kN ? m
2m
2m
Solution
Support Reactions.
shown in Fig. a,
a + ΣMA = 0;
Referring to the FBD of the entire beam,
NB(6) - 12(2) - 18(4) - 18 = 0 NB = 19.0 kN
Internal Loadings. Referring to the FBD of the right segment of
the beam sectioned through point C, Fig. b,
+
S ΣFx = 0;
NC = 0
Ans.
+ c ΣFy = 0;
a + ΣMC = 0;
VC + 19.0 - 18 = 0 VC = - 1.00 kN
19.0(4) - 18(2) - MC = 0 MC = 40.0 kN # m
Ans.
Ans.
Ans.
NC = 0
VC = - 1.00 kN
MC = 40.0 kN # m
132
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*4–8. Determine the internal normal force, shear force, and
moment at points E and D of the compound beam.
800 N
200 N m
A
D
2m
2m
4m
B
E
2m
2m
C
Solution
Segment EC:
+
S Σ Fx = 0;
NE = 0
+ c ΣFy = 0;
VE + 50 = 0
a + Σ ME = 0;
VE = - 50 N
Ans.
Ans.
- 200 + 50(2) - ME = 0
ME = - 100 N # m
Segment DB:
+
S Σ Fx = 0;
ND = 0
+ c ΣFy = 0;
VD - 800 + 50 = 0
a + ΣMD = 0;
VD = 750 N
Ans.
Ans.
- 800(2) + 6(50) - MD = 0
MD = - 1300 N # m = -1.30 kN # m Ans.
Ans.
NE = 0
VE = - 50 N
ME = - 100 N # m
VD = 750 N
MD = - 1.30 kN # m
133
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4–9. Determine the internal normal force, shear force, and
moment at point C of the beam.
400 N>m
200 N>m
A
B
C
3m
3m
Solution
Beam:
a + Σ MB = 0;
+
S ΣFx = 0;
600(2) + 1200(3) - Ay(6) = 0
Ay = 800 N
Ax = 0
Segment AC:
+
S ΣFx = 0;
NC = 0
+ c ΣFy = 0;
800 - 600 - 150 - VC = 0
a + ΣMC = 0;
VC = 50 N
Ans.
Ans.
- 800(3) + 600(1.5) + 150(1) + MC = 0
MC = 1350 N # m = 1.35 kN # m
Ans.
Ans.
NC = 0
VC = 50 N
MC = 1.35 kN # m
134
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4–10. Determine the internal normal force, shear force, and
moment in the beam at points C and D. Point D is just to the
right of the 5-kip load.
5 kip
0.5 kip>ft
A
6 ft
B
D
C
6 ft
6 ft
6 ft
Solution
Entire beam:
a + ΣMA = 0;
5162 + 61182 - Ay 1242 = 0
Ay = 5.75 k c
+ ΣF = 0;
S
x
Ax = 0
Segment AC:
+
S
ΣFx = 0;
NC = 0
+ c ΣFy = 0;
5.75 - 3 - VC = 0
a + ΣMC = 0;
VC = 2.75 k
Ans.
Ans.
MC + 3132 - 5.75162 = 0
MC = 25.5 k # ft
Ans.
Segment AD:
+
S
ΣFx = 0;
ND = 0
Ans.
+ c ΣFy = 0;
5.75 - 6 - 5 - VD = 0
a + ΣMD = 0;
VD = - 5.25 k
Ans.
MD + 61122 - 5.751182 = 0
MD = 31.5 k # ft
Ans.
Ans.
NC = 0
VC = 2.75 k
MC = 25.5 k # ft
ND = 0
VD = - 5.25 k
MD = 31.5 k # ft
135
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4–11. Determine the internal normal force, shear force, and
bending moment in the beam at point C.
3 k>ft
B
A
9 ft
9 ft
C
9 ft
Solution
Support Reactions.
shown in Fig. a,
Referring to the FBD of the entire beam,
a + ΣMA = 0;
By(18) +
+ ΣF = 0;
S
x
Bx = 0
1
1
(3)(9)(3) - (3)(18)(6) = 0 By = 6.75 k
2
2
Internal Loadings. Referring to the FBD of the right segment of the beam sectioned through point C, Fig. b,
+ ΣF = 0;
S
x
NC = 0
+ c ΣFy = 0;
VC + 6.75 -
a + ΣMC = 0;
6.75(9) -
Ans.
1
(1.5)(9) = 0 VC = 0
2
Ans.
1
(1.5)(9)(3) - MC = 0 MC = 40.5 k # ft Ans.
2
Ans.
NC = 0
VC = 0
MC = 40.5 k # ft
136
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600 N
*4–12. Determine the shear and moment throughout the
beam as a function of x.
A
B
x
4m
2m
Solution
Support Reactions. Referring to the FBD of the entire beam
shown in Fig. a,
a + ΣMB = 0;
a + ΣMA = 0;
+ ΣF = 0;
S
x
600(2) - NA(6) = 0
NA = 200 N
By(6) - 600(4) = 0
By = 400 N
Bx = 0
Internal Loadings. For 0 … x 6 4 m, refer to the FBD of the
left segment of the beam, Fig. b.
+ c ΣFy = 0;
a + ΣMO = 0;
200 - V = 0
M - 200x = 0
V = {200} N
Ans.
M = {200x} N # m
Ans.
For 4m 6 x … 6 m, refer to the FBD of the right segment of
the beam, Fig. c.
+ c ΣFy = 0;
a + ΣMO = 0;
V + 400 = 0
400(6 - x) - M = 0
V = { - 400} N
N = {400(6 - x)} N # m
Ans.
Ans.
Ans.
For 0 … x 6 4 m,
V = {200} N
M = {200x} N # m
For 4 m 6 x … 6 m,
V = { -400} N
N = {400(6 - x)} N # m
137
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4–13. Determine the shear and moment throughout the
beam as a function of x.
15 kN>m
B
A
x
4m
2m
Solution
Support Reactions.
shown in Fig. a,
a + ΣMA = 0;
a + ΣMB = 0;
+ ΣF = 0;
S
x
Referring to the FBD of the entire beam,
NB(6) - 15(4)(2) = 0
NB = 20.0 kN
15(4)(4) - Ay(6) = 0
Ay = 40.0 kN
Ax = 0
Internal Loadings. For 0 … x 6 4 m, refer to the FBD of the
left segment of the beam, Fig. b.
+ c ΣFy = 0;
a + ΣMO = 0;
40.0 - 15x - V = 0
V = 540.0 - 15x 2 6 kN
x
M + 15xa b - 40.0x = 0
2
M = 540.0x - 7.50x 2 6 kN # m Ans.
Ans.
For 4 m 6 x … 6 m, refer to the FBD of the right segment of
the beam, Fig. c.
+ c ΣFy = 0;
a + ΣMO = 0;
V + 20.0 = 0
20.0(6 - x) - M = 0
V = 5 - 20.06 kN
M = 520.0(6 - x)6 kN # m
Ans.
Ans.
Ans.
For 0 … x 6 4 m,
V = 540.0 - 15x 2 6 kN
M = 540.0x - 7.50x 2 6 kN # m
For 4 m 6 x … 6 m,
V = 5 -20.06 kN
M = 520.0(6 - x)6 kN # m
138
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4–14. Determine the shear and moment throughout the
beam as a function of x.
4m
2m
B
A
12 kN ? m
x
Solution
Support Reactions.
shown in Fig. a,
a + ΣMA = 0;
a + ΣMB = 0;
+ ΣF = 0;
S
x
Referring to the FBD of the entire beam,
12 - NB(6) = 0 NB = 2.00 kN
12 - Ay(6) = 0
Ay = 2.00 kN
Ax = 0
Internal Loadings. For 0 … x 6 4 m, refer to the FBD of the
left segment of the beam, Fig. b.
+ c ΣFy = 0;
a + ΣMO = 0;
2.00 - V = 0
V = 52.006 kN
Ans.
M - 2.00x = 0 M = 52.00x6 kN # m
Ans.
For 4 m 6 x … 6 m, refer to the FBD of the right segment of
the beam, Fig. c.
+ c ΣFy = 0;
a + ΣMO = 0;
V - 2.00 = 0
V = 52.006 kN
Ans.
- M - 2.00(6 - x) = 0 M = 52.00(x - 6)6 kN # m Ans.
Ans.
For 0 … x 6 4 m,
V = 52.006 kN
M = 52.00x6 kN # m
For 4 m 6 x … 6 m,
V = 52.006 kN
M = 52.00(x - 6)6 kN # m
139
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4–15. Determine the shear and moment in the floor girder as
a function of x. Assume the support at B is a pin and A is a
roller.
6 kN
4 kN
A
B
1m
x
2m
1m
Solution
Support Reactions. Referring to the FBD of the entire beam
in Fig. a,
a + ΣMA = 0;
a + ΣMB = 0;
By(4) - 4(1) - 6(3) = 0
By = 5.50 kN
6(1) + 4(3) - Ay(4) = 0
Ay = 4.50 kN
+ ΣF = 0;
S x
Ax = 0
Internal Loadings. For 0 … x 6 1 m, referring to the FBD of
the left segment of the beam in Fig. b,
+ c ΣFy = 0;
a + ΣMO = 0;
4.50 - V = 0 V = 4.50 kN
Ans.
M - 4.50x = 0
M = 54.50x6 kN # m
Ans.
For 1 m 6 x 6 3 m, referring to the FBD of the left segment
of the beam in Fig. c,
+ c ΣFy = 0;
4.50 - 4 - V = 0
a + ΣMO = 0;
V = 0.500 kN
Ans.
M + 4(x - 1) - 4.50x = 0
M = 50.5x + 46 kN # m
Ans.
For 3 m 6 x … 4 m, referring to the FBD of the right segment
of the beam in Fig. d,
+ c ΣFy = 0;
a + ΣMO = 0;
V + 5.50 = 0 V = - 5.50 kN
Ans.
Ans.
For 0 … x 6 1 m,
V = 4.50 kN
M = 54.50x6 kN # m
For 1 m 6 x 6 3 m,
V = 0.500 kN
M = 50.5x + 46 kN # m
For 3 m 6 x … 4 m,
V = - 5.50 kN
M = 5- 5.50x + 226 kN # m
5.50(4 - x) - M = 0
M = 5- 5.50x + 226 kN # m
Ans.
140
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*4–16. Determine the shear and moment throughout the
beam as a function of x.
8 kN>m
A
B
x
3m
3m
Solution
Support Reactions. Referring to the FBD of the entire beam
in Fig. a,
a + ΣMA = 0;
a + ΣMB = 0;
By(6) - 8(3)(4.5) -
1
(8)(3)(2) = 0
2
By = 22 kN
8(3)(1.5) +
1
(8)(3)(4) - Ay(6) = 0
2
Ay = 14 kN
+ ΣF = 0;
S
x
Ax = 0
Internal Loadings. For 0 … x 6 3 m, refer to the FBD of the
left segment of the beam in Fig. b.
+ c ΣFy = 0;
a + ΣMO = 0;
14 -
1 8
a xbx - V = 0
2 3
V = 5 - 1.33x 2 + 146 kN
M +
Ans.
1 8
x
a xb(x) a b - 14x = 0
2 3
3
M = 5 - 0.444x 3 + 14x6 kN # m
Ans.
For 3 m 6 x … 6 m, refer to the FBD of the right segment of
the beam in Fig. c.
+ c ΣFy = 0;
a + ΣMO = 0;
V + 22 - 8(6 - x) = 0
V = 5 - 8x + 266 kN
22(6 - x) - 8(6 - x) a
Ans.
6 - x
b - M = 0
2
M = 5 - 4x 2 + 26x - 126 kN # m
Ans.
Ans.
For 0 … x 6 3 m,
V = 5 -1.33x 2 + 146 kN
M = 5 -0.444x 3 + 14x6 kN # m
For 3 m 6 x … 6 m,
V = 5 -8x + 266 kN
M = 5 -4x 2 + 26x - 126 kN # m
141
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–17. Determine the shear and moment in the beam as a
function of x.
Solution
Internal Loadings. Referring to the FBD of the left segment
of the beam in Fig. a,
+ c ΣFy = 0;
a + ΣMO = 0;
- 800 -
1 200
a
xb(x) - V = 0
2 10
V = 5 -10x 2 - 8006lb
M +
Ans.
1 200
x
a
xb(x) a b + 800x + 1200 = 0
2 10
3
M = 5- 3.33x 3 - 800x - 12006 lb # ft
Ans.
Ans.
V = 5 -10x 2 - 8006 lb
M = 5 -3.33x 3 - 800x - 12006 lb # ft
142
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4–18. Determine the shear and moment throughout the
beam a function of x.
2 k>ft
10 k
8k
40 k ? ft
x
6 ft
4 ft
Solution
Support Reactions. As shown on FBD.
Shear and Moment Functions.
For 0 … x 6 6 ft,
+ c ΣFy = 0;
a + ΣMNA = 0;
30.0 - 2x - V = 0
V = 530.0 - 2x6 k
Ans.
x
M + 216 + 2xa b - 30.0x = 0
2
M = 5 - x 2 + 30.0x - 2166 k # ft
For 6 ft 6 x … 10 ft,
+ ΣF = 0;
S
V - 8 = 0
y
a + ΣMNA = 0;
V = 8.00 k
- M - 8(10 - x) - 40 = 0
M = 58.00x - 1206 k # ft
Ans.
Ans.
Ans.
Ans.
For 0 … x 6 6 ft, V = 530.0 - 2x6 k;
M = 5 -x 2 + 30.0x - 2166 k # ft;
For 6 ft 6 x … 10 ft, V = 8.00 k;
M = 58.00x - 1206 k # ft
143
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4–19. Determine the shear and moment throughout the
beam a function of x.
250 lb
250 lb
150 lb>ft
A
B
x
4 ft
6 ft
4 ft
Solution
Support Reactions. As shown on FBD.
Shear and Moment Functions.
For 0 … x 6 4 ft,
+ c ΣFy = 0;
a + ΣMNA = 0;
- 250 - V = 0
Ans.
V = - 250 lb
M = 5 - 250x6 lb # ft
M + 250x = 0
Ans.
For 4 ft 6 x 6 10 ft,
+ c ΣFy = 0;
-250 + 700 - 150(x - 4) - V = 0
V = 51050 - 150x6 lb
a + ΣMNA = 0;
M + 150(x - 4) a
Ans.
x - 4
b + 250x - 700(x - 4) = 0
2
M = 5 - 75x 2 + 1050x - 40006 lb # ft
Ans.
For 10 ft 6 x … 14 ft,
+ c ΣFy = 0;
a + ΣMNA = 0;
V - 250 = 0
Ans.
V = 250 lb
-M - 250(14 - x) = 0
M = 5250x - 35006 lb # ft
Ans.
Ans.
For 0 … x 6 4 ft, V = -250 lb;
M = 5 -250x6 lb # ft;
For 4 ft 6 x 6 10 ft, V = 51050 - 150x6 lb;
M = 5 -75x 2 + 1050x - 40006 lb # ft
For 10 ft 6 x … 14 ft, V = 250 lb;
M = 5250x - 35006 lb # ft
144
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*4–20. Determine the shear and moment in the beam a
function of x.
w0
A
B
x
L
__
2
L
__
2
Solution
Support Reactions. As shown on FBD.
Shear and Moment Functions.
For 0 … x 6 L>2,
+ c ΣFy = 0;
a + ΣMNA = 0;
3woL
wo
- w ox - V = 0 V =
(3L - 4x)
4
4
Ans.
7woL2
3woL
x
x + woxa b + M = 0
24
4
2
M =
wo
( - 12x 2 + 18Lx - 7L2)
24
Ans.
For L>2 6 x … L,
1 2wo
c
(L - x) d (L - x) = 0
2 L
wo
V =
(L - x)2
L
+ c ΣFy = 0; V -
a + ΣMNA = 0;
Ans.
L - x
1 2w0
-M - c
1L - x2 d 1L - x2 a
b = 0
2 L
3
wo
M = (L - x)3
Ans.
3L
Ans.
For 0 … x 6 L>2,
wo
V =
(3L - 4x)
4
wo
M =
( -12x 2 + 18Lx - 7L2)
24
For L>2 6 x … L,
wo
V =
(L - x)2
L
wo
M = (L - x)3
3L
145
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–21. Determine the shear and moment in the beam a
function of x.
150 lb> ft
200 lb ft
A
200 lb ft
B
x
4 ft
6 ft
4 ft
Solution
+ c ΣFy = 0;
a + ΣMs = 0;
1 w
a xbx - P = 0
2 30
wx 2
V = - P
60
-V -
M +
Ans.
1 w
x
a xbxa b + Px = 0
2 30
3
M = -
wx 3
- Px
180
Ans.
Ans.
146
V = -
wx 2
- P;
60
M = -
wx 3
- Px
180
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–22. Draw the shear and moment diagrams for the beam
and determine the shear and moment in the beam a function
of x, where 4 ft 6 x 6 10 ft.
150 lb> ft
200 lb ft
A
200 lb ft
B
x
4 ft
6 ft
4 ft
Solution
+ c ΣFy = 0;
a + ΣM = 0;
- 150(x - 4) - V + 450 = 0
V = 1050 - 150x
- 200 - 150(x - 4)
(x - 4)
2
Ans.
- M + 450(x - 4) = 0
M = - 75x 2 + 1050x - 3200
Ans.
Ans.
V = 1050 - 150x
M = -75x 2 + 1050x - 3200
147
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4–23. Draw the shear and moment diagrams for the beam
and determine the shear and moment a function of x.
400 N>m
200 N> m
A
B
x
3m
3m
Solution
Support Reactions. As shown on FBD.
Shear and Moment Functions.
For 0 … x 6 3 m:
+ c ΣFy = 0;
a + ΣMNA = 0;
200 - V = 0
V = 200 N
Ans.
M - 200x = 0
M = 5200x6 N # m
Ans.
For 3 m 6 x … 6 m:
+ c ΣFy = 0;
200 - 200(x - 3) V = e-
1 200
c
(x - 3) d (x - 3) - V = 0
2 3
100 2
x + 500 f N
3
Ans.
Set V = 0, x = 3.873 m.
a + ΣMNA = 0;
M +
1 200
x - 3
b
c
(x - 3) d (x - 3) a
2 3
3
+ 200(x - 3) a
M = e-
x - 3
b - 200x = 0
2
100 3
x + 500x - 600 f N # m
9
Ans.
Substitute x = 3.87m, M = 691 N # m.
Ans.
For 0 … x 6 3 m: V = 200 N, M = 5200x6 N # m,
For 3 m 6 x … 6 m: V = e M = e148
100 2
x + 500 f N,
3
100 3
x + 500x - 600 f N # m
9
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–24.
500 lb
Draw the shear and moment diagrams for the beam.
300 lb
200 lb
B
A
8 ft
8 ft
8 ft
Solution
Ans.
V max = - 400 lb
M max = - 2400 lb # ft
149
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–25. Draw the shear and moment diagrams for the beam.
2k
2k
2k
2k
A
4 ft
4 ft
4 ft
4 ft
4 ft
Solution
Ans.
Vmax = {4 k
Mmax = 24 k # ft
150
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4–26. The beam is subjected to the uniformly distributed
moment m (moment/length). Draw the shear and moment
diagrams for the beam.
m
A
B
L
Solution
Support Reactions. As shown on FBD.
Shear and Moment Function.
+ c ΣFy = 0:
a + ΣMS = 0:
-m - Y = 0
V = -m
M + m(x) - mx = 0
M = 0
Ans.
Ans.
Shear and Moment Diagram.
Ans.
V = -m
M = 0
151
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4–27. The flooring system for a building consists of a girder
that supports laterally running floor beams, which in turn
support the longitudinal simply supported floor slabs. Draw
the shear and moment diagrams for the girder. Assume the
girder is simply supported.
4k
2 k>ft
2 ft
B
A
5 ft
5 ft
5 ft
5 ft
5 ft
Solution
Ans.
Vmax = - 12.1 k
Mmax = 71.2 k # ft
152
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*4–28.
Draw the shear and moment diagrams for the beam.
12 kN>m
30 kN ? m
A
B
5m
5m
Solution
12 kN/m
30 kN ? m
A
B
5m
48 kN
5m
12 kN
Ans.
V max = 48 kN
M max = 96 kN # m
153
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4–29.
Draw the shear and moment diagrams for the beam.
10 kN
10 kN
5 kN>m
A
3m
B
6m
3m
Solution
Ans.
Vmax = {15.0 kN
154
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4–30.
10 kN
Draw the shear and moment diagrams for the beam.
10 kN>m
A
1m
2m
Solution
Ans.
V max = - 30.0 kN;
M max = - 50.0 kN # m
155
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–31. The concrete beam supports the wall, which subjects
the beam to the uniform loading shown. The beam itself has
cross-sectional dimensions of 12 in. by 26 in. and is made from
concrete having a specific weight of g = 150 lb>ft3. Draw the
shear and moment diagrams for the beam and specify the
maximum moment in the beam. Neglect the weight of the steel
reinforcement in the beam.
800 lb>ft
800 lb>ft
10 ft
3 ft
8 ft
Solution
Weight of Beam.
12(26)
= 2.1667 ft2
144
wconc = 150(2.1667) = 325 lb>ft
V max = 10.7 k
M max = 51.1 k # ft
A =
Ans.
Ans.
Ans.
V max = 10.7 k
M max = 51.1 k # ft
156
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*4–32. Draw the shear and moment diagrams for the compound
beam.
5 kN>m
A
B
2m
C
1m
D
1m
Solution
Support Reactions.
From the FBD of segment AB,
a + ΣMA = 0;
By(2) - 10.0(1) = 0
By = 5.00 kN
+ c ΣFy = 0;
Ay - 10.0 + 5.00 = 0
Ay = 5.00 kN
From the FBD of segment BD,
a + ΣMC = 0;
5.00(1) + 10.0(0) - Dy(1) = 0
Dy = 5.00 kN
+ c ΣFy = 0;
Cy - 5.00 - 5.00 - 10.0 = 0
Cy = 20.0 kN
+ ΣF = 0;
S
x
Bx = 0
From the FBD of segment AB,
+ ΣF = 0;
S
A = 0
x
x
Shear and Moment Diagram.
Ans.
V max = {10.0 kN
M max = - 7.50 kN # m
157
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4–33.
Draw the shear and moment diagrams for the beam.
600 lb>ft
B
A
6 ft
12 ft
Solution
Referring to the FBD of the left segment of the beam section through x = 6 ft, Fig. a,
a + ΣM = 0;
1
(600)(6)(2) + M = 0
2
M = - 3600 lb # ft
Referring to the FBD of the right segment of the beam sectioned at an arbitrary distance
of x¿ from the right support, Fig. b, and set V = 0,
1 x′
c (600) d x′ = 0
2 12
+ c ΣFy = 0;
900 -
a + ΣM = 0;
900(6.00) -
x′ = 6.00 ft
1 6.00
6.00
b - M = 0
c
(600) d (6.00) a
2 12
3
M = 3600 lb # ft
Ans.
Vmax = 2700 lb
Mmax = {3600 lb # ft
158
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4–34.
w
Draw the shear and moment diagrams for the beam.
A
C
B
L
––
2
L
Solution
Support Reactions.
From FBD (a),
a + ΣMA = 0;
Cy(L) -
+ c ΣFy = 0;
Ay +
wL 3L
3wL
b = 0 Cy =
a
2
4
8
3wL
wL
wL
b = 0 Ay =
- a
8
2
8
Shear and Moment Funtions.
+ c ΣFy = 0;
a + ΣM = 0;
For
For 0 … x 6
L
[FBD (b)],
2
wL
wL
- V = 0 V =
8
8
M-
Ans.
wL
wL
(x) = 0 M =
(x)
8
8
Ans.
L
6 x … L [FBD (c)],
2
+ c ΣFy = 0;
a + ΣM = 0;
V +
3wL
- w(L - x) = 0
8
w
V =
(5L - 8x)
8
Ans.
3wL
L - x
b - M = 0
(L - x) - w(L - x) a
8
2
w
M =
( - L2 + 5Lx - 4x 2)
8
Ans.
Ans.
For 0 … x 6 L>2,
wL
V =
8
wL
M =
(x)
8
For L>2 6 x … L,
w
V =
(5L - 8x)
8
w
M =
( -L2 + 5Lx - 4x 2)
8
159
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4–35.
Draw the shear and moment diagrams for the beam.
200 lb>ft
C
D
E
F
G
A
B
x
4 ft
4 ft
4 ft
4 ft
Solution
V max = {1200 lb
M max = 6400 lb # ft
Ans.
Ans.
Ans.
V max = {1200 lb
M max = 6400 lb # ft
160
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150 lb>ft
*4–36. Draw the shear and moment diagrams for the
compound beam.
150 lb>ft
A
B
6 ft
C
3 ft
Solution
Support Reactions.
From the FBD of segment AB,
a + ΣMB = 0;
+ c ΣFy = 0;
+ ΣF = 0;
S
x
450(4) - Ay(6) = 0
By - 450 + 300.0 = 0
Ay = 300.0 lb
By = 150.0 lb
Bx = 0
From the FBD of segment BC,
a + ΣMC = 0;
+ c ΣFy = 0;
+ ΣF = 0;
S
225(1) + 150.0(3) - MC = 0
MC = 675.0 lb # ft
Cy - 150.0 - 225 = 0
Cy = 375.0 lb
Cx = 0
x
Shear and Moment Diagram. The maximum positive moment occurs when V = 0.
+ c ΣFy = 0;
a + ΣMNA = 0;
150.0 - 12.5x 2 = 0
x = 3.464 ft
150(3.464)
- 12.5(3.4642) a
3.464
b - M max = 0
3
M max = 346.4 lb # ft
Ans.
V max = - 375 lb
M max = - 675 lb # ft
161
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4–37.
Draw the shear and moment diagrams for the beam.
6 kN>m
6 kN
A
B
3m
1.5 m
1.5 m
Solution
Referring to the FBD of the left segment of the beam sectioned
at an arbitrary distance of x from the left support, Fig. a, and
setting V = 0,
+ c ΣFy = 0;
7.50 -
a + ΣM = 0;
M +
1 6
a xbx = 0
2 3
x = 2.739 m
1 6
2.739
c (2.739) d (2.739) a
b - 7.50(2.739) = 0
2 3
3
M = 13.69 kN # m
= 13.7 kN # m
When x = 3 m,
a + ΣM = 0;
M +
1 6
3
c (3) d (3) a b - 7.50(3) = 0
2 3
3
M = 13.5 kN # m
Ans.
Vmax = {7.5 kN
Mmax = 13.7 kN # m
162
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50 lb> ft
4–38. Draw the shear and bending-moment diagrams for the
beam.
200 lb?ft
A
C
B
20 ft
10 ft
Solution
Support Reaction at A:
a + ΣMB = 0; 1000(10) - 200 - Ay(20) = 0 Ay = 490 lb
Shear and Moment Functions.
+ c ΣFy = 0;
a + ΣM = 0;
For 0 … x 6 20 ft [FBD (a)],
490 - 50x - V = 0
V = 5490 - 50.0x6 lb
Ans.
x
M + 50xa b - 490x = 0
2
M = (490x - 25.0x 2) lb # ft
Ans.
For 20 ft 6 x … 30 ft [FBD (b)],
+ c ΣFy = 0; V = 0
a + ΣM = 0;
- 200 - M = 0 M = - 200 lb # ft
Ans.
Ans.
Ans.
For 0 … x 6 20 ft,
V = 5490 - 50.0x6 lb
M = (490x - 25.0x 2) lb # ft
For 20 ft 6 x … 30 ft,
V = 0
M = -200 lb # ft
163
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–39. Draw the shear and moment diagrams for each of the
three members of the frame. Assume the frame is pin connected
at A, C, and D and there is a fixed joint at B.
50 kN
1.5 m
40 kN
2m
1.5 m
B
C
15 kN>m
4m
6m
Solution
A
D
Ans.
V max = 83.0 kN
M max = -180 kN # m
164
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–40. Draw the shear and moment diagrams for each member of
the frame.
6 kN>m
B
C
1.5 m
3m
6 kN
1.5 m
Solution
a + ΣMA = 0;
+ ΣF - 0;
S
x
6 kN
MA - 4 ft (0.8 k) - 8 ft (0.8 k) - 12 ft (1.2 k) - 4 ft (1.6 k) = 0
MA = 30.4 k # ft
1.5 m
A
Ax - 1.6 k = 0
Ax - 1.6 k
+ c ΣFy = 0;
Ay - 0.8 k - 0.8 k - 1.2 k = 0
+ c ΣFy = 0;
2.8 k - 0.8 k - 0.8 k - 1.2 k + VB = 0
Ay = 2.8 k
+ ΣF = 0;
S
x
a + ΣMB = 0;
VB = 0
1.6 k - NB = 0
NB = 1.6 k
MB + 30.4 k # ft + 4 ft (0.8 k) + 8 ft (0.8 k) - 12 ft (2.8 k) = 0
MB = - 6.40 k # ft
Ans.
V max = 2.8 k
M max = - 30.4 k # ft
165
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4–41. Draw the shear and moment diagrams for each member
of the frame. Assume the frame is pin connected at A and C is
a roller.
4 k>ft
B
15 k
C
4 ft
10 k
4 ft
A
10 ft
Solution
Ans.
Vmax = -36 k
Mmax = 162 k # m
166
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4–42. Draw the shear and moment diagrams for each member
of the frame. The joint at B is fixed connected.
6 kN
A
Solution
Frame:
B
3m
2m
3m
a + ΣMC = 0;
2m
20 kN (2.5 m) - Ay(8 m) = 0
9 kN
Ay = 6.25 kN
+ c ΣFy = 0;
C
1.5 m 1.5 m
- 20 kN + 6.25 kN + Cy = 0
Cy = 13.75 kN
+ ΣF = 0;
S
x
Cx = 0
Ans.
V max = - 13.75 kN
M max = 23.6 kN # m
167
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–43. Draw the shear and moment diagrams for each member
of the frame. Assume A is fixed, the joint at B is a pin, and
support C is a roller.
6 kN>m
B
A
4m
1.5 m
12 kN
1.5 m
C
Solution
V max = 20.0 k
M max = -144 k # ft
Ans.
Ans.
Ans.
V max = 20.0 k
M max = - 144 k # lb
168
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*4–44. Draw the shear and moment diagrams for each
member of the frame. Assume the frame is roller supported at
A and pin supported at C.
1.5 k>ft
B
A
10 ft
6 ft
2k
6 ft
C
Solution
169
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–45. Draw the shear and moment diagrams for each member
of the frame. Assume the joint at A is a pin and support C is a
roller. The joint at B is fixed. The wind load is transferred to the
members at the girts and purlins from the simply supported
wall and roof segments.
300 lb>ft
C
3.5 ft
308
500 lb>ft
3.5 ft
B
7 ft
7 ft
Solution
A
Support Reactions.
a + ΣMA = 0;
- 7(7) - 4.2 cos 30° (7 cos 30°)
- 4.2 sin 30°(14 + 3.5) + Cx(21) = 0
Cx = 5.133 kN = 5.13 kN d
+ ΣF = 0; 7 + 4.2 sin 30° - 5.133 - A = 0
S
x
x
Ax = 3.967 kN = 3.97 kN d
+ c ΣFy = 0; Ay - 4.20 cos 30° = 0
Ay = 3.64 kN c
Ans.
Vmax = 2.22 k
Mmax = 15.5 k # m
170
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8k
4–46. Draw the shear and moment diagrams for each member
of the frame. Assume A is a pin and D is a roller.
8k
4 ft
4 ft
B
A
10 ft
Solution
D
2 k>ft C
Ans.
V max = 16.0 kN
M max = 26.7 kN # m
171
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–47. Draw the shear and moment diagrams for each member
of the frame. Assume joint B is rigid and C is pin connected.
12 kN
3m
2m
B
C
A
D
8 kN
3m
10 kN
Solution
3m
a + ΣMA = 0;
0.8 k(15 ft) - 16 k(4 ft) - 1.25 k(12 ft) + Dy(12 ft) = 0
Dy = 5.58 k
+
S ΣFx = 0;
- 0.8 k + Dx = 0
Dx = 0.8 k
+ c ΣFy = 0;
5.58 k - 1.25 k - 16 k + Ay = 0
Ay = 11.67 k
172
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–47.
(Continued)
Ans.
V max = 11.7 k
M max = 34.0 k # ft
173
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–48. Draw the shear and moment diagrams for each
member of the frame.
2 k>ft
B
4 ft
C
12 kN
4 ft
A
D
6 ft
Solution
12 ft
6 ft
Ans.
V max = 4.10 kN
M max = 18.8 kN # m
174
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4–49. Leg BC on the framework can be designed to extend
either outward as shown, or inward with the support C
positioned below the center 2-k load. Draw the moment
diagrams for the frame in each case, to make a comparison of
the two designs.
2k
2k
2k
A
B
4 ft
4 ft
4 ft
8 ft
C
4 ft
Solution
Ans.
Outward: Mmax = 12.0 k # ft
Inward: Mmax = - 24.0 k # ft
175
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4–50. Draw the shear and moment diagrams for each member
of the frame.
5 kN
10 kN
B
5 kN
2 kN>m
C
4m
D
A
Solution
3m
a + ΣMD = 0; 10(2.5) + 5(3) + 10(5) + 5(7) - Ay(10) = 0
2m
2m
3m
Ay = 12.5 kN
+ ΣF = 0;
S
x
4
- 10a b + Dx = 0
5
Dx = 8 kN
+ c ΣFy = 0;
3
12.5 - 5 - 10 - 5 - 10a b + Dy = 0
5
Dy = 13.5 kN
Ans.
Vmax = -14.5 kN
Mmax = 52.5 kN # m
176
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4–51. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be simply
supported at A and B as shown.
200 lb>ft
100 lb?ft
100 lb?ft
A
B
20 ft
Solution
Ans.
Mmax = 9900 lb # ft
177
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*4–52. Draw the moment diagrams for the beam using the
method of superposition.
80 lb/ft
12 ft
12 ft
600 lb
Solution
178
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800 lb
4–53. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be simply
supported. Assume A is a pin and B is a roller.
50 lb/ft
A
B
750 lb ? ft
10 ft
10 ft
Solution
Ans.
Mmax = 6875 lb # ft
179
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4–54. Solve Prob. 4–53 by considering the beam to be
cantilevered from the support at A.
800 lb
50 lb/ft
A
B
750 lb ? ft
10 ft
10 ft
Solution
Ans.
Mmax = 6875 lb # ft
180
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4–55. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be cantilevered
from the pin support at C.
8 kN/m
A
C
B
18 kN ? m
2m
4m
Solution
Ans.
Mmax = - 34 kN # m
181
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*4–56. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be cantilevered
from the roller at B.
8 kN/m
A
C
B
18 kN ? m
2m
Solution
182
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8 kN/m
4–57. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be cantilevered
from end A.
A
C
B
18 kN ? m
2m
4m
Solution
Ans.
Mmax = - 34 kN # m
183
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4–58. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be cantilevered
from end C.
30 kN
4 kN/m
80 kN ? m
C
A
B
8m
4m
Solution
Ans.
Mmax = -200 kN # m
184
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4–59. Draw the moment diagrams for the beam using the
method of superposition. Consider the beam to be cantilevered
from the support at B.
12 kN
4 kN
A
B
3m
3m
12 kN ? m
C
3m
Solution
Ans.
Mmax = - 24 kN # m
185
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4–1P. The balcony located on the third floor of a motel is
shown in the photo. It is constructed using a 4-in.-thick concrete
(plain stone) slab which rests on the four simply supported
floor beams, two cantilevered side girders AB and HG, and the
front and rear girders. The idealized framing plan with average
dimensions is shown in the adjacent figure. According to local
code, the balcony live load is 45 psf. Draw the shear and
moment diagrams for the front girder BG and a side girder AB.
Assume the front girder is a channel that has a weight of
25 lb>ft and the side girders are wide flange sections that have
a weight of 45 lb>ft. Neglect the weight of the floor beams and
front railing. For this solution treat each of the five slabs as­
two-way slabs.
H
A
6 ft
B
C
4 ft
D
4 ft
E
4 ft
G
F
4 ft
4 ft
Solution
Dead load = (4 in.)(12 lb>ft2 # in.) = 48 psf
Live load = 45 psf
Total load = 93 psf
L2
6
= = 1.5 6 2
L1
4
Two@way slab
Floor beam load:
+ c ΣFy = 0;
1
2R - 372(2) - 2a b(372)(2) = 0
2
R = 744 lb
Front girder:
+ c ΣFy = 0;
1
2R′ - 4(744) - 5a b (25 + 211)(4) = 0
2
R′ = 2668 lb
Maximum moment is at center of girder.
a + ΣMA = 0;
M + 186(0.667) + 744(2) + 744(6) + 372(4) + 372(8)
+ 250(5) - 2668(10) = 0
Mmax = 14 890 lb # ft = 14.9 k # ft
Ans.
Side girder:
Maximum moment at support.
a + ΣMA = 0;
Mmax - 1758(3) - 5336(6) = 0
M = 37 290 lb # ft = 37.3 k # ft
Ans.
Roof load on intermediate joist is
(102 lb>ft3) a
R =
4
ft b(1.5 ft) = 51 lb>ft
12
1
31020 + 1354 = 577.5 lb
2
The loading on the girder:
RC = 577.5 + 190 + 230 = 997.5 lb
RD = 577.5 + 250 + 290 = 1117.5 lb
RE = 577.5 + 310 + 350 = 1237.5 lb
RF = 577.5 + 370 + 410 = 1357.5 lb
Ans.
Front girder: M max = 14.9 k # ft
Side girder: M max = 37.3 k # ft
186
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4–2P. The canopy shown in the photo provides shelter for the
entrance of a building. Consider all members to be simply
supported. The bar joists at C, D, E, F each have a weight of 135 lb
and are 20 ft long. The roof is 4 in. thick and is to be plain
lightweight concrete having a density of 102 lb>ft3. Live load
caused by drifting snow is assumed to be trapezoidal, with 60 psf
at the right (against the wall) and 20 psf at the left (overhang).
Assume the concrete slab is simply supported between the joists.
Draw the shear and moment diagrams for the side girder AB.
Neglect its weight.
A
C
1.5 ft 1.5 ft
D
E
1.5 ft
F
1.5 ft
B
1.5 ft
Solution
Floor loading:
Reinforced concrete stone slab = (12.5 lb>ft2 # in.)(4 in.) = 50 psf
Lift lobby live load = 100 psf
Total loading = 150 psf
Concrete block wall:
4
(105)(10) = 350 lb>ft
12
From slab ADIB:
w = (4)(105)′ = 600 lb>ft
Beam EF:
F1 =
1
1
(350)(8) + (600)(8) = 2600 lb
2
2
Beam HG:
F2 =
1
1
(350)(8) + 450(2) + 2a b(450)(3) = 2525 lb
2
2
Equilibrium for entire beam AB:
a + ΣMA = 0; By (20) - 7200(4) - 2600(8) - 5700(11)
- 2525(14) - 4950(17) = 0
By = 11 590 lb = 11.59 k
+ c ΣFY = 0; Ay + 11 590 - 7200 - 2600 - 5700 - 2525 - 4950 = 0
Ay = 11 385 lb = 11.385 k
For beam segment:
a + ΣM = 0; M + 7.20(4) - 11.385(8) = 0
M = 62.28 k # ft
Ans.
V max = - 2475 lb
M max = 5389 lb # ft
187
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5–1. Determine the tension in each segment of the cable and
the distance yD.
D
yD
6m
A
C
3 kN
3m
B
20 kN
Solution
2m
3m
4m
Joint C. Referring to the FBD in Fig. a,
+
S
ΣFx = 0;
+ c ΣFy = 0;
Solving,
2
1
b - TBC a
b = 0
113
12
3
1
TCD a
b - TBC a
b - 3 = 0
113
12
TCD a
TCD = 3113 kN = 10.8 kN
Ans.
TBC = 612 kN = 8.49 kN Ans.
Joint B. Referring to the FBD in Fig. b,
+
S
ΣFx = 0;
+ c ΣFy = 0;
Solving,
1
b - TAB cos u = 0
12
1
TAB sin u + (612) a
b - 20 = 0
12
612a
u = 66.80°
TAB = 15.23 kN = 15.2 kN
Ans.
From the geometry of the cable,
tan 66.80° =
9 - yD
2
yD = 4.333 m = 4.33 m
Ans.
Ans.
TCD = 10.8 kN
TBC = 8.49 kN
TAB = 15.2 kN
yD = 4.33 m
188
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5–2. The cable supports the loading shown. Determine the
magnitude of the vertical force P so that yC = 10 ft.
10 ft
4 ft
8 ft
A
D
5 ft
B
yC
P
C
4k
Solution
Joint C.
Referring to the FBD shown in Fig. a,
+ ΣF = 0;
S
x
TCD a
+ c ΣFy = 0;
TCD a
Solving,
4
2
b - TBC a
b = 0
141
15
5
1
b + TBC a
b - 4 = 0
141
15
4
141 kN
7
8
TBC = 15 kN
7
TCD =
Joint B.
Referring to the FBD shown in Fig. b,
+ ΣF = 0;
S
x
8
2
4
15a
b - TAB a
b = 0
7
15
141
TAB =
+ c ΣFy = 0;
a
4
141 k
7
4
5
8
1
141b a
b - a 15b a
b - P = 0
7
7
141
15
P =
12
k = 1.71 k
7
Ans.
Ans.
P = 1.71 k
189
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5–3. Determine the forces P1 and P2 needed to hold the cable
in the position shown, i.e., so segment BC remains horizontal.
E
A
D
2m
3m
B
C
P2
P1
4m
8 kN
5m
4m
1.5 m
Solution
Write the moment equation of equilibrium about E by referring to the FBD in Fig. a,
a + ΣME = 0;
3
4
8(1.5) + TCD a b(1.5) - TCD a b(2) = 0
5
5
TCD = 17.14 kN
Consider the equilibrium of joint C by referring to the FBD
shown in Fig. b,
+ ΣF = 0;
S
x
+ c ΣFy = 0;
4
17.14a b - TBC = 0
5
3
17.14a b - P2 = 0
5
TBC = 13.71 kN
P2 = 10.29 kN = 10.3 kN
Ans.
Next, joint B by referring to the FBD in Fig. c,
+ ΣF = 0;
S
x
13.71 - TAB a
+ c ΣFy = 0;
21.95a
4
b = 0
141
5
b - P1 = 0
141
TAB = 21.95 kN
P1 = 17.14 kN = 17.1 kN
Ans.
Ans.
P2 = 10.3 kN;
P1 = 17.1 kN
190
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*5–4. The cable supports the loading shown. Determine the
distance y and the tension in cable BC. Set P = 3 kN.
2m
4m
1m
D
A
y
B
1m
P
C
8 kN
Solution
Method of Sections. Referring to the FBD of the left segment of the
cable system sectioned through cable BC, Fig. a,
a + ΣMA = 0;
TBC a
4
1
b(y) - TBC a
b(2) - 3(2) = 0
117
117
(1)
Referring to the FBD of the right segment, Fig. b,
a + ΣMD = 0;
8(1) - TBC a
Solving Eqs. (1) and (2),
4
1
b(y + 1) - TBC a
b(1) = 0
117
117
2
117 kN = 1.18 kN
7
y = 5.75 m
(2)
Ans.
TBC =
Ans.
Ans.
TBC = 1.18 kN
y = 5.75 m
191
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5–5. The cable supports the loading shown. Determine the
magnitude of the vertical force P so that y = 4 m.
2m
4m
1m
D
A
y
B
1m
P
C
8 kN
Solution
Method of Sections. Referring to the FBD of the right segment of the cable
system sectioned through cable BC, Fig. a,
a + ΣMD = 0;
8(1) - (TBC 2 a
TBC =
Method of Joints.
+ ΣF = 0;
S
x
+ c ΣFy = 0;
1
4
b(1) - TBC a
b(5) = 0
117
117
8
117 kN
21
Referring to the FBD of joint B, Fig. b,
8
4
1
117b a
b - TAB a
b = 0
21
117
115
32
TAB =
15 kN
21
32
2
8
1
a
15b a
b - a
117b a
b - P = 0
21
21
15
117
8
Ans.
P =
kN = 2.67 kN
3
a
Ans.
P = 2.67 kN
192
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5–6. The cable segments support the loading shown.
Determine the magnitude of the vertical force P so that
yC = 6 ft.
2 ft
3 ft
5 ft
8 ft
A
D
B
yC
C
6k
P
Solution
a + ΣMA = 0;
TDC cos 39.81°(10) + TDC sin 39.81°(6) - 6(2) - P(10) = 0
11.523TDC - 10P = 12
(1)
Joint C:
+ c ΣFy′ = 0;
TDC cos 19.25° - P cos 20.56° = 0(2)
Solving Eqs. (1) and (2) yields:
P = 8.40 k
Ans.
TDC = 8.33 k
tan - 1
5
= 39.81°
6
f = tan - 1
8
= 69.44°
3
Ans.
P = 8.40 k
193
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5–7. The cable supports the three loads shown. Determine the
magnitude of P1 if P2 = 2 k, yB = 6 ft, and yC = 10 ft. Also find
the sag yD.
3 ft
E
yB
A
yD
B
yC
C
P2
10 ft
P1
15 ft
12 ft
D
P2
8 ft
Solution
Method of Sections. Referring to the FBD of the left segment of
the cable system sectioned through cable BC, Fig. a,
a + ΣMA = 0;
TBC a
15
4
b(6) - TBC a
b(10) - 2(10) = 0
1241
1241
TBC =
2
1241 k
5
Referring to the FBD of the right segment of the cable system sectioned through cable CD, Fig. b,
a + ΣME = 0;
2(8) + TCD sin u(8) - TCD cos u(yD + 3) = 0
TCD cos u(yD + 3) - 8TCD sin u = 16
Method of Joints.
+ ΣF = 0;
S
x
Referring to the FBD of joint C, shown in Fig. c,
TCD cos u - a
2
15
1241b a
b = 0
5
1241
TCD cos u = 6
+ c ΣFy = 0;
(1)
TCD sin u + a
2
4
1241b a
b - P1 = 0
5
1241
P1 = TCD sin u +
8
5
(2)
(3)
Dividing Eq. (1) by (2),
cos u(yD + 3) - 8 sin u
cos u
8 tan u - yD =
16
6
1
3
10 - yD
. Then
12
However, tan u =
8a
Subsequently,
=
10 - yD
1
b - yD =
12
3
Ans.
yD = 3.80 ft
tan u =
10 - 3.80
; u = 27.32°
12
Then from Eq. (2),
TCD cos 27.32° = 6 TCD = 6.754 k
Substitute the results of TCD and u into Eq. (3),
P1 = 6.754 sin 27.32° +
8
= 4.70 k
5
Ans.
194
Ans.
yD = 3.80 ft
P1 = 4.70 k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–8. The cable supports the uniform load of w0 = 600 lb>ft.
Determine the tension in the cable at each support A and B.
B
A
15 ft
10 ft
w0
25 ft
Solution
y =
wo 2
x
2FH
15 =
600 2
x
2FH
10 =
600
(25 - x)2
2FH
600 3
600
x =
(25 - x)2
2(15)
2(10)
x 2 = 1.5(625 - 50x + x 2 2
0.5x 2 - 75x + 937.50 = 0
Choose root 6 25 ft.
x = 13.76 ft
FH =
wo 2
600
x =
(13.76)2 = 3788 lb
2y
2(15)
At B:
y =
wo 2
600
x =
x2
2FH
2(3788)
dy
= tan uB = 0.15838x `
= 2.180
dx
x = 13.76
uB = 65.36°
TB =
FH
3788
=
= 9085 lb = 9.09 kip
cos uB
cos 65.36°
Ans.
At A:
y =
wo 2
600
x =
x2
2FH
2(3788)
dy
= tan uA = 0.15838x `
= 1.780
dx
x = (25 - 13.76)
uA = 60.67°
TA =
FH
3788
=
= 7734 lb = 7.73 kip
cos uA
cos 60.67°
Ans.
195
Ans.
TB = 9.09 kip
TA = 7.73 kip
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–9. Determine the maximum and minimum tension in the
cable.
y
60 ft
60 ft
A
B
12 ft
x
2 k>ft
Solution
The minimum tension in the cable occurs when u = 0°. Thus,
Tmin = FH . With wo = 2 k>ft, L = 60 ft and h = 12 ft,
Tmin = FH =
(2 k>ft)(60 ft)2
w oL 2
=
= 300 k
2h
2(12 ft)
Ans.
And
Tmax = woL
A
1 + a
L 2
b
2h
= (2 k>ft)(60 ft)
A
1 + c
= 323.11 k = 323 k
2
60 ft
d
2(12 ft)
Ans.
Ans.
Tmin = 300 k
Tmax = 323 k
196
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5–10. The cable is subjected to a uniform loading of
w = 500 lb>ft. Determine the maximum and minimum tension
in the cable.
100 ft
20 ft
w
Solution
The minimum tension in the cable occurs at its lowest point.
Thus, Tmin = FH . With wo = 500 lb>ft = 0.5 k>ft, L = 50 ft
and h = 20 ft,
Tmin = FH =
(0.5 k>ft)(50 ft)2
w oL 2
=
= 31.25 k
2h
2(20 ft)
Ans.
And
Tmax = woL
B
1 + a
L 2
b
2h
= (0.5 k>ft)(50 ft)
B
= 40.02 k = 40.0 k
1 + c
2
50 ft
d
2(20 ft)
Ans.
Ans.
Tmin = 31.25 k
Tmax = 40.0 k
197
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5–11. The cable is subject to the uniform loading. Determine
the equation y = f(x) which defines the cable shape AB and
the maximum tension in the cable.
y
50 ft
A
50 ft
B
20 ft
x
150 lb>ft
Solution
From Eq. 5–9,
y =
h 2
20 2
x =
x
2
L
(50)2
y = 0.008x 2
Ans.
From Eq. 5–11,
Tmax = woL
A
1 + a
Tmax = 150(50)
A
L 2
b
2h
1 + a
50 2
b = 12.0 k
2(20)
Ans.
Ans.
y = 0.008x 2
Tmax = 12.0 k
198
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–12. The cable will break when the maximum tension
reaches 50 k. Determine the maximum uniform distributed
load w that can be supported by the cable.
120 ft
15 ft
w
Solution
Eqs. 5–7:
y =
w 2
x
2FH
At x = 7.5 m, y = 6m,
FH = 4.688 w
T =
FH
p
for 0 … u …
cos u
2
T max will occur when u is maximum.
dy
w
`
= tan (umax 2 =
x`
dx max
FH x = 7.5 m
u max = tan - 1 a
T max =
7.5
b = 57.99°
4.688
FH
4.688 w
=
= 12
cos (u max 2
cos 57.99°
w = 1.36 kN>m
Ans.
Ans.
w = 1.36 kN>m
199
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5–13. The trusses are pin connected and suspended from the
parabolic cable. Determine the maximum force in the cable
when the structure is subjected to the loading shown. The
support at A is a pin and C is a rocker.
D
E
14 ft
6 ft
K
J
I
16 ft
A
G H B
5k
4 @ 12 ft 5 48 ft
C
F
4k
4 @ 12 ft 5 48 ft
Solution
Entire Structure:
a + ΣMC = 0;
4(36) + 5(72) + FH(36) - FH(36) - (Ay + Dy 2(96) = 0
Section ABD:
(Ay + Dy 2 = 5.25
Using Eq. (1):
FH(14) - (Ay + Dy 2(48) + 5(24) = 0
a + ΣMB = 0;
(1)
FH = 9.42857 k
From Eq. 5–8:
wo =
2FHh
L
2
=
2(9.42857)(14)
482
= 0.11458 k>ft
From Eq. 5–11:
Tmax = woL
A
1 + a
2
L 2
48
d = 10.9 k
b = 0.11458(48) 1 + c
2h
A
2(14)
Ans.
Ans.
Tmax = 10.9 k
200
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5–14. Determine the maximum and minimum tension in the
parabolic cable and the force in each of the hangers. The girder
is subjected to the uniform load and is pin connected at B.
D
Solution
10 ft
E
15 ft
3 k>ft
2.5 ft
A
Referring to the geometry shown in Fig. a, the equation of the
profile for the cable is
h
y = 2 x2
L
C
B
40 ft
20 ft
Here, h = 10 ft. Then
y =
10 2
x
L2
For point E, x = 60 - L and y = 2.5 ft
2.5 =
10
(60 - L)2
L2
7.5L2 - 1200L + 36,000 = 0
Solving and choosing L 6 60 ft,
L = 40 ft
Referring to the FBD of member BC, Fig. c,
+ ΣF = 0;
S
B = 0
x
x
Then, from the FBD of member AB, Fig. b,
+ ΣF = 0;
S
A = 0
x
x
Referring to the FBDs of the left and right segments
of the system sectioned through the lowest point of the
cable, Fig. d and e respectively,
a + ΣMD = 0;
By(40) + FH(10) - 3(40)(20) = 0
40By + 10FH = 2400
a + ΣME = 0;
(1)
3(20)(10) + By(20) - FH(2.5) = 0
2.5FH - 20By = 600
(2)
Solving Eqs. (1) and (2),
FH = 240 k By = 0
FH =
w oL 2
;
2h
240 k =
wo(40 ft)2
2(10 ft)
wo = 3.00 k>ft
The minimum tension in the cable occurs
at its lowest point, where
T min = FH = 240 k
Ans.
The maximum tension in the cable occurs at point D,
where
Tmax = woL
B
1 + a
L 2
b
2h
= (3.00 k>ft)(40 ft)
B
= 268.33 k = 268 k
1 + c
2
40 ft
d
2(10 ft)
Ans.
Ans.
T min = 240 k
T max = 268 k
Th = 30.0 k
Each hanger supports 10 ft of wo. Thus,
T = (3.00 k>ft)(10 ft) = 30.0 k
Ans.
201
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5–15. Draw the shear and moment diagrams for the pin
connected girders AB and BC. The cable has a parabolic shape.
D
10 ft
E
15 ft
3 k>ft
2.5 ft
A
C
B
40 ft
20 ft
Solution
Here, each hanger supports 10 ft of wo. Thus,
T = (3.00 k>ft)(10 ft) = 30.0 k
Again, referring to the FBDs of members AB, Fig. b,
a + ΣMB = 0;
3(40)(20) - 30.0(10) - 30.0(20) - 30.0(30) - Ay(40) = 0
Ay = 15.0 k
and member BC, Fig. c,
a + ΣMB = 0;
NC(20) + 30.0(10) - 3(20)(10) = 0
NC = 15.0 k
Ans.
Vmax = {15.0 k
Mmax = 37.5 k # ft
202
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*5–16. The cable will fail when the maximum tension reaches
Tmax = 300 k. Determine the maximum uniform distributed
load w that can be supported by the cable.
300 ft
50 ft
w
Solution
Here, T max = 300 k, L =
300 ft
= 150 ft, h = 50 ft and wo = w.
2
Applying Eq. 5–11,
Tmax = woL
A
1 + a
300 k = w(150 ft)
A
L 2
b
2h
1 + c
2
150
d
2(50 ft)
w = 1.109 k>ft = 1.11 k>ft
Ans.
Ans.
w = 1.11 k>ft
203
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5–17. The cable is subjected to a uniform loading of
w = 1.5 k>ft. Determine the maximum and minimum tension
in the cable.
300 ft
50 ft
w
Solution
The minimum tension in the cable occurs at its lowest point.
300 ft
= 150 ft,
Thus, T min = FH . With wo = 1.5 k>ft, L =
2
and h = 50 ft,
Tmin = FH =
(1.5 k>ft)(150 ft)2
w oL 2
=
= 337.5 k
2h
2(50 ft)
Ans.
And
Tmax = woL
B
1 + a
L 2
b
2h
= (1.5 k>ft)(150 ft)
B
= 405.62 k = 406 k
1 + c
150 ft 2
d
2(50 ft)
Ans.
Ans.
Tmin = 337.5 k
Tmax = 406 k
204
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–18. The beams AB and BC are supported by the cable that
has a parabolic shape. Determine the tension in the cable at
points D and E.
E
D
2m
3m
A
B
80 kN
C
40 kN
2m 2m 2m 2m 2m 2m 2m 2m
Solution
Referring to the FBD of member BC, Fig. b,
+ ΣF = 0;
S
Bx = 0
x
Then from the FBD of member AB, Fig. a,
+ ΣF = 0;
S
Ax = 0
x
Referring to the FBDs of left and right segments of the system sectioned through the lowest point of the cable, Figs. c and
d, respectively,
a + ΣMD = 0;
FH(2) + By(8) - 80(4) = 0
2FH + 8By = 320(1)
a + ΣME = 0;
By(8) - FH(2) + 40(4) = 0
2FH - 8By = 160(2)
Solving Eqs. 1 and 2,
FH = 120 kN
By = 10.0 kN
Then
FH =
w oL 2
;
2h
120 kN =
wo(8 m)2
2(2 m)2
wo = 7.50 kN>m
At points D and E, the cable is subjected to maximum tension.
Applying Eq. 5–11,
TD = TE = T max = woL
B
1 + a
L 2
b
2h
= (7.50 kN>m)(8 m)
B
= 134.16 kN = 134 kN
1 + c
2
8m
d
2(2 m)
Ans.
Ans.
TD = TE = 134 kN
205
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5–19. The beams AB and BC are supported by the cable that
has a parabolic shape. Draw the shear and moment diagrams
for members AB and BC.
E
D
2m
3m
A
B
80 kN
C
40 kN
2m 2m 2m 2m 2m 2m 2m 2m
Soultion
Here, each hanger supports 2 m of wo. Thus,
T = (7.50 kN>m)(2 m) = 15.0 kN
Again, referring to the FBDs of member AB, Fig. a,
a + ΣMB = 0;
80(4) - 15.0(2) - 15.0(4) - 15.0(6) - Ay(8) = 0
Ay = 17.5 kN
And member BC, Fig. b,
a + ΣMB = 0;
15.0(6) + 15.0(4) + 15.0(2) - 40(4) - NC(8) = 0
NC = 2.50 kN
Ans.
Vmax = {32.5 kN
Mmax = 100 kN # m
206
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*5–20. The cable AB is subjected to a uniform loading of
300 lb>ft. If the weight of the cable is neglected and the slope
angles at points A and B are 30° and 45°, respectively, determine
the curve that defines the cable shape and the maximum
tension developed in the cable.
y
B
458
A
308
x
300 lb>ft
20 ft
Solution
Applying Eq. 5–2,
d(T sin u)
d(T sin u)
= wnS
= 300
dx
dx
Integrate this equation.
T sin u = 300x + C1
Divide this result by Eq. 5–4 and then substitute with Eq. 5–3.
dy
1
=
(300x + C1 2
dx
FH
dy
= tan 30° at x = 0,
Applying the boundary condition
dx
1
tan 30° =
c 300 (0) + C1 d
FH
C1 = FH tan 30°
Then
dy
300
=
x + tan 30°
dx
FH
(1)
Integrating Eq. (1),
150 2
x + (tan 30°)x + C2
y =
FH
Applying boundary condition y = 0 at x = 0,
150 2
0 =
(0 2 + (0) tan 30° + C2 1 C2 = 0
FH
Thus,
y =
150 2
x + (tan 30°)x
FH
(2)
dy
= tan 45° = 1 at x = 20 ft. Then Eq. (1) gives
dx
300
(20) + tan 30°
I =
FH
FH = 14196.15 lb
Also,
Substitute this result into Eq. 2.
y = 0.0106x 2 + 0.577x
Ans.
Here, u max = 45°, then
FH
14196.15 lb
T max =
=
= 20 076.39 lb = 20.1 k
cos u max
cos 45°
Ans.
207
Ans.
y = 0.0106x 2 + 0.577x
T max = 20.1 k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–21. The power transmission cable weighs 10 lb>ft. If the
resultant horizontal force on tower BD is required to be zero,
determine the sag h of cable BC.
300 ft
A
200 ft
B
10 ft
h
C
D
Solution
The origin of the x, y coordinate system is set at the lowest point of the
cables. Here, w0 = 10 lb>ft. Using Eq. 4 of Example 7–13,
wo
FH
c cosh a
xb - 1 d
wo
FH
FH
10
y =
c cosh a
xb - 1 d ft
10
FH
y =
Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft,
10 =
(FH)AB
10
c cosh a
10(150)
(FH)AB
Solving by trial and error yields
b - 1d
(FH 2 AB = 11266.63 lb
Since the resultant horizontal force at B is required to be zero,
(FH 2 BC = (FH 2 AB = 11266.62 lb. Applying the boundary condition of
cable BC, y = h at x = - 100 ft, to Eq. (1), we obtain
10( - 100)
11266.62
e cosh c
d - 1f
10
11266.62
= 4.44 ft
h =
Ans.
Ans.
h = 4.44 ft
208
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5–22. The power transmission cable weighs 10 lb>ft. If
h = 10 ft, determine the resultant horizontal and vertical
forces the cables exert on tower BD.
300 ft
A
200 ft
B
10 ft
h
C
D
Solution
The origin of the x, y coordinate system is set at the lowest point of the cables.
Here, w0 = 10 lb>ft. Using Eq. 4 of Example 7–13,
y =
y =
Fh
w0
c cosh a
xb - 1 d
w0
Fh
Fh
10
c cosh a xb - 1 d ft
10
Fh
Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft,
10 =
(Fh 2 AB
10
c cosh a
10(150)
(Fh)AB
Solving by trial and error yields
b - 1d
(Fh 2 AB = 11266.63 lb
Applying the boundary condition of cable BC, y = 10 ft at x = - 100 ft, to
Eq. (2), we have
10 =
(Fh 2 BC
10
c cosh a
10(100)
(Fh)BC
Solving by trial and error yields
b - 1d
(Fh 2 BC = 5016.58 lb
Thus, the resultant horizontal force at B is
(Fh 2 R = (Fh 2 AB - (Fh 2 BC = 11266.63 - 5016.58 = 6250 lb = 6.25 kip
Using Eq. (1), tan (uB 2 AB = sinhc
tan (uB 2 BC = sinhc
10( - 100)
5016.58
and BC acting on point B are
10(150)
11266.63
Ans.
d = 0.13353 and
d = 0.20066. Thus, the vertical force of cables AB
(Fy 2 AB = (Fh 2 AB tan (uB 2 AB = 11266.63(0.13353) = 1504.44 lb
(Fy 2 BC = (Fh 2 BC tan (uB 2 BC = 5016.58(0.20066) = 1006.64 lb
The resultant vertical force at B is therefore
(Fy 2 R = (Fy 2 AB + (Fy 2 BC = 1504.44 + 1006.64
Ans.
= 2511.07 lb = 2.51 kip
209
Ans.
(Fh 2 R = 6.25 kip
(Fy 2 R = 2.51 kip
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–23. The cable has a mass of 0.5 kg>m and is 25 m long.
Determine the vertical and horizontal components of force it
exerts on the top of the tower.
A
B
308
15 m
Solution
x =
L
ds
1
51 + F12 (wods)2 6 2
h
Performing the integration yields:
Fh
1
x =
e sinh - 1 c
(4.905s + C1 2 d + C2 f (1)
4.905
Fh
From Eq. 7–13,
dy
1
=
w ds
dx
Fh L o
dy
1
=
(4.905s + C1 2
dx
Fh
At s = 0,
dy
= tan 30°. Hence, C1 = Fh tan 30°
dx
dy
4.905s
=
+ tan 30°
dx
Fh
(2)
Applying boundary conditions at x = 0, s = 0 to Eq. (1) and using the result,
C1 = Fh tan 30°, yields C2 = - sinh - 1(tan 30°). Hence,
x =
Fh
1
e sinh - 1 c
(4.905s + Fh tan 30°) d - sinh - 1(tan 30°) f 4.905
Fh
(3)
At x = 15 m, s = 25 m. From Eq. (3),
15 =
Fh
1
e sinh - 1 c
(4.905(25) + Fh tan 30°) d - sinh - 1(tan 30°) f
4.905
Fh
By trial and error, Fh = 73.94 N.
At point A, s = 25 m. From Eq. (2),
tan uA =
4.905(25)
dy
`
=
+ tan 30° uA = 65.90°
dx s = 25 m
73.94
(Fv 2 A = Fh tan uA = 73.94 tan 65.90° = 16.5 N
Ans.
(Fh 2 A = Fh = 73.9 N
Ans.
210
Ans.
(Fy 2 A = 165 N
(Fh 2 A = 73.9 N
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–24. The cable of length ℒ = 50 ft is suspended between
two points A and B a distance of 15 ft apart. If the minimum
tension in the cable is 200 lb, determine the total weight of the
cable and the maximum tension developed in the cable.
15 ft
A
B
Solution
T min = FH = 200 lb
From Example 7–13:
w0x
FH
s =
sinh a
b
w0
FH
w0 15
50
200
=
sinh a
a bb
2
w0
200 2
Solving,
w0 = 79.9 lb>ft
Total weight = w0 l = 79.9(50) = 4.00 kip
Ans.
dy
w0s
`
= tan u max =
dx max
FH
Then,
u max = tan - 1 c
T max =
79.9(25)
200
d = 84.3°
FH
200
=
= 2.01 kip
cos u max
cos 84.3°
Ans.
Ans.
Total weight = 4.00 kip
T max = 2.01 kip
211
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5–25. Show that the deflection curve of the cable discussed in
Example 5.4 reduces to Eq. 5–9 when the hyperbolic cosine
function is expanded in terms of a series and only the first two
terms are retained. (The answer indicates that the catenary
may be replaced by a parabola in the analysis of problems in
which the sag is small. In this case, the cable weight is assumed
to be uniformly distributed along the horizontal.)
L
A
B
h
Solution
cosh x = 1 +
x2
+ g
21
Substituting into
y =
=
=
w0
FH
c cosh a
xb - 1 d
w0
FH
w20x 2
FH
c1 +
+ g - 1d
w0
2F 3H
w0x 2
2FH
Using Eq. (3) in Example 7–12,
We get
FH =
w0L2
8h
y =
4h 2
x
L2
Ans.
y =
212
4h 2
x
L2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–26. The cable stretches between two points A and B which
are L = 150 ft apart and at the same elevation. The line sags
h = 5 ft and the cable has a weight of 0.3 lb>ft. Determine the
length of the cable and the maximum tension in the cable.
L
A
h
B
Solution
w = 0.3 lb>ft
From Example 7–13,
s =
y =
FH
w
sinh a
xb
w
FH
FH
w
c cosh a
xb - 1 d
w
FH
At x = 75 ft, y = 5 ft, w = 0.3 lb>ft,
s =
FH
75w
c cosh a
b - 1d
w
FH
FH = 169.0 lb
dy
w
`
= tan umax = sinh a
xb `
dx max
FH
x = 75 ft
u max = tan - 1 c sinh a
T max =
s =
75(0.3)
169
b d = 7.606°
FH
169
=
= 170 lb
cos umax
cos 7.606°
169.0
0.3
sinh c
(75) d = 75.22
0.3
169.0
L = 2s = 150 ft
Ans.
Ans.
Ans.
T max = 170 lb
L = 150 ft
213
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–27. The cable has a weight of 2 lb>ft. If it can span
L = 100 ft and has a sag of h = 12 ft, determine the length of
the cable. The ends of the cable are supported from the same
elevation.
L
A
h
B
Solution
From Eq. (5) of Example 7–13:
h =
12 =
w0L
FH
c cosh a
b - 1d
w0
2FH
2(100)
FH
c cosh a
b - 1d
2
2FH
24 = FH c cosh a
FH = 212.2 lb
100
b - 1d
FH
From Eq. (3) of Example 7–13:
s =
w0
FH
sinh a
xb
w0
FH
2(50)
l
212.2
b
=
sinh a
2
2
212.2
l = 104 ft
Ans.
Ans.
L = 104 ft
214
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*5–28. The cable has a weight of 3 lb>ft and is supported at
points A and B that are 500 ft apart and at the same elevation.
If it has a length of 600 ft, determine the sag h.
L
A
h
B
Solution
w0 = 3 lb>ft
From Example 7–15,
s =
w0
FH
sinh a
xb
w0
FH
At x = 250 ft, s = 300 ft,
300 =
3(250)
FH
sinh a
b
3
FH
FH = 704.3 lb
y =
h =
w0
FH
c cosh
x - 1d
w0
FH
3(250)
704.3
b - 1d
c cosh a
3
704.3
h = 146 ft
Ans.
Ans.
h = 146 ft
215
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–29. The cable has a weight of 5 lb>ft. If it can span
L = 300 ft and has a sag of h = 15 ft, determine the length of
the cable. The ends A and B of the cable are supported at the
same elevation.
L
A
h
B
Solution
w0 = 5 lb>ft
From Example 7–15,
y =
At x = 150 ft, y = 15 ft,
w0
FH
c cosh a
xb - 1 d
w0
FH
15w0
150w0
= cosh a
b - 1
FH
FH
FH = 3762 lb
s =
w0
FH
sinh a
xb
w0
FH
s = 151.0 ft
L = 2s = 302 ft
Ans.
Ans.
L = 302 ft
216
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–30. The 10 kg>m cable is suspended between the supports
A and B. If the cable can sustain a maximum tension of 1.5 kN
and the maximum sag is 3 m, determine the maximum distance
L between the supports.
L
A
3m
B
Solution
The origin of the x, y coordinate system is set at the lowest point
of the cable. Here, w0 = 10(9.81) N>m = 98.1 N>m. Using Eq. (4)
of Example 7–13,
y =
y =
w0
FH
c cosh a
xb - 1 d
w0
FH
FH
98.1x
c cosh a
b - 1d
98.1
FH
Applying the boundary equation y = 3 m at x =
3 =
L
, we have
2
FH
49.05L
c cosh a
b - 1d
98.1
FH
The maximum tension occurs at either points A or B where the
cable makes the greatest angle with the horizontal. From Eq. (1),
tan u max = sinh a
49.05L
b (1)
FH
By referring to the geometry shown in Fig. b, we have
cos umax =
Thus,
T max =
1
49.05L
1 + sinh2 a
b
B
FH
=
1
(2)
49.05L
cosh a
b
FH
FH
cos u max
1500 = FH cosh a
49.05L
b
FH
(3)
Solving Eqs. (2) and (3) yields
Ans.
L = 16.8 m
FH = 1205.7 N
Ans.
L = 16.8 m
217
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5–31. Determine the horizontal and vertical components of
reaction at A, B, and C of the three-hinged arch. Assume A, B,
and C are pin connected.
5k
B
A
5
8k
3
4
10 ft
20 ft
15 ft
C
6 ft 6 ft
24 ft
8 ft
Solution
Member AB:
a + ΣMA = 0;
- By(12) + Bx(10) - 5(6) = 0
- 6By + 5Bx - 15 = 0(1)
Member BC:
a + ΣMC = 0;
- By(32) - Bx(25) +
4
3
(8)(20) + (8)(8) = 0
5
5
- 32By - 25Bx + 166.4 = 0
Solving Eqs. (1) and (2) yields:
By = 1.474 k = 1.47 k
Bx = 4.769 k = 4.77 k
Member AB:
+ ΣF = 0;
S
x
+ c ΣFy = 0;
Member BC:
+ ΣF = 0;
S
x
+ c ΣFy = 0;
Ax - 4.769 = 0
Ax = 4.77 k S
(2)
Ans.
Ans.
Ans.
Ay - 5 - 1.474 = 0
Ay = 6.47 k c
Ans.
4
(8) = 0
5
Cx = 1.63 k S
Ans.
4.769 + Cx -
1.474 + Cy Cy = 3.33 k c
3
(8) = 0
9
Ans.
Ans.
Bx = 4.77 k
By = 1.47 k
Ax = 4.77 k S
Ay = 6.47 k c
Cx = 1.63 k S
Cy = 3.33 k c
218
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*5–32. Determine the magnitudes of the resultant forces at
the pins A, B, and C of the three-hinged arched roof truss.
6k
4k
3k
2k
B
20 ft
A
10 ft
C
6 ft
12 ft
8 ft
10 ft
10 ft
Solution
Equations of Equilibrium. Write the moment equation of equilibrium about points A and C by referring to the FBDs of truss AB,
Fig. a, and truss BC, Fig. b, respectively.
a + ΣMA = 0;
a + ΣMC = 0;
Bx(20) - By(28) - 4(16) - 2(10) = 0
(1)
4(10) + 6(20) - Bx(20) - By(28) = 0
(2)
Solving Eq. (1) and (2),
Bx = 6.10 k
By = 1.357 k
Using these results to write the force equations of equilibrium
by referring to the FBD of truss AB, Fig. a,
+ ΣF = 0;
S
x
Ax - 6.10 = 0
Ax = 6.10 k
+ c ΣFy = 0;
Ay - 2 - 4 - 1.357 = 0
Ay = 7.357 k
And referring to the FBD of truss BC, Fig. b,
+ ΣF = 0;
S
6.10 - Cx = 0
Cx = 6.10 k
x
+ c ΣFy = 0;
Cy + 1.357 - 6 - 4 = 0
Cy = 8.643 k
Thus, the magnitudes of the forces on pins A, B and C are
FA = 2A2x + A2y = 26.102 + 7.3572 = 9.557 k = 9.56 k
Ans.
2
2
Ans.
2
2
Ans.
FB =
FC =
2B2x + B2y =
2C 2x + C 2y =
26.10 + 1.357 = 6.249 k = 6.25 k
26.10 + 8.643 = 10.58 k = 10.6 k
Ans.
FA = 9.56 k
FB = 6.25 k
FC = 10.6 k
219
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5–33. The three-hinged spandrel arch is subjected to the
loading shown. Determine the internal moment in the arch at
point D.
8 kN 8 kN
6 kN 6 kN
3 kN
3 kN
2m 2m 2m
4 kN
4 kN
2m 2m 2m
B
D
A
3m
3m
5m
5m
C
8m
Solution
Member AB:
a + ΣMA = 0;
Bx(5) + By(8) - 8(2) - 8(4) - 4(6) = 0
Bx + 1.6By = 14.4
(1)
Member CB:
a + ΣMC = 0;
B(y)(8) - Bx(5) + 6(2) + 6(4) + 3(6) = 0
- Bx + 1.6By = - 10.8
(2)
Solving Eqs. (1) and (2) yields:
By = 1.125 kN
Bx = 12.6 kN
Segment BD:
a + ΣMD = 0;
- MD + 12.6(2) + 1.125(5) - 8(1) - 4(3) = 0
MD = 10.825 kN # m = 10.8 kN # m
Ans.
Ans.
MD = 10.8 kN # m
220
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5–34. The tied three-hinged truss arch is subjected to the
loading shown. Determine the components of reaction at A
and C, and the tension in the tie rod.
80 kN
60 kN
B
8m
C
A
2m
6m
4m
4m
Solution
Equations of Equilibrium. First, we will consider the equilibrium of the FBD of the entire truss shown in Fig. a,
a + ΣMA = 0;
a + ΣMC = 0;
+ ΣF = 0;
S
x
NC(16) - 60(2) - 80(12) = 0 NC = 67.5 kN
Ans.
80(4) + 60(14) - Ay(16) = 0 Ay = 72.5 kN
Ans.
Ax = 0
Ans.
Then, using the result of NC, the equilibrium of the FBD of
truss BC shown in Fig. b gives
a + ΣMB = 0;
Ans.
67.5(8) - 80(4) - T(8) = 0 T = 27.5 kN
Ans.
NC = 67.5 kN
Ay = 72.5 kN
Ax = 0
T = 27.5 kN
221
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5–35. The bridge is constructed as a three-hinged trussed arch.
Determine the horizontal and vertical components of reaction
at the hinges (pins) at A, B, and C. The dashed member DE is
intended to carry no force.
60 k
40 k 40 k
20 k 20 k
D 10 ft E
B
100 ft
A
h1
h2
h3
C
30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft
Solution
Member AB:
a + ΣMA = 0;
Bx(90) + By(120) - 20(90) - 20(90) - 60(30) = 0
9Bx + 12By = 480
Member BC:
a + ΣMC = 0;
(1)
- Bx(90) + By(120) + 40(30) + 40(60) = 0
- 9Bx + 12By = - 360
(2)
Solving Eqs. (1) and (2) yields:
Bx = 46.67 k = 46.7 k
Member AB:
+ ΣF = 0;
S
x
Ans.
By = 5.00 k
Ax - 46.67 = 0
Ans.
Ax = 46.7 k
+ c ΣFy = 0;
Ay - 60 - 20 - 20 + 5.00 = 0
Ans.
Ay = 95.0 k
Member BC:
+ ΣF = 0;
S
x
- Cx + 46.67 = 0
Ans.
Cx = 46.7 k
+ c ΣFy = 0;
Cy - 5.00 - 40 - 40 = 0
Ans.
Cy = 85 k
Ans.
Bx = 46.7 k; By = 5.00 k;
Ax = 46.7 k; Ay = 95.0 k;
Cx = 46.7 k; Cy = 85 k
222
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*5–36. Determine the design heights h1, h2, and h3 of the
bottom cord of the truss so the three-hinged trussed arch
responds as a funicular arch.
60 k
40 k 40 k
20 k 20 k
D 10 ft E
B
100 ft
A
h1
h2
h3
C
30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft
Solution
y = - Cx 2
- 100 = -C(120)2
C = 0.0069444
Thus,
y = - 0.0069444x 2
y1 = - 0.0069444(90 ft)2 = - 56.25 ft
y2 = - 0.0069444(60 ft)2 = - 25.00 ft
y3 = - 0.0069444(30 ft)2 = - 6.25 ft
h1 = 100 ft - 56.25 ft = 43.75 ft
Ans.
h2 = 100 ft - 25.00 ft = 75.00 ft
Ans.
h3 = 100 ft - 6.25 ft = 93.75 ft
Ans.
Ans.
h1 = 43.75 ft;
h2 = 75.0 ft;
h3 = 93.75 ft
223
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5–37. The laminated-wood three-hinged arch is subjected to
the loading shown. Determine the horizontal and vertical
components of reaction at the pins A, B, and C.
B
15 kN
20 kN
2m
3m
2.5 m
A
15 kN
1.5 m
4m
4m
C
Solution
Equations of Equilibrium. Write the moment equation of equilibrium about
points A and C by referring to the FBDs of member AB, Fig. a, and member
BC, Fig. b, respectively.
a + ΣMA = 0;
a + ΣMC = 0;
Bx(3) - By(4) - 20(2.5) = 0
(1)
15(1.5) + 15(3.5) - By(4) - Bx(3) = 0
(2)
Solving Eqs. (1) and (2),
Bx = 20.83 kN = 20.8 kN
Ans.
By = 3.125 kN
Using these results to write the force equation of equilibrium by referring to
the FBD of member AB, Fig. a,
+ ΣF = 0;
S
x
+ c ΣFy = 0;
3
Ax + 20a b - 20.83 = 0 Ax = 8.833 kN = 8.83 kN
5
4
Ay - 20a b - 3.125 = 0 Ay = 19.125 kN = 19.1 kN
5
Ans.
Ans.
And referring to the FBD of member BC, Fig. b,
+ ΣF = 0;
S
x
+ c ΣFy = 0;
3
20.83 - 2(15) a b - Cx = 0 Cx = 2.833 kN = 2.83 kN
5
4
Cy + 3.125 - 2(15) a b = 0 Cy = 20.875 kN = 20.9 kN
5
Ans.
Ans.
Ans.
Bx = 20.8 kN
Ax = 8.83 kN
Ay = 19.1 kN
Cx = 2.83 kN
Cy = 20.9 kN
224
By = 3.125 kN
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–38. The three-hinged truss arch is subjected to the loading
shown. Determine the horizontal and vertical components of
reaction at the pins A, B, and C.
2k
10 ft
4k
10 ft
10 ft
1k
10 ft
2k
B
15 ft
A
C
Solution
a + ΣMA = 0;
a + ΣMC = 0;
Bx(15 ft) + By(20 ft) - 4 k(10 ft) = 0
By(20 ft) - Bx(15 ft) + 1 k(10 ft) = 0
Bx = 1.67 k
Ans.
By = 0.75 k
Ans.
+ ΣF = 0;
S
x
Ax - 1.67 k = 0;
Ax = 1.67 k S
Ans.
+ c ΣFy = 0;
+ ΣF = 0;
S
Ay - 2 k - 4 k + 0.75 k = 0;
Ay = 5.25 k c
Ans.
- Cx + 1.67 k = 0;
Cx = 1.67 k d
Ans.
+ c ΣFy = 0;
Cy - 2 k - 1 k - 0.75 k = 0;
Cy = 3.75 k c
Ans.
x
Ans.
Bx = 1.67 k
By = 0.75 k
Ax = 1.67 k S
Ay = 5.25 k c
Cx = 1.67 k d
Cy = 3.75 k c
225
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5–39. The arch structure is subjected to the loading shown.
Determine the horizontal and vertical components of reaction
at A and C, and the force in member AC.
6k
B
8k
10 ft
C
A
8 ft
10 ft
5 ft
5 ft
Solution
Equations of Equilibrium. First, we will consider the equilibrium of
the FBD of the entire structure shown in Fig. a.
a + ΣMA = 0;
a + ΣMC = 0;
1
1
b(23) + 8a
b(5) = 0
12
12
Ans.
1
1
b(5) + 8a
b(5) + 6(20) - Ay(28) = 0
12
12
Ans.
NC(28) - 6(8) - 8a
NC = 5.351 k = 5.35 k
8a
Ay = 6.306 k = 6.31 k
+ ΣF = 0;
S
x
Ax - 8a
1
b = 0 Ax = 5.657 k = 5.66 k
12
Ans.
Then, using the result of NC, the equilibrium of the FBD of member BC,
shown in Fig. b, gives
1
1
a + ΣMB = 0;
5.351(10) - 8a
b(5) - 8a
b(5) + FAC(10) = 0
12
12
FAC = 0.3060 k = 0.306 k
Ans.
Ans.
NC = 5.35 k
Ay = 6.31 k
Ax = 5.66 k
FAC = 0.306 k
226
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6–1. Draw the influence lines for (a) the moment at C,
(b) the vertical reaction at A, and (c) the shear at C. Assume A
is a fixed support. Solve this problem using the basic method of
Sec. 6.1.
B
A
1m
C
2m
Solution
Ans.
(MC)max = - 2
Ay = 1
(VC)max = 1
227
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6–2.
Solve Prob. 6–1 using the Müller-Breslau principle.
B
A
1m
Solution
228
C
2m
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6–3. Draw the influence lines for (a) the vertical reaction at
B, (b) the moment at C, and (c) the shear just to the right of the
support at A. Solve this problem using the basic method of
Sec. 6.1.
A
B
C
6 ft
9 ft
9 ft
9 ft
Solution
Ans.
(VB)max = 1.5
(MC)max = {4.5
(VA+) m ax = 1
229
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*6–4.
Solve Prob. 6–3 using the Müller-Breslau principle.
A
B
C
6 ft
Solution
230
9 ft
9 ft
9 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–5. Draw the influence lines for (a) the vertical reaction at
B, (b) the shear at C, and (c) the moment at C. Solve this
problem using the basic method of Sec. 6–1.
B
A
10 ft
10 ft
C
10 ft
Solution
Ans.
(By)max = 1
(VC)max = {0.5
(MC)max = {5
231
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6–6.
Solve Prob. 6–5 using the Müller-Breslau principle.
B
A
10 ft
Solution
232
10 ft
C
10 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–7. Draw the influence lines for (a) the moment at B,
(b) the shear at B, and (c) the vertical reaction at A. Solve this
problem using the basic method of Sec. 6.1. Hint: The support at
C resists only a horizontal force and a bending moment.
C
A
B
2m
2m
2m
Solution
Ans.
(MB)max = 2
(VB)max = 1
Ay = 1
233
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*6–8.
Solve Prob. 6–7 using the Müller-Breslau principle.
C
A
B
2m
Solution
234
2m
2m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–9. Draw the influence lines for (a) the moment at C,
(b) the vertical reaction at A, and (c) the vertical reaction at B.
There is a short link at E. Solve this problem using the basic
method of Sec. 6–1.
A
C
5 ft
5 ft
B
D
E
10 ft
10 ft
Solution
Ans.
(MC)max = - 5
(Ay)max = {1
(By)max = 2
235
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6–10.
Solve Prob. 6–9 using the Müller-Breslau principle.
A
C
5 ft
Solution
236
5 ft
B
D
E
10 ft
10 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–11. Draw the influence lines for (a) the vertical reaction at
B, and (b) the moment at E. Assume the supports at B and D
are rollers. There is a short link at C. Solve this problem using
the basic method of Sec. 6–1.
A
E
10 ft
B
10 ft
D
C
15 ft
15 ft
Solution
Ans.
(By)max = 1.75
(ME)max = - 7.5
237
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*6–12. Solve Prob. 6–11 using the Müller-Breslau principle.
A
E
10 ft
Solution
238
B
10 ft
D
C
15 ft
15 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–13. Draw the influence lines for (a) the moment at C,
(b) the shear just to the right of the support at B, and (c) the
vertical reaction at B. Solve this problem using the basic
method of Sec. 6.1. Assume A is a pin and B is a roller.
A
1m
B
C
1m
3m
1m
Solution
Ans.
(MC)max = {0.75
(VB +) m ax = 1
(By)max = 1.25
239
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6–14.
Solve Prob. 6–13 using the Müller-Breslau principle.
A
1m
Solution
240
B
C
1m
3m
1m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–15. The beam is subjected to a uniform dead load of
200 lb>ft and a single live load of 5 k. Determine (a) the
maximum moment created by these loads at C, and (b) the
maximum positive shear at C. Assume A is a pin, and B is a
roller.
8 ft
A
4 ft
C
B
5k
Solution
Referring to the influence line for the moment at C shown in
Fig. a, the maximum moment at C is
(MC) max = 2.6667(5) +
1
(12)(2.6667)(0.2)
2
= 16.53 k # ft = 16.5 k # ft
Ans.
Referring to the influence line for shear at C shown in Fig. b,
the maximum positive shear at C is
1
1
(VC) max = c (8)( - 0.6667) d (0.2) + c (12.8)(0.3333) d (0.2) + 5(0.333)
2
2
Ans.
= 1.2667 k = 1.27 k
241
Ans.
(MC) max = 16.5 k # ft;
(VC) max = 1.27 k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–16. Draw the influence lines for (a) the force in the cable
AC, (b) the vertical reaction at B, and (c) the moment at D.
C
12 ft
B
A
10 ft
Solution
242
D
6 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–17. A uniform live load of 300 lb>ft and a single live
concentrated force of 1500 lb are to be placed on the beam.
The beam has a weight of 150 lb>ft. Determine (a) the
maximum vertical reaction at support B, and (b) the maximum
negative moment at B. Assume the support at A is a pin and B
is a roller.
A
B
20 ft
10 ft
Solution
Referring to the influence line for the vertical reaction at B
shown in Fig. a, the maximum reaction is
1
(By) max ( + ) = 1.5(1500) + c (30 - 0)(1.5) d (300 + 150)
2
Ans.
= 12375 lb = 12.4 k
Referring to the influence line for the moment at B shown in
Fig. b, the maximum negative moment is
1
(MB) max (-) = -10(1500) + c (30 - 20)( - 10) d (300 + 150)
2
= -37500 lb # ft = -37.5 k # ft
Ans.
Ans.
(By) max ( + ) = 12.4 k;
(MB) max (-) = - 37.5 k # ft
243
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–18. The beam supports a uniform dead load of 6 kN>m,
a live load of 20 kN>m, and a single live concentrated force of
40 kN. Determine (a) the maximum positive moment at C,
and (b) the maximum positive shear just to the right of the
support at A. Assume A is a roller and B is a pin.
A
3m
C
2m
B
4m
Solution
Referring to the influence line for the moment at C shown in
Fig. a, the maximum positive moment at C is
1
1
(MC) max ( + ) = (40)(1.3333) + c (3)( -2) d (6) + c (9 - 3)(1.333) d (6)
2
2
1
+ c (9 - 3)(1.333) d (20)
2
= 139.33 kN # m = 139 kN # m
Ans.
Referring to the influence line for the shear at a point just to
the right of support A shown in Fig. b, the maximum positive
shear at this point is
1
1
(VA + ) max ( + ) = c (3)(0.5) d (6) + c (9 - 3)(1) d (6)
2
2
1
1
+ c (3)(0.5) d (20) + c (9 - 3)(1) d (20) + 40(1)
2
2
= 137.5 kN
244
Ans.
(MC) max ( + ) = 139 kN # m;
Ans.
(VA + ) max ( + ) = 137.5 kN
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–19. Where should the beam ABC be loaded with a
300-lb>ft uniform distributed live load so it causes (a) the
largest live moment at point A and (b) the largest live shear at
D? Calculate the values of the moment and shear. Assume the
support at A is fixed, B is pinned and C is a roller.
A
B
C
D
8 ft
8 ft
20 ft
Solution
(a) (MA) max =
1
(36)( - 16)(0.3) = -86.4 k # ft
2
(b) (VD) max = c (1)(8) +
Ans.
1
(1)(20) d (0.3) = 5.40 k
2
Ans.
Ans.
(MA) max = - 86.4 k # ft;
(VD) max = 5.40 k
245
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–20. The compound beam is subjected to a uniform dead
load of 200 lb>ft and a single live load of 2 k. Determine (a) the
maximum negative moment created by these loads at C, and
(b) the maximum negative shear at B. Assume C is a fixed
support, B is a pin, and A is a roller.
A
B
18 ft
C
9 ft
Solution
Referring to the influence line for the moment at C shown in
Fig. a, the maximum negative moment at C is
1
(MC) max (-) = c (27)( -9) d (0.2) + 2( - 9) = - 42.3 k # ft
2
Ans.
Referring to the influence line for the shear at B shown in
Fig. b, the maximum negative shear at B is
1
(VB) max (-) = c (18)( - 1) d (0.2) + 2( - 1) = - 3.80 k
2
Ans.
Ans.
(MC) max (-) = - 42.3 k # ft;
(VB) max (-) = - 3.80 k
246
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–21. The compound beam is subjected to a uniform dead
load of 200 lb>ft and a uniform live load of 150 lb>ft. Determine
(a) the maximum negative moment these loads develop at A,
and (b) the maximum positive shear at D. Assume B is a pin
and C is a roller.
A
B
5 ft
C
D
5 ft
5 ft
Solution
1
a) 1MA 2 max 1-2 = 1200 + 1502 a b 1 - 521152
2
= - 13125 lb # ft
Ans.
1
1
b) 1VD 2 max 1+2 = 12002 a b 1521 - 0.52 + 1200 + 1502 a b 15210.52
2
2
= 188 lb
Ans.
Ans.
(MA) max (-) = - 13,125 lb # ft;
(VD) max ( + ) = 188 lb
247
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–22. The beam is subjected to a uniform live load of
1.2 kN>m, a dead load of 0.5 kN>m, and a single live load of
40 kN. Determine (a) the maximum positive moment created
by these loads at E, and (b) the maximum positive shear at E.
Assume A and C are rollers, and B is a short link.
A
B
4m
C
4m
D
E
4m
4m
Solution
Referring to the influence line for the moment at E, shown in
Fig. a, the maximum positive moment at this point is
1
1
(ME) max ( + ) = 40(2.00) + c (16 - 8)(2.00) d (1.2 + 0.5) + c (8 - 0)( -2.00) d (0.5)
2
2
= 89.6 kN # m
Ans.
Referring to the influence line for the shear at E, shown in
Fig. b, the maximum positive shear at this point is
1
1
(VE) max ( + ) = 40(0.5) + c (8 - 0)(0.5) d (1.2 + 0.5) + c (16 - 12)(0.5) d (1.2 + 0.5)
2
2
1
+ c (12 - 8)( - 0.5) d (0.5)
2
= 24.6 kN
Ans.
Ans.
(ME) max ( + ) = 89.6 kN # m;
(VE) max ( + ) = 24.6 kN
248
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–23. The beam supports a uniform dead load of 500 lb>ft
and a single live concentrated force of 5 k. Determine
(a) the maximum negative moment at E, and (b) the maximum
positive shear at E. Assume the support at D is a pin, A and C
are rollers, and B is a pin.
A
C
B
9 ft
9 ft
D
E
9 ft
9 ft
Solution
Referring to the influence line for the moment at E shown in
Fig. a, the maximum negative moment at E is
1
1
(ME) max (-) = c (18)( - 4.50) d (0.5) + c (36 - 18)(4.50) d (0.5) + 5( - 4.50)
2
2
= -22.5 k # ft
Ans.
Referring to the influence line for the shear at E shown in
Fig. b, the maximum positive shear at E is
1
1
1
(VE) max ( + ) = c (18)(0.5) d (0.5) + c (27 - 18)(0.5) d (0.5) + c (36 - 27)(0.5) d (0.5) + 5(0.5)
2
2
2
= 4.75 k
Ans.
Ans.
(ME) max (-) = - 22.5 k # ft;
(VE) max ( + ) = 4.75 k
249
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–24. The beam supports a uniform dead load of 500 N>m
and single live concentrated force of 3000 N. Determine (a) the
maximum positive moment that can be developed at point C,
and (b) the maximum positive shear that can be developed at
point C. Assume the support at A is a pin and B is a roller.
A
C
B
10 m
10 m
Solution
1
a) (MC) max = 500a b(5)(20) + 3000(5)
2
Ans.
1
1
b) (VC) max = 500a b(0.5)(10) + 3000(0.5) - 500a b(0.5)(10)
2
2
Ans.
= 40 000 N # m = 40.0 kN # m
= 1500 N = 1.50 kN
Ans.
(MC) max = 40.0 kN # m;
(VC) max = 1.50 kN
250
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–25. Draw the influence lines for (a) the vertical reaction at
A, (b) the shear just to the right of A, and (c) the moment at C.
Assume C is fixed, A is a roller, and B is a pin.
A
8 ft
B
16 ft
C
10 ft
Solution
Ans.
(Ay)max = 1.5
(VA+)max = 1
(MC)max = - 10
251
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–26. A uniform live load of 0.4 k>ft and a concentrated live
force of 2 k are to be placed on the floor slabs. Determine
(a) the maximum live shear in panel CD, and (b) the maximum
live moment at B.
A
B
5 ft
C
5 ft
D
5 ft
E
5 ft
Solution
1
a) (VCD) max ( + ) = 2(1) + 0.4c (1)(5) + (1)(5) d
2
= 5k
Ans.
1
b) (MB) max (-) = 2( - 15) + 0.4a b( -15)(15)
2
Ans.
= -75 k # ft
Ans.
(VCD) max ( + ) = 5 k;
(MB) max (-) = - 75 k # ft
252
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–27. Draw the influence line for the moment at E in the
girder. Determine the maximum positive moment in the girder
at E if a single concentrated live force of 1200 lb and a uniform
live load of 200 lb>ft can be placed on the floor beams. Assume
A is a pin and D is a roller.
8 ft
A
8 ft
B
4 ft
C
D
E
Solution
Referring to the influence line for the moment at point E
shown in Fig. a,
1
(ME) max ( + ) = c (20)(3.20) d (200) + 1200(3.20)
2
= 10.24(103) lb # ft = 10.2 k # ft
Ans.
Ans.
(ME) max ( + ) = 10.2 k # ft
253
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–28. A uniform live load of 20 kN>m and a single
concentrated live force of 30 kN are placed on the floor beams.
If the beams also support a uniform dead load of 4 kN>m,
determine (a) the maximum positive shear in panel CD of the
girder and (b) the maximum negative moment in the girder at
C. Assume the support at B is a roller and E is a pin.
A
B
2m
D
C
2m
2m
E
2m
Solution
Referring to the influence line for the shear in panel CD shown in
Fig. a, the maximum positive shear within this panel is
1
1
(VCD) max ( + ) = c (2)(0.3333) d (20 + 4) + c (8 - 5)(0.3333) d (20 + 4)
2
2
1
+ c (5 - 2)( - 0.3333) d (4) + 30(0.3333)
2
Ans.
= 28.0 kN
Referring to the influence line for the moment at C shown in
Fig. b, the maximum negative moment at this point is
1
1
(MC) max (-) = c (2)( - 1.333) d (20 + 4) + c (8 - 2)(1.3333) d (4) + 30( - 1.333)
2
2
= -56.0 kN # m
Ans.
Ans.
(VCD) max ( + ) = 28.0 kN;
(MC) max (-) = -56.0 kN # m
254
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–29. Draw the influence lines for (a) the shear in
panel BC of the girder, and (b) the moment at D.
A
C
B
2m
2m
2m
F
E
D
2m
2m
Solution
Ans.
(VBC)max = 0.6
(MD)max = 2.4
255
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–30. A uniform live load of 250 lb>ft and a single
concentrated live force of 1.5 k are to be placed on the floor
beams. Determine (a) the maximum positive shear in
panel AB, and (b) the maximum moment at D. Assume only
vertical reactions occur at the supports.
15 ft
10 ft
A
E
10 ft
B
G
5 ft
C
D
5 ft
15 ft
F
H
5 ft
Solution
1
1
(25 - 18.33)(0.5)(0.250) + (0.5 + 0.25)(35 - 25)(0.250)
2
2
1
+ (50 - 35)(0.25)(0.250) + 0.5(1.5) = 2.57 k
Ans.
2
1
1
1
(MD) max = c (3.75)(15) + (3.75 + 7.5)(10) + (7.5 + 6.25)(10)
2
2
2
(VAB) max =
+
1
(6.25)(15) d (0.250) + 7.5(1.5)
2
= 61.25 k # ft
Ans.
Ans.
(VAB) max = 2.57 k;
(MD) max = 61.25 k # ft
256
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–31. A uniform live load of 200 lb>ft and a single
concentrated live force of 1500 lb are placed on the top beams.
If the beams also support a uniform dead load of 50 lb>ft,
determine (a) the maximum negative shear in panel BC of the
girder and (b) the maximum positive moment in the girder at B.
Assume C is a roller and A is a pin.
A
C
B
16 ft
8 ft
D
8 ft
Solution
Referring to the influence line for the shear within panel BC
shown in Fig. a, the maximum negative shear within this panel is
(VBC) max (-)
= c
1
(24)( - 0.6667) d (0.2 + 0.05)
2
1
+ c ( - 0.3333)(32 - 24) d (0.2 + 0.05) + 1.5( -0.6667)
2
= - 3.333 k = -3.33 k
Ans.
Referring to the influence line for the moment at B shown in
Fig. b, the maximum positive moment at B is
1
1
(MB) max ( + ) = c (24)(5.333) d (0.2 + 0.05) + c (32 - 24)( - 5.333) d (0.05) + 1.5(5.333)
2
2
= 22.93 k # ft = 22.9 k # ft
Ans.
Ans.
(VBC) max (-) = - 3.33 k;
(MB) max ( + ) = 22.9 k # ft
257
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–32. Draw the influence lines for (a) the moment at D in
the girder, and (b) the shear in panel BC.
A
B
4m
Solution
258
C
4m
E
D
4m
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–33. A uniform live load of 300 lb>ft and a single
concentrated live force of 2 k are to be placed on the floor
beams. Determine (a) the maximum negative shear in
panel AB, and (b) the maximum negative moment at B.
Assume the supports at A and E are pins and the pipe columns
only exert vertical reactions on the beams.
4 ft
A
4 ft
B
4 ft
C
4 ft
4 ft
D
E
Solution
Referring to the influence line of the shear within panel AB
shown in Fig. a, the maximum negative shear within this panel is
1
(VAB) max (-) = 2( - 1.00) + c (16 - 4)( - 1.00) d (0.3) = -3.80 k
2
Ans.
Referring to the influence line of moment at B, shown in Fig. b,
the maximum negative moment at this point is
1
(MB) max (-) = 2( - 4.00) + c (16 - 4)( - 4.00) d (0.3)
2
= - 15.2 k # ft
Ans.
Ans.
(VAB)max (-) = - 3.80 k;
(MB) max (-) = - 15.2 k # ft
259
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–34. Draw the influence line for the moment at B in the
girder. Determine the maximum positive live moment in the
girder at B if a single concentrated live force of 10 k moves
across the top beams. Assume the supports for these beams can
exert both upward and downward forces on the beams.
8 ft
A
8 ft
B
8 ft
E
8 ft
C
8 ft
D
Solution
(MB) max (+) = 10(12) = 120 k # ft
Ans.
Ans.
(MB) max (+) = 120 k # ft
260
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–35. Determine the maximum positive live shear in
panel CD due to a uniform live load of 20 kN>m acting on the
top beams.
A
B
3m
D
C
3m
3m
E
3m
F
3m
Solution
Referring to the influence line for the shear within panel CD
shown in Fig. a, the maximum positive shear within this panel is
1
1
(VCD) max ( + ) = c (3)(0.25) d (20) + c (15 - 7)(0.5) d (20)
2
2
= 47.5 kN
Ans.
Ans.
(VCD) max ( + ) = 47.5 kN
261
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
A
*6–36. A uniform live load of 3 k>ft and a single concentrated
live force of 45 k are placed on the floor beams. If the beams
also support a uniform dead load of 0.5 k>ft, determine (a) the
maximum negative shear in panel CD of the girder and (b) the
maximum positive moment in the girder at F.
B
D
C
E
F
12 ft
12 ft
12 ft
12 ft
Solution
Referring to the influence line for the shear within panel CD
shown in Fig. a, the maximum shear within this panel is
1
1
(VCD) max (-) = c (36)( - 0.6667) d (3 + 0.5) + c (48 - 36)( - 0.3333) d (3 + 0.5) + 45( - 0.6667)
2
2
= 79.0 k
Ans.
Referring to the influence line for the moment at F shown in
Fig. b, the maximum moment at this point is
1
1
(MF) max ( + ) = c (36)(8) d (3 + 0.5) + c (48 - 36)( - 8) d (0.5) + 45(8)
2
2
= 840 k # ft
Ans.
Ans.
(VCD) max (-) = 79.0 k;
(MF) max ( + ) = 840 k # ft
262
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–37. A uniform live load of 1.75 kN>m and a single
concentrated live force of 8 kN are placed on the floor beams.
If the beams also support a uniform dead load of 250 N>m,
determine (a) the maximum negative shear in panel BC of the
girder and (b) the maximum positive moment at B.
3m
1.5 m
1.5 m
D
A
B
C
Solution
By referring to the influence line for the shear in panel BC
shown in Fig. a, the maximum negative shear is
(VBC) max (-) = - 0.6667(8)
1
+ c (4.5 - 0)( -0.6667) d (1.75 + 0.25)
2
1
+ c (6 - 4.5)(0.6667) d (0.25)
2
= -8.21 kN
Ans.
By referring to the influence line for the moment at B shown in
Fig. b, the maximum positive moment is
1
(MB) max ( + ) = 1(8) + c (4.5 - 0)(1) d (1.75 + 0.25)
2
1
+ c (6 - 4.5)( -1) d (0.25)
2
= 12.3 kN # m
Ans.
Ans.
(VBC) max (-) = - 8.21 kN;
(MB) max ( + ) = 12.3 kN # m
263
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
L
6–38. Draw the influence line for the force in (a) member KJ
and (b) member DK.
K
J
I
H
4m
G
A
C
B
3m
3m
D
3m
F
E
3m
3m
3m
Solution
Ans.
(FKJ)max = - 1.125
(FDK)max = 0.625
264
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–39. Draw the influence line for the force in (a) member HI,
(b) member FI, and (c) member EF.
L
K
J
I
H
4m
G
A
C
B
3m
3m
D
3m
F
E
3m
3m
3m
Solution
Ans.
(FHI)max = - 0.625
(FFI)max = - 0.833
(FEF)max = 1
265
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–40.
L
Draw the influence line for the force in member KJ.
K
J
I
H
8 ft
A
B
8 ft
Solution
266
C
8 ft
D
8 ft
E
8 ft
G
F
8 ft
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–41.
Draw the influence line for the force in member JE.
L
K
J
I
H
8 ft
A
B
8 ft
C
8 ft
D
8 ft
E
8 ft
G
F
8 ft
8 ft
Solution
Ans.
(FJE)max = - 0.707
267
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–42.
8 ft
Draw the influence line for the force in member IJ.
8 ft
8 ft
L
A
8 ft
K
8 ft
J
8 ft
I
H
G
6 ft
B
C
D
E
F
Solution
Ans.
(FIJ)max = -2
268
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–43.
Draw the influence line for the force in member DI.
8 ft
8 ft
8 ft
L
A
8 ft
K
8 ft
J
8 ft
I
H
G
6 ft
B
C
D
E
F
Solution
Ans.
(FDI)max = 0.833
269
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–44.
Draw the influence line for the force in member EF.
8 ft
8 ft
8 ft
L
A
8 ft
K
8 ft
J
8 ft
I
H
G
6 ft
B
Solution
270
C
D
E
F
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–45. Draw the influence line for the force in (a) member EF
and (b) member CE.
G
E
F
4m
D
A
B
3m
3m
C
3m
3m
3m
3m
Solution
Ans.
(FEF)max = - 1
(FCE)max = 0.833
271
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–46.
G
Draw the influence line for the force in member BF.
E
F
4m
D
A
B
3m
3m
C
3m
3m
3m
3m
Solution
Ans.
(FBF)max = {0.417
272
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6–47.
Draw the influence line for the force in member BC.
G
E
F
4m
D
A
B
3m
3m
C
3m
3m
3m
3m
Solution
Ans.
(FBC)max = 0.75
273
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–48. Draw the influence line for the force in member BC of
the Warren truss. Indicate numerical values for the peaks. All
members have the same length.
G
608
A
B
20 ft
Solution
274
E
F
C
20 ft
20 ft
608
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–49. Draw the influence line for the force in (a) member KJ
and (b) member CJ.
L
K
J
I
H
8 ft
G
A
C
B
6 ft
6 ft
D
6 ft
F
E
6 ft
6 ft
6 ft
Solution
Ans.
(FKJ)max = - 1
(FCJ)max = - 0.625
275
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6–50. Draw the influence line for the force in (a) member JI,
(b) member IE, and (c) member EF.
L
K
J
I
H
8 ft
G
A
C
B
6 ft
6 ft
D
6 ft
F
E
6 ft
6 ft
6 ft
Solution
Ans.
(FJI)max = - 1
(FIE)max = 0.667
(FEF)max = 1
276
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–51. Draw the influence line for the force in member RQ
of the Baltimore truss.
T
W
S
O
A
R
Q
N
P
25 ft
V
M
B
C
D
E F G H I
12 @25 ft 5 300 ft
J
K
25 ft
L
Solution
Ans.
(FRQ)max = - 1.75
277
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*6–52. Draw the influence line for the force in member TC
of the Baltimore truss.
T
W
S
O
A
278
Q
N
P
25 ft
V
M
B
Solution
R
C
D
E F G H I
12 @25 ft 5 300 ft
J
K
L
25 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–53. Draw the influence line for the force in member NP
of the Baltimore truss.
T
W
S
O
A
R
Q
N
P
25 ft
V
M
B
C
D
E F G H I
12 @25 ft 5 300 ft
J
K
25 ft
L
Solution
Ans.
(FNP)max = 0.707
279
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–54. Draw the influence line for the force in member RN
of the Baltimore truss.
T
W
S
O
A
R
Q
N
P
25 ft
V
M
B
C
D
E F G H I
12 @25 ft 5 300 ft
J
K
L
Solution
Ans.
(FRN)max = 0.825
280
25 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–55. Draw the influence line for the force in member NG
of the Baltimore truss.
T
W
S
O
A
R
Q
N
P
25 ft
V
M
B
C
D
E F G H I
12 @25 ft 5 300 ft
J
K
L
Solution
Ans.
(FNG)max = 0.707
281
25 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–56. Draw the influence line for the force in member CO
of the Baltimore truss.
T
W
S
O
A
282
Q
N
P
25 ft
V
M
B
Solution
R
C
D
E F G H I
12 @25 ft 5 300 ft
J
K
L
25 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–57. Draw the influence line for the force in member CD.
L
K
J
I
H
6 ft
G
A
B
6 ft
C
6 ft
D
6 ft
E
6 ft
F
6 ft
6 ft
Solution
Ans.
(FCD)max = 1.50
283
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6–58.
Draw the influence line for the force in member CJ.
L
K
J
I
H
6 ft
G
A
B
6 ft
C
6 ft
D
6 ft
E
6 ft
F
6 ft
6 ft
Solution
Ans.
(FCJ)max = - 0.707
284
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6–59. Draw the influence line for the force in member HC,
then determine the maximum live force (tension or
compression) that can be developed in this member due to a
uniform live load of 800 lb>ft that acts on the bridge deck along
the bottom cord of the truss.
H
G
F
20 ft
E
A
C
B
20 ft
20 ft
D
20 ft
20 ft
Solution
(FHC) max (T) = 0.8a
(FHC) max (C) = 0.8a
1
b(0.7071)(53.33) = 15.1 k (T)
2
Ans.
1
b( -0.3536)(26.67) = - 3.77 k = 3.77 k (C)
2
Ans.
Ans.
(FHC) max (T) = 15.1 k (T);
(FHC) max (C) = 3.77 k (C)
285
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H
*6–60. Draw the influence line for the force in member CD,
and then determine the maximum force (tension or compres­
sion) that can be developed in this member due to a uniform
live load of 24 kN>m which acts along the bottom cord of the
truss.
G
F
4m
E
A
B
3m
C
3m
D
3m
3m
Solution
Referring to the influence line for the force in member CD
shown in Fig. a, we notice that member CD is subjected to only
tensile force. Thus, the maximum force in this member is
1
(FCD) max = c (12)(0.75) d (24) = 108 kN (T)
2
Ans.
Ans.
(FCD) max = 108 kN (T)
286
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6–61. Draw the influence line for the force in member DG,
and then determine the maximum force (tension or compression)
that can be developed in this member due to a uniform live load
of 24 kN>m which is transmitted to the truss along the bottom
cord.
H
G
F
4m
E
A
B
3m
C
3m
D
3m
3m
Solution
Referring to the influence line for the force in member DG
shown in Fig. a, we notice that member DG could be subjected
to tensile or compression force. The maximum tensile force and
compressive force are
1
(FDG) max (T) = c (12.8)(0.3125) d (24) = 15.0 kN (T)
2
1
(FDG) max (C) = c (8)( -0.625) d (24) = - 60.0 kN = 60.0 kN (C) (Max!)
2
Ans.
Ans.
(FDG) max = 60.0 kN (C)
287
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6–62. Determine the maximum moment at point C on the
single girder caused by the moving dolly that has a weight
of 3 k and a center of gravity at G.
G
2 ft 4 ft
A
B
C
12 ft
12 ft
12 ft
Solution
The vertical reactions of the wheels on the girder are shown in
Fig. a. The maximum moment at point C occurs when the moving
loads are at the position shown in Fig. b.
(MC) max ( + ) = 6(2) + 3(1) = 15.0 k # ft
Ans.
Ans.
(MC) max ( + ) = 15.0 k # ft
288
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6–63. Determine the maximum moment in the suspended
rail at point B if the rail supports the load of 2.5 k on the trolley.
8 ft
6 ft
A
6 ft
B
8 ft
C
1 ft 2 ft
2.5 k
Solution
Check maximum positive moment:
h
3
= ;
3
6
h = 1.5 ft
(MB) max = 1.667(3) + (0.833)(1.5) = 6.25 k # ft
Check maximum negative moment:
h
4
= ;
5
8
h = 2.5 ft
(MB) max = 1.667( - 4) + (0.833)( -2.5) = -8.75 k # ft
Ans.
Ans.
(MB)max = - 8.75 k # ft
289
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*6–64. Determine the maximum live moment at C caused by
the moving loads.
80 kN
60 kN
30 kN
30 kN
A
B
C
6m
10 m
2m
1m
Solution
Moving the 60-kN force 2 m to the left of point C, the change
in moment is
∆M = 30a -
3.75
3.75
3.75
3.75
b(2) + 60a b(2) + 80a
b(2) + 30a
b(2)
6
6
10
10
= -30 kN # m
Since ∆M is negative, the case where the 60-kN force at point
C will generate the maximum positive moment, Fig. a.
(MC) max ( + ) = 3.125(30) + 3.75(60) + 3.00(80) + 30(2.625)
= 637.5 kN # m
Ans.
Ans.
(MC) max ( + ) = 637.5 kN # m
290
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6–65. Determine the maximum live shear at C caused by the
moving loads.
80 kN
60 kN
30 kN
30 kN
A
B
C
6m
10 m
2m
1m
Solution
By inspecting the influence line for the shear at C, the maximum shear at this point is positive and occurs when the moving
loads are at the position shown in Fig. a.
(VC) max ( + ) = 30(0.625) + 60(0.5625) + 80(0.4375) + 30(0.375)
= 98.75 kN
Ans.
Ans.
(VC) max ( + ) = 98.75 kN
291
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8 kN
6–66. Determine the maximum positive moment at the splice
C on the side girder caused by the moving load which travels
along the center of the bridge.
4 kN
A
C
B
4m
8m
8m
8m
Solution
The maximum positive moment at point C occurs when the
moving loads are at the position shown in Fig. a.
(MC) max ( + ) = 4(4) + 2(2) = 20.0 kN # m
Ans.
Ans.
(Mc)max = 20.0 kN # m
292
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6–67. Determine the maximum moment at C caused by the
moving load.
12 kN
1m
0.5 m
A
B
C
8m
4m
Solution
The vertical reactions of the wheels on the beam are shown
in Fig. a. The maximum moment at point C occurs when the
moving loads are at the position shown in Fig. b.
(MC) max ( + ) = 8(2.6667) + 4(2.1667) = 30.0 kN # m
Ans.
Ans.
(MC) max ( + ) = 30.0 kN # m
293
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–68. The truck and trailer exerts the wheel reactions shown
on the deck of the girder bridge. Determine the largest moment
it exerts in the splice at C. Assume the truck travels left to right
along the center of the deck, and therefore transfers half of the
load shown to each of the two side girders. Assume the splice is
a fixed connection and, like the girder, can support both shear
and moment.
6 kN
4 kN
2 kN
1 kN
1.5 m 2 m
1m
A
10 m
C
5m
B
Solution
Worst case is 6 kN at P = 10 m.
1
c 1 kN(2.5) + 2 kN(3) + 6 kN(3.33) + 4 kN(2) d
2
Ans.
= 18.25 kN # m
(MC) max =
Ans.
(MC) max = 18.25 kN # m
294
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–69. Draw the influence line for the force in member CD of
the bridge truss. Determine the maximum live force (tension or
compression) that can be developed in the member due to a
15-kN truck having the wheel loads shown. Assume the truck
can travel in either direction along the center of the deck, so
that half the load shown is transferred to each of the two side
trusses. Also assume the members are pin connected at the
gusset plates.
3m
6 kN
9 kN
I
H
J
G
F
4.5 m
B
A
6m
C
6m
D
6m
E
6m
Solution
Refering to the influence for the force in member CD shown
in Fig. a, we notice that member CD will be subjected to tensile force only, and the maximum occurs when the moving load,
which is half that shown in the problem figure, is at the position
shown.
(FCD)max = 3(1.000) + 4.5(1.333) = 9.00 kN(T)
Ans.
By symmetry, the result would be the same if the truck were
traveling in the other direction.
Ans.
(FCD)max = 9.00 kN (T)
295
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6–70. The 9-k truck exerts the wheel reactions shown on the
deck of a girder bridge. Determine (a) the largest live shear it
creates in the splice at C, and (b) the largest moment it exerts
in the splice. Assume the truck travels in either direction along
the center of the deck, and therefore transfers half of the load
shown to each of the two side girders. Assume the splice is a
fixed connection and, like the girder, can support both shear
and moment.
6k
3k
A
C
B
15 ft
20 ft
40 ft
Solution
(a)
(VC)max = 3(0.6667) + 1.5(0.4167) = 2.62 k
Ans.
(b)
(MC)max = 3(13.33) + 1.5(8.333) = 52.5 k # ft
Ans.
Ans.
(VC)max = 2.62 k;
(MC)max = 52.5 k # ft
296
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–71. The truck has a mass of 4 Mg and mass center at G1,
and the trailer has a mass of 1 Mg and mass center at G2.
Determine the absolute maximum live moment developed in
the bridge.
G2
G1
1.5 m 1.5 m
0.75 m
A
B
8m
Solution
Placement is shown in FBD (1). Using segment (2):
M max = 17 780.625(3.625) = 64.5 kN # m
Ans.
Ans.
Mabs = 64.5 kN # m
max
297
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G
*6–72. The truck has a weight of 12 k and mass center at G.
Determine the absolute maximum live moment developed in
the bridge.
4 ft
8 ft
A
B
30 ft
Solution
The vertical reactions of the wheels on the girder are shown in
Fig. a. By inspection, the absolute maximum moment occurs
under the 8-k force, Fig. b.
a + ΣMB = 0;
- Ay(30) + 8(17) + 4(5) = 0
Ay = 5.20 k
Referring to Fig. c,
a + ΣMS = 0;
MS - 5.20(13) = 0
MS = 67.6 k # ft (Abs. Max.)
Ans.
Ans.
M abs = 67.6 k # ft
max
298
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6–73. Determine the absolute maximum live moment in the
girder bridge due to the loading shown. The load is applied
directly to the girder.
3k
2k
10 ft
A
B
15 ft
D
C
30 ft
15 ft
Solution
Referring to Fig. a, the location of FR for the moving load is
+ TFR = ΣFy;
a +FRx = ΣMD;
FR = 3 + 2 = 5 k
- 5x = - 21102
x = 4 ft
By observation, the maximum positive moment occurs under
the 3-k force, Fig. b.
a + ΣMC = 0;
2(7) + 3(17) - By(30) = 0 By = 2.1667 k
Referring to Fig. c,
a + ΣMS = 0;
MS( + ) - 2.1667(13) = 0
MS( + ) = 28.17 k # ft
The maximum negative moment occurs when the 3-k force of
the moving load is at the edge of the overhang, Fig. d.
a + ΣMS = 0;
MS(-) + 2(5) + 3(15) = 0
MS(-) = -55.0 k # ft (Abs. Max.)
Ans.
Ans.
Mabs = - 55.0 k # ft
max
299
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–74. Determine the absolute maximum live shear and
absolute maximum live moment in the jib beam AB due to the
crane loading. The end constraints require 0.1 m … x … 3.9 m.
4m
A
B
x
12 kN
Solution
V max = + 12 kN
Ans.
(position at 0.1 m … x … 3.9 m)
M max = -12(3.9) = - 46.8 kN # m
Ans.
(position at x = 3.9 m)
Ans.
Vabs = +12 kN;
max
Mabs = -46.8 kN # m
max
300
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–75. The maximum wheel loadings for the wheels of a crane
that is used in an industrial building are given. The crane travels
along the runway girders that are simply supported on columns.
Determine (a) the absolute maximum shear in intermediate
girder AB, and (b) the absolute maximum moment in the
girder.
12 kN
8 kN
A
B
1m
6m
Solution
The absolute maximum shear occurs when the moving loads
are positioned with the 12-kN force just to the left of the support at B, Fig. a.
a + ΣMA = 0;
By(6) - 8(5) - 12(6) = 0
By = 18.67 kN
Therefore, the absolute maximum shear stress is,
Ans.
Vabs = By = 18.67 kN = 18.7 kN
max
Referring to Fig. b, the location of FR for the moving load is
+ TFR = ΣFy;
FR = 8 + 12 = 20 kN
a +FRx = ΣMC; 20x = 8(1) x = 0.4 m
By observation, the absolute maximum moment occurs under
the 12-kN force, Fig. c.
a + ΣMA = 0;
By(6) - 8(2.2) - 12(3.2) = 0
By = 9.333 kN
Referring to Fig. d,
a + ΣMS = 0;
9.333(2.8) - MS = 0
MS = 26.13 kN # m = 26.1 kN # m (Abs. Max.)
Ans.
Ans.
Vabs = 18.7 kN;
max
M abs = 26.1 kN # m
max
301
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–76. Determine the maximum positive live moment in the
girder bridge due to the loading shown. The load is applied
directly to the girder.
2400 lb
1800 lb
10 ft
A
B
15 ft
C
15 ft
D
15 ft
Solution
Maximum occurs when wheel A is at C.
M max =
1
c 11.77 kN(1) + 7.848 kN(0.375) d = 7.36 kN # m
2
Ans.
M max = 7.36 kN # m
302
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–77. Determine the absolute maximum live moment in the
girder bridge due to the loading shown. The load is applied
directly to the girder.
2k
1k
A
B
8 ft
30 ft
Solution
x =
2(8)
3
= 5.333 ft
a + ΣMS = 0;
- MS + 1.367(13.667) = 0
MS = 18.7 k # ft
FPO
Ans.
Mabs = 18.7 k # ft
max
303
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–78. Determine the absolute maximum moment in the
beam due to the loading shown.
10 k
6k
16 ft
4k
8 ft A
B
45 ft
Solution
Referring to Fig. a, the location of FR for the moving load is
+ TFR = ΣFy;
FR = 10 + 6 + 4 = 20 k
a +FRx = ΣMC; - 20x = - 6(16) - 4(24) x = 9.60 ft
Assuming that the absolute maximum moment occurs under
the 10-k force, Fig. b,
a + ΣMB = 0;
4(3.3) + 6(11.3) + 10(27.3) - Ay(45) = 0 Ay = 7.8667 k
Referring to Fig. c,
a + ΣMS = 0;
MS - 7.8667(17.7) = 0 MS = 139.24 k # ft = 139 k # ft (Abs. Max.)
Ans.
Assuming that the absolute maximum moment occurs under
the 6-k force, Fig. d,
a + ΣMA = 0;
By(45) - 10(9.7) - 6(25.7) - 4(33.7) = 0 By = 8.5778 k
Referring to Fig. e,
a + ΣMS = 0;
8.5778(19.3) - 4(8) - MS = 0
MS = 133.55 k # ft
Ans.
Mabs = 139 k # ft
max
304
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6–79. Determine the absolute maximum shear in the bridge
girder due to the loading shown.
10 kN
3m
5 kN
B
A
12 m
Solution
By inspection, the absolute maximum shear occurs when the
moving loads are in the position with the 10-kN force just to
the right of the roller support at A, Fig. a.
a + ΣMB = 0;
5(9) + 10(12) - NA(12) = 0
NA = 13.75 kN
Therefore, the absolute maximum shear is
abs = NA = 13.75 kN
V max
Ans.
Ans.
Vabs = 13.75 kN
max
305
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*6–80. Determine the absolute maximum moment in the
bridge girder due to the loading shown.
10 kN
3m
5 kN
B
A
12 m
Solution
Referring to Fig. a, the location of FR for the moving loads is
+ c FR = ΣFy;
a +FRx = ΣMC;
-FR = - 10 - 5
FR = 15 kN
15x = 10(3)
x = 2m
By inspection, the absolute maximum moment occurs under
the 10-kN force, Fig. b. Write the moment equation of equilibrium about B.
a + ΣMB = 0;
15(5.5) - NA(12) = 0
NA = 6.875 kN
Referring to the FBD of the left segment of the beam sectioned through x = 5.5 m, Fig. c,
a + ΣMS = 0;
M abs = - 6.875(5.5) = 0 MS = M abs = 37.8 kN # m
max
max
Ans.
Ans.
Mabs = 37.8 kN # m
max
306
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6–81. Determine the absolute maximum shear in the beam
due to the loading shown.
6k
3k
2 k4 k
5 ft 3 ft 3 ft
30 ft
Solution
The maximum shear occurs when the moving loads are positioned either with the 3-k force just to the right of the support
at A, Fig. a, or with the 4-k force just to the left of the support
at B. Referring to Fig. a,
a + ΣMB = 0;
4(19) + 2(22) + 6(25) + 3(30) - Ax(30) = 0
Ax = 12.0 k
Referring to Fig. b,
a + ΣMA = 0;
By(30) - 3(19) - 6(24) - 2(27) - 4(30) = 0
By = 12.5 k
Therefore, the absolute maximum shear occurs for the case in
Fig. b.
V abs = - By = - 12.5 k
max
Ans.
Ans.
Vabs = - 12.5 k
max
307
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6–82. Determine the absolute maximum moment in the
beam due to the loading shown.
20 kN
10 kN
5 kN
1m 2m
12 m
Solution
Referring to Fig. a, the location of FR for the moving load is
+ TFR = ΣFy;
FR = 5 + 10 + 20 = 35 kN
a +FRx = ΣMC; 35x = 10(2) + 5(3) x = 1.00 m
Assuming that the absolute maximum moment occurs under
the 10-kN force, Fig. b,
a + ΣMB = 0;
20(4.5) + 10(6.5) + 5(7.5) - Ay(12) = 0 Ay = 16.0417 kN
Referring to Fig. c,
a + ΣMS = 0;
MS + 5(1) - 16.0417(5.5) = 0
MS = 83.23 kN # m
Assuming that the absolute maximum moment occurs under
the 20-kN force, Fig. d,
a + ΣMA = 0;
By(12) - 5(3.5) - 10(4.5) - 20(6.5) = 0 By = 16.0417 kN
Referring to Fig. c,
a + ΣMS = 0;
16.0417(5.5) - MS = 0
MS = 88.23 kN # m = 88.2 kN # m (Abs. Max.)
Ans.
Ans.
M abs = 88.2 kN # m
max
308
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–83. Determine the absolute maximum moment in the
beam due to the loading shown.
20 kN
25 kN
40 kN
4m
A
1.5 m
B
12 m
Solution
Referring to Fig. a, the location of FR for the moving load is
+ TFR = ΣFy;
a + FRx = ΣMC;
FR = 40 + 25 + 20 = 85 kN
- 85x = - 25(4) - 20(5.5)
x = 2.4706 m
Assuming that the absolute maximum moment occurs under
40 kN force, Fig. b,
a + ΣMB = 0;
20(1.7353) + 25(3.2353) + 40(7.2353) - Ay(12) = 0
Ay = 33.75 kN
Referring to Fig. c,
a + ΣMS = 0;
MS - 33.75(4.7647) = 0
MS = 160.81 kN # m
Assuming that the absolute moment occurs under 25 kN force,
Fig. d,
a + ΣMA = 0;
By(12) - 40(2.7647) - 25(6.7647) - 20(8.2647) = 0
By = 37.083 kN
Referring to Fig. e,
a + ΣMS = 0;
37.083(5.2353) - 20(1.5) - MS = 0
MS = 164.14 kN # m = 164 kN # m (Abs. Max.)
Ans.
Ans.
Mabs = 164 kN # m
max
309
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–84. The trolley rolls at C and D along the bottom and top
flanges of beam AB. Determine the absolute maximum
moment developed in the beam if the load supported by the
trolley is 2 k. Assume the support at A is a pin and at B a roller.
20 ft
D
A
B
C
1 ft
0.5 ft
Solution
Referring to the FBD of the trolley in Fig. a,
a + ΣMC = 0;
a + ΣMD = 0;
ND(1) - 2(1.5) = 0
ND = 3.00 k
NC(1) - 2(0.5) = 0
NC = 1.00 k
Referring to Fig. b, the location of FR is
+ TFR = ΣFy;
FR = 3.00 - 1.00 = 2.00 kT
+ TFRx = ΣMC;
- 2.00 (x) = - 3.00(1)
x = 1.5 ft
The absolute maximum moment occurs under the 3.00-k
force, Fig. c.
a + ΣMA = 0;
By(20) + 1.00(8.75) - 3.00(9.75) = 0
By = 1.025 k
Referring to Fig. d,
a + ΣMS = 0;
1.025(10.25) - MS = 0
MS = 10.5 k # ft (Abs. Max.)
Ans.
Ans.
MS = 10.5 k # ft (Abs. Max.)
310
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6–1P. The chain hoist on the wall crane can be placed
anywhere along the boom (0.1 m 6 x 6 3.4 m) and has a
rated capacity of 28 kN. Use an impact factor of 0.3 and
determine the absolute maximum bending moment in the
boom and the maximum force developed in the tie rod BC. The
boom is pinned to the wall column at its left end A. Neglect the
size of the trolley at D.
C
0.75 m
B
A
x
0.1 m
D
28 kN
3m
0.5 m
Solution
Absolute maximum moment occurs when the trolley is at x = 1.5 m.
Load = 28 + 0.3(28) = 36.4 kN
a + ΣMA = 0; T sin 12.23°(3) + T cos 12.23°(0.1) - 36.4(1.5) = 0
T = 74.49 kN
a + c ΣFy = 0; Ay - 36.4 + 74.49 sin 12.23° = 0
Ay = 20.63 kN
a + ΣMS = 0; MS - 20.63(1.5) = 0
M max = 30.9 kN # m;
Ans.
Absolute maximum tension occurs in the tie rod when trolley x = 3.4 m.
a + ΣMA = 0; T max sin 12.23°(3) + T max cos 12.23°(0.1) - 36.4(3.4) = 0
T max = 169 kN
Ans.
Ans.
M max = 30.9 kN # m;
T max = 169 kN
311
H is protected
G
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underFall copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1.25 m
A
6–2P. A simply supported pedestrian bridge is to be
constructed in a city park and two designs have been proposed
as shown in case a and case b. The truss members are to be
made from timber. The deck consists of 1.5-m-long planks that
have a mass of 20 kg>m2. A local code states the live load on the
deck is required to be 5 kPa with an impact factor of 0.2.
Consider the deck to be simply supported on stringers. Floor
beams then transmit the load to the bottom joints of the truss.
(See Fig. 6–23.) In each case find the member subjected to the
largest tension and largest compression load and suggest why
you would choose one design over the other. Neglect the
weights of the truss members.
E
H
B
1.25 m 1.25 m
A
G
C
1.25 m
B
1.25 m
F
D
1.25 m
case a
C
1.25 m
E
1.25 m
E
D
1.25 m
E
1.25 m
case a
H
G
F
H
B
G
C
F
D
1.25 m
A
E
Solution
1.25 m 1.25 m
A
Dead load:
wd = 20(9.81)(1.5) = 294.3 N>m
case a
B
Live load:
1.25 m
w1 = 5000(1 + 0.2)(1.5) = 9000 N>m
1.25 m
case b
C
1.25 m
1.25 m
1.25 m
E
D
1.25 m
case b
For each truss:
wd =
1.25 m
294.3
= 147.15 N>m
2
w1 = 4500 N>m
Case a:
Largest compression members are AH or EF.
1
FAH = FEF = (147.15 + 4500) a b(1.061)(5) = 12.3 kN (C)
2
Ans.
Largest tension members are AB, BC, CD, DE.
1
FAB = FBC = FCD = FDE = (147.15 + 4500) a b(0.75)(5) = 8.71 kN (T)
2
Ans.
Case b:
Largest compression members are AH and EF.
case b
1
FAH = FEF = (147.15 + 4500) a b(1.061)(5) = 12.3 kN (C)
2
Ans.
Largest tension members are BC and CD.
1
FBC = FCD = (147.15 + 4500) a b(0.1)(5) = 11.6 kN (T)
2
Ans.
Choose case a to avoid the larger tension forces in member
BC and CD. In timber construction, joints subjected to large
tension should be avoided.
Ans.
Ans.
Case a: F max (C) = 12.3 kN; F max (T) = 8.71 kN;
Case b: F max (C) = 12.3 kN; F max (T) = 11.6 kN;
Choose case a.
312
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7–1. Determine the equation of the elastic curve using the
coordinate x, and specify the slope at point A and the deflection
at point C. EI is constant.
w
B
A
C
x
L
__
2
L
__
2
Solution
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(b).
Slope and Elastic Curve:
EI
d 2y
= M(x)
dx 2
EI
d 2y
wL
w 2
=
x x
2
2
dx 2
EI
dy
wL 2
w 3
=
x x + C1
dx
4
6
EI y =
(1)
wL 3
w 4
x x + C1x + C2(2)
12
24
Boundary Conditions: Due to symmetry,
dy
L
= 0 at x =
.
dx
2
Also, y = 0 at x = 0.
wL L 2 w L 3
wL3
a b a b + C1 C1 = 4
2
6 2
24
From Eq. [1],
0 =
From Eq. [2],
0 = 0 + 0 + 0 + C2 C2 = 0
The Slope: Substituting the value of C1 into Eq. [1],
dy
w
=
( - 4x 3 + 6Lx 2 - L3)
dx
24EI
uA =
dy
wL3
wL3
`
= =
A
dx x = 0
24EI
24EI
Ans.
The negative sign indicates clockwise rotation.
The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2],
y =
wx
( - x 3 + 2Lx 2 - L3)
24EI
yc = yx =
L
2
= -
5wL4
5wL4
=
T
384EI
384EI
Ans.
Ans.
The negative sign indicates downward displacement.
wL3
24EI
5wL4
yc =
T
384EI
uA =
313
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–2. The bar is supported by a roller constraint at B, which allows
vertical displacement but resists axial load and moment. If the bar
is subjected to the loading shown, determine the slope at A and the
deflection at C. EI is constant.
200 lb
A
8 ft
Solution
Slope and Elastic Curve. The support reactions and the elastic
curve are shown in the FBD of the bar, Fig. a. Also indicated are
the two position coordinates x1 and x2. The moment functions written based on x1 and x2 are shown in Figs. b and c, respectively.
For M(x1) = 200x1,
EI
EI
d 2y1
dx 21
d 2y1
dx 21
EI
= M(x1)
= 200x1
dy1
= 100x 21 + C1(1)
dx1
EI y1 =
100 3
x + C1x1 + C2(2)
3 1
For M(x2) = 1600,
EI
EI
d 2y2
dx 22
d 2y2
dx 22
EI
= M(x2)
= 1600
dy2
= 1600x2 + C3(3)
dx2
EIy2 = 800x 22 + C3x2 + C4(4)
Boundary Condition. At x1 = 0, y1 = 0. Eq. (2) gives C2 = 0.
At x2 = 0,
dy2
= 0. Eq. (3) gives C3 = 0.
dx2
Continuity Condition. At x1 = x2 = 8 ft,
Using Eqs. (1) and (3),
dy1
dy2
= .
dx1
dx2
100(82) + C1 = - 31600(8)4 + 0
C1 = - 19 200 lb # ft2
314
B
C
8 ft
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7–2.
(Continued)
Then, the equations of slope and elastic curve for x1 are
dy1
1
=
(100x 21 - 19 200) lb # ft2
dx1
EI
y1 =
1 100 3
a
x - 19 200x1 b lb # ft3
EI
3 1
At point A, x1 = 0. Thus,
uA =
dy1
19 200 lb # ft2
19 200 lb # ft2
`
= =
dx1 x1 = 0
EI
EI
Ans.
At point C, x1 = 8 ft. Thus,
yC =
1 100 3
c
(8 ) - 19 200(8) d EI 3
= =
409 600 lb # ft3
3EI
409 600 lb # ft3
T
3EI
Ans.
Ans.
19 200 lb # ft2
EI
409 600 lb # ft3
yC =
T
3EI
uA =
315
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–3. Determine the slope at C, and the deflection at B of the
bar in Prob. 7–2.
200 lb
A
8 ft
Solution
Slope and Elastic Curve. The support reactions and the elastic
curve are shown in the FBD of the bar, Fig. a. Also indicated are
the two position coordinates x1 and x2. The moment functions written based on x1 and x2 are shown in Figs. b and c, respectively.
For M(x1) = 200x1,
EI
EI
d 2y1
dx 21
d 2y1
dx 21
EI
= M(x1)
= 200x1
dy1
= 100x 21 + C1(1)
dx1
EIy1 =
100 3
x + C1x1 + C2(2)
3 1
For M(x2) = 1600,
EI
EI
EI
d 2y2
dx 22
d 2y2
dx 22
= M(x2)
= 1600
dy2
= 1600x2 + C3
dx2
(3)
EIy2 = 800x 22 + C3x2 + C4(4)
Boundary Condition. At x1 = 0, y1 = 0. Eq. (2) gives C2 = 0.
At x2 = 0,
dy2
= 0. Eq. (3) gives C3 = 0.
dx2
Continuity Condition. At x1 = x2 = 8 ft, y1 = y2. Using
Eqs. (2) and (4),
100 3
(8 ) + ( - 19 200)(8) + 0 = 800(82) + 0 + C4
3
563 200
C4 = lb # ft3
3
316
B
C
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–3.
(Continued)
Then, the equations of the slope and elastic curve for x2 are
dy2
1600
=
x
dx2
EI 2
y2 =
1
563 200
1
a800x 22 (2400x 22 - 563 200)
b =
EI
3
3EI
At point C, x2 = 8 ft. Then
uC =
dy2
1600
12 800 lb # ft2
`
=
(8) =
dx2 x2 = 8 ft
EI
EI
Ans.
1
563 200 lb # ft3
563 200 lb # ft3
(0 - 563 200) = =
T
3EI
3EI
3EI
Ans.
At point B, x2 = 0. Then
y2 =
Ans.
uC =
yB =
317
12 800 lb # ft2
EI
563 200 lb # ft3
T
3EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–4. Determine the equations of the elastic curve using the
coordinates x1 and x2 and specify the slope and deflection at B.
EI is constant.
2 k>ft
C
A
B
x1
8 ft
x2
12 ft
Solution
Slope and Elastic Curve. The support reactions and the elastic
curve are shown in the FBD of the beam, Fig. a. The moment
function written based on x1 is shown in Fig. b.
For M(x1) = -x 21 + 16x1 - 64,
EI
EI
d 2y1
dx 21
d 2y1
dx 21
EI
= M(x1)
= - x 21 + 16x1 - 64
dy1
x 31
= + 8x 21 - 64x1 + C1(1)
dx1
3
EIy1 = -
x 41
8
+ x 31 - 32x 21 + C1x1 + C2(2)
12
3
dy1
= 0. Then Eq. (1) gives
dx1
C1 = 0. Also, y1 = 0 at x1 = 0. Then Eq. 2 gives C2 = 0. Thus,
the equation of the elastic curve becomes
Boundary Condition. At x1 = 0,
x 41
1
8
a+ x 31 - 32x 21 b k # ft3
EI
12
3
1
=
( - x 41 + 32x 31 - 384x 21) k # ft3
Ans.
12EI
The moment function written based on x2 is shown in Fig. c.
y1 =
For M(x2) = 0,
EI
EI
d 2y2
dx 22
d 2y2
dx 22
EI
= M(x2)
= 0
dy2
= C3(3)
dx2
EIy2 = C3x2 + C4(4)
dy1
dy2
Continuity Condition. At x1 = x2 = 8 ft,
` =
.
dx1
dx2
Using Eqs. (1) and (3),
-
83
512 # 2
+ 8(82) - 64(8) + 0 = C3 C3 = k ft
3
3
Also, y1 = y2 at x1 = x2 = 8 ft. Using Eqs. (2) and (4),
-
x3
84
8
512
+ (83) - 32(82) + 0 + 0 = a b(8) + C4
12
3
3
1024 # 3
C4 =
k ft
3
318
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–4.
(Continued)
Thus, the equations of the slope and elastic curve are
u2 =
y2 =
dy2
512 k # ft2
= dx2
3EI
1
512
1024
1
ax +
b =
( - 512x2 + 1024) k # ft3
EI
3 2
3
3EI
Ans.
At point B, x2 = 12 ft. Then
uB = u2 0 x2 = 12 ft = yB = y2 0 x2 = 12 ft =
=-
512 k # ft2
512 k # ft2
=
3EI
3EI
Ans.
1
3 - 512(12) + 10244
3EI
5120 k # ft3
5120 k # ft3
=
T
3EI
3EI
Ans.
Ans.
1
( -x 41 + 32x 31 - 384x 21) k # ft3
12EI
1
y2 =
( -512x2 + 1024) k # ft3
3EI
512 k # ft2
uB =
3EI
5120 k # ft3
yB =
T
3EI
y1 =
319
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–5. Determine the equations of the elastic curve using the
coordinates x1 and x3 and specify the slope and deflection at
point B. EI is constant.
2 k>ft
C
A
B
x1
8 ft
x2
12 ft
Solution
Slope and Elastic Curve. The support reactions and the elastic
curve are shown in the FBD of the beam, Fig. a. The moment
function written based on x1 is shown in Fig. b.
For M(x1) = -x 21 + 16x1 - 64,
EI
EI
d 2y1
dx 21
d 2y1
dx 21
EI
= M(x1)
= - x 21 + 16x1 - 64
dy1
x 31
= + 8x 21 - 64x1 + C1(1)
dx1
3
EIy1 = -
x 41
8
+ x 31 - 32x 21 + C1x1 + C2(2)
12
3
dy1
= 0. Then Eq. (1) gives
dx1
C1 = 0. Also, y1 = 0 at x1 = 0. Then Eq. 2 gives C2 = 0. Thus,
the equation of the elastic curve becomes
Boundary Condition. At x1 = 0,
x 41
1
8
a+ x 31 - 32x 21 b k # ft3
EI
12
3
1
=
( -x 41 + 32x 31 - 384x 21) k # ft3
12EI
The moment function written based on x3 is shown in Fig. c.
y1 =
Ans.
For M(x3) = 0,
EI
EI
d 2y3
dx 23
d 2y3
dx 23
EI
= M(x3)
= 0
dy3
= C3(3)
dx3
EIy3 = C3x3 + C4(4)
dy3
dy1
Continuity Condition. At x1 = 8 ft and x3 = 4 ft,
= .
dx1
dx3
Using Eqs. (1) and (3),
-
83
512
+ 8(82) - 64(8) + 0 = -C3 C3 = a
b k # ft2
3
3
Also, y1 = y3, at x1 = 8 ft and x3 = 4 ft. Using Eqs (2) and (4),
-
x3
84
8
512
+ (83) - 32(82) + 0 + 0 =
(4) + C4
12
3
3
5120 # 3
C4 = k ft
3
320
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–5.
(Continued)
Thus, the equations of slope and elastic curve are
u3 =
y3 =
dy3
512 k # ft2
=
dx3
3EI
1 512
5120
1
a
x b =
(512x3 - 5120) k # ft3
EI
3 3
3
3EI
Ans.
At point B, x3 = 0. Then
uB = u3 `
yB = y3 `
x3 = 0
x2 = 0
=
512 k # ft2
3EI
Ans.
=
1
5120 k # ft3
5120 k # ft3
(0 - 5120) = =
T
3EI
EI
3EI
Ans.
Ans.
y1 =
1
( -x 41 + 32x 31 - 384x 21) k # ft3
12EI
y3 =
1
(512x3 - 5120) k # ft3
3EI
uB =
yB =
321
512 k # ft2
3EI
5120 k # ft3
T
3EI
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7–6. Determine the equation of the elastic curve for the
beam using the x coordinate. Specify the slope at A and the
maximum deflection of the beam. EI is constant.
8 k ? ft
8 k ? ft
A
B
x
20 ft
Solution
Slope and Elastic Curve. The support reactions and the elastic
curve are shown in the FBD of the beam, Fig. a. The moment
function written based on x is shown in Fig. b. For M(x) = 8 k ft,
EI
d 2y
= M(x)
dx 2
EI
d 2y
= 8
dx 2
EI
dy
= 8x + C1(1)
dx
EIy = 4x 2 + C1x + C2(2)
Boundary Conditions. At x = 0, y = 0. Eq (2) gives
C2 = 0
Also, y = 0 at x = 20 ft. Using Eq. (2),
u = 4(202) + C1(20) + 0 C1 = - 80 k # ft2
Then
u =
dy
1
=
(8x - 80) k # ft2
dx
EI
y =
1
(4x 2 - 80x) k # ft3
EI
Ans.
At point A, x = 0. Then
1
80 k # ft2
80 k # ft2
(0 - 80) = =
EI
EI
EI
uA = u `
x=0
y max =
1
400 k # ft3
400 k # ft3
34(102) - 80(10)4 = =
T
EI
EI
EI
=
Ans.
Due to the symmetrical loading and beam system, the maximum deflection of the beam occurs at mid-span, x = 10 ft.
Ans.
Ans.
1
(4x 2 - 80x) k # ft3
EI
80 k # ft2
uA =
EI
400 k # ft3
y max =
T
EI
y =
322
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–7. Determine the equations of the elastic curve for the
beam using the xl and x2 coordinates. Specify the beam’s
maximum deflection. EI is constant.
P
A
B
x1
L
Solution
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(b) and (c).
Slope and Elastic Curve:
EI
For M(x1) = -
d 2y
= M(x)
dx 2
P
x,
2 1
d 2y1
P
x
2 1
dy1
P
EI
= - x 21 + C1
(1)
dx1
4
P 2
EIy1 =
x + C1x1 + C2(2)
12 1
EI
dx 21
= -
For M(x2) = -Px2,
d 2y2
= - Px1
dx 22
dy2
P
EI
= - x 22 + C3
dx2
2
P 2
EIy2 = - x 2 + C3x2 + C4
6
EI
(3)
(4)
Boundary Conditions:
y1 = 0 at x1 = 0. From Eq. [2], C2 = 0
y1 = 0 at x1 = L. From Eq. [2],
0 = y2 = 0 at x2 =
PL3
+ C1L
12
C1 =
PL2
12
L
. From Eq. [4],
2
PL3
L
0 = +
C + C4 48
2 3
(5)
Continuity Conditions:
At x1 = L and x2 =
-
L
,
2
dy1
dy2
= . From Eqs. [1] and [3],
dx1
dx2
PL2
PL2
PL2
+
= -a+ C3 b
4
12
8
C3 =
7PL2
24
323
x2
L
—
2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–7.
(Continued)
From Eq. [5], C4 = -
PL3
8
The Slope: Substitute the value of C1 into Eq. [1].
dy1
P
=
= (L2 - 3x 21)
dx1
12EI
dy1
P
= 0 =
(L2 - 3x 21)
dx1
12EI
x1 =
L
13
The Elastic Curve: Substitute the values of C1, C2, C3, and C4
into Eqs. [2] and [4], respectively.
Px1
y1 =
( - x 21 + L2)
Ans.
12EI
yD = y1 †
y2 =
=
x
x1 = 23
2
b
L2
0.0321PL3
13
a+ L2 b =
12EI
3
EI
Pa
P
( - 4x 32 + 7L2x2 - 3L3)
24EI
yC = y2 `
x2 = 0
= -
Ans.
PL3
8EI
Hence,
y max = yC =
PL3
T
8EI
Ans.
Ans.
Px1
( -x 21 + L2)
12EI
P
y2 =
( -4x 32 + 7L2x2 - 3L3)
24EI
PL3
y max =
T
8EI
y1 =
324
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–8. Determine the equations of the elastic curve using the
coordinates x1 and x2 and specify the slope at C and
displacement at B. EI is constant.
w
C
A
B
a
x1
x2
x3
Solution
Support Reactions and Elastic Curve: As shown in FBD(a).
Moment Function: As shown in FBD(b) and (c).
Slope and Elastic Curve:
EI
d 2y
= M(x)
dx 2
3wa 2
,
2
For M(x1) = wax1 EI
d 2y1
dx 21
EI
= wax1 -
dy1
wa 2 3wa 2
=
x x1 + C1
dx1
2 1
2
EIy1 =
For M(x2) = EI
3wa 2
2
(1)
wa 3 3wa 2 2
x x 1 + C1x1 + C2
6 1
4
(2)
w 2
x,
2 2
d 2y2
dx 22
= -
w 2
x
2 2
dy2
w
= - x 32 + C3
dx2
6
w
EIy2 = - x 42 + C3x2 + C4
24
(3)
EI
(4)
Boundary Conditions:
dy1
= 0 at x1 = 0. From Eq. [1], C1 = 0
dx1
y1 = 0 at x1 = 0. From Eq. [2], C2 = 0
Continuity Conditions:
At x1 = a and x2 = a,
dy1
dy2
= . From Eqs. [1] and [3],
dx1
dx2
wa 3
3wa 3
wa 3
= -(+ C3)
2
2
6
C3 =
At x1 = a and x2 = a,
From Eqs. [2] and [4],
4
4
y1 = y2.
7wa 3
6
4
wa
3wa
wa
5wa 4
41wa 4
= +
+ C4 C4 = 6
4
24
6
8
325
a
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–8.
(Continued)
The Slope: Substituting into Eq. [1],
dy1
wax1
=
(x - 3a)
dx1
2EI 1
uC =
dy2
wa 2
`
= dx2 x1 = a
EI
Ans.
The Elastic Curve: Substituting the values of C1, C2, C3, and C4
into Eqs. [2] and [4], respectively,
wax1
(2x 21 - 9ax1)
12EI
w
( - x 42 + 28a 3x2 - 41a 4)
y2 =
24EI
y1 =
yB = y2 `
x2 = 0
= -
41wa 4
24EI
Ans.
Ans.
Ans.
Ans.
wa 2
EI
wax1
y1 =
(2x 21 - 9ax1)
12EI
w
y2 =
( -x 42 + 28a 3x2 - 41a 4)
24EI
41wa 4
yB = 24EI
uC = -
326
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–9. Determine the equations of the elastic curve using the
coordinates x1 and x3 and specify the slope at B and deflection
at C. EI is constant.
w
C
A
B
a
x1
x2
x3
Solution
Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Functions: As shown on FBD(b) and (c).
Slope and Elastic Curve:
EI
d 2y
= M(x)
dx 2
3wa 2
,
2
d 2y1
3wa 2
EI 2 = wax1 2
dx 1
dy1
wa 2
3wa 2
=
x x1 + C1
EI
dx1
2 1
2
wa 3
3wa 2 2
x1 x 1 + C1x1 + C2
EIy1 =
6
4
w 2
x - 2wa 2,
For M(x3) = 2wax3 2 3
d 2y3
w 2
x - 2wa 2
EI 2 = 2wax3 2 3
dx 3
dy3
w 3
EI
= wax 23 x - 2wa 2x3 + C3
dx3
6 3
wa 3
w 4
x x - wa 2x 23 + C3x3 + C4
Ely3 =
3 3
24 3
For M(x1) = wax1 -
(1)
(2)
(3)
(4)
Boundary Conditions:
dy1
= 0 at x1 = 0. From Eq. [1], C1 = 0
dx1
y1 = 0 at x1 = 0. From Eq. [2], C2 = 0
Continuity Conditions:
dy3
dy1
=
. From Eqs. [1] and [3],
dx1
dx3
wa 3
3wa 3
wa 3
wa 3
= wa 3 - 2wa 3 + C3
C3 =
2
2
6
6
At x1 = a and x3 = a,
At x1 = a and x3 = a,
y1 = y3. From Eqs. [2] and [4],
wa 4
3wa 4
wa 4
wa 4
wa 4
wa 4
=
- wa 4 +
+ C4 C4 = 6
4
3
24
6
24
327
a
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–9.
(Continued)
The Slope: Substituting the value of C3 into Eq. [3],
dy3
w
=
(6ax 23 - x 33 - 12a 2x3 + a 3)
dx3
2EI
dy3
7wa 3
uB =
`
= dx3 x3 = 2a
6EI
The Elastic Curve: Substituting the values of C1, C2, C3, and C4
into Eqs. [2] and [4], respectively,
wax1
y1 =
(2x 21 - 9ax1)
12EI
7wa 4
yC = y1 `
= 12EI
x1 = a
w
y3 =
( - x 43 + 8ax 33 - 24a 2x 23 + 4a 3x3 - a 4)
24EI
Ans.
Ans.
Ans.
Ans.
Ans.
7wa 3
;
6EI
wax1
y1 =
(2x 21 - 9ax1);
12EI
7wa 4
yC = 12EI
w
y3 =
( -x 43 + 8ax 33 - 24a 2x 33 + 4a 3x3 - a 4)
24EI
uB = -
328
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–10. Determine the slope at B and the maximum
displacement of the beam. Use the moment-area theorems.
Take E = 200 GPa, I = 550(106) mm4.
30 kN
A
C
B
2m
1m
Solution
M
diagram and the elastic curve shown in Fig. a
EI
and b, respectively, Theorems 1 and 2 give
Using the
1 - 60 kN # m
60 kN # m2
60 kN # m2
b(2m) = a
=
2
EI
EI
EI
1 - 60 kN # m
2
b(2m) d c 1m + (2m) d
tC>A = c a
2
EI
3
140 kN # m3
140 kN # m3
= =
T
EI
EI
uB>A =
Here,
uB = uB>A =
60(103) N # m2
60 kN # m2
=
= 0.545(10 - 3) rad
EI
[200(109) N>m2][550(10 -6) m4]
∆ max = ∆ C = tC>A =
Ans.
140(103) N # m3
140 kN # m3
=
9
EI
[200(10 ) N>m2][550(10 - 6) m4]
= 1.2727(10 - 3) m
Ans.
= 1.27 mmT
Ans.
uB = 0.545(10 - 3) rad
∆ max = ∆ C = 1.27 mmT
329
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–11.
30 kN
Solve Prob. 7–10 using the conjugate-beam method.
A
C
B
2m
1m
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
+ c ΣFy = 0;
1 60 kN # m
b(2m) = 0
a
2
EI
60 kN # m2
uB = V′B = EI
- VB′ -
=
60 kN # m2
EI
60(103) N # m2
=
9
Referring to Fig d,
a + ΣMC = 0; M′C + c
2
c 200(10 ) N>m d c 550(10
= 0.545(10 - 3) rad
-6
4
)m d
Ans.
1 60 kN # m
2
b(2 m) d c 1 m + (2 m) d = 0
a
2
EI
3
∆ max = ∆ C = M′C = =
=
140 kN # m3
EI
140 kN # m3
T
EI
140(103) N # m3
c 200(109) N>m2 d c 550(10 - 6) m4 d
= 1.2727(10-3) m = 1.27 mmT
Ans.
Ans.
uB = 0.545(10 - 3) rad
∆ max = ∆ C = 1.27mmT
330
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–12. Use the moment-area theorems and determine the
slope at A and displacement at C. EI is constant.
6 kN
1.5 m
A
B
C
3m
3m
Solution
M
diagram and the elastic curve shown in
EI
Figs. a and b, respectively, theorems 1 and 2 give
Referring to the
tB>A = c
1 6.75 kN # m
1
b(1.5 m) d c 4.5 m + (1.5 m) d +
a
2
EI
3
c
=
70.875 kN # m3
EI
tC>A = c
1 6.75 kN # m
1
b(1.5 m) d c 1.5 m + (1.5 m) d +
a
2
EI
3
c
=
Then,
∆′ =
uA =
1 6.75 kN # m
2
b(4.5 m) d c (4.5 m) d
a
2
EI
3
1 2.25 kN # m
2
4.5 kN # m
1
b(1.5 m) d c (1.5 m) d + c a
b(1.5 m) d c (1.5 m) d
a
2
EI
3
EI
2
16.875 kN # m3
EI
3
1 70.875 kN # m3
35.4375 kN # m3
b =
tB>A = a
6
2
EI
EI
tB>A
LAB
=
70.875 kN # m3 >EI
∆ C = ∆ ′ - tC>A =
=
6m
=
11.8 kN # m2
EI
Ans.
35.4375 kN # m3
16.875 kN # m3
EI
EI
18.6 kN # m3
T
EI
Ans.
Ans.
uA =
∆C =
331
11.8 kN # m2
EI
18.6 kN # m3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–13.
Solve Prob. 7–12 using the conjugate-beam method.
6 kN
1.5 m
A
B
C
3m
3m
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to the FBD of the right segment of the
conjugate beam sectioned through C, Fig. c,
a + ΣMC = 0;
c
1 4.5 kN # m
1
8.4375 kN # m
a
b(3 m) d c (3 m) d - a
b(3 m) - M′C = 0
2
EI
3
EI
∆ C = M′C = Also,
uA = V′A = -
18.5625 kN # m3
18.6 kN # m3
=
T
EI
EI
11.8125 kN # m2
11.8 kN # m2
=
EI
EI
Ans.
Ans.
Ans.
∆C =
uA =
332
18.6 kN # m3
T
EI
11.8 kN # m2
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–14. Determine the value of a so that the slope at A is equal
to zero. EI is constant. Use the moment-area theorems.
P
P
C
A
B
D
a
L
__
2
L
__
2
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uA>B =
=
1 PL
1
Pa
b(L) + a b(a + L)
a
2 4EI
2
EI
PL2
Pa 2
PaL
8EI
2EI
2EI
tD>B = c
=
1 PL
L
1
Pa
L
b(L) d a b + c a - b(L) d a b
a
2 4EI
2
2
EI
3
PL3
PaL2
16EI
6EI
Then
uB =
tD>B
L
=
PL2
PaL
16EI
6EI
Here, it is required that
uB = uA>B
2
PL
PaL
PL2
Pa 2
PaL
=
16EI
6EI
8EI
2EI
2EI
24a 2 + 16La - 3L2 = 0
Choose the positive root.
a = 0.153 L
Ans.
Ans.
a = 0.153 L
333
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–15.
Solve Prob. 7–14 using the conjugate-beam method.
P
P
C
A
B
D
a
L
__
2
L
__
2
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. d,
a + ΣMB = 0; D′y (L) + c
D′y =
PL2
PaL
16EI
3EI
1 Pa
2
1 PL
1
a b(L) d a L b - c a
b(L) d a b = 0
2 EI
3
2 4EI
2
It is required that V′A = uA = 0. Referring to Fig. c,
+ c ΣFy = 0;
PL2
PaL
Pa 2
= 0
16EI
3EI
2EI
24a 2 + 16La - 3L2 = 0
Choose the positive root.
a = 0.153 L
Ans.
Ans.
a = 0.153 L
334
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–16. Determine the value of a so that the displacement at
C is equal to zero. EI is constant. Use the moment-area
theorems.
P
P
C
A
B
D
a
L
__
2
L
__
2
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorem 2 gives
Using the
tD>B = c
1 PL
L
1
Pa
L
b(L) d a b + c a - b(L) d a b
a
2 4EI
2
2
EI
3
TC>B = c
L
1 L
L
1 L
1 PL
1
Pa
ba bdc a bd + c aba bdc a bd
a
2 4EI
2
3 2
2
2EI
2
3 2
=
=
PL3
PaL2
16EI
6EI
PL3
PaL2
96EI
48EI
It is required that
tC>B =
1
t
2 D>B
PL3
PaL2
1 PL3
PaL2
= c
d
96EI
48EI
2 16EI
6EI
a =
L
3
Ans.
Ans.
a =
335
L
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–17.
P
Solve Prob. 7–16 using the conjugate-beam method.
P
C
A
B
D
a
L
__
2
L
__
2
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + ΣMD = 0;
1 PL
L
1 Pa
L
b(L) d a b - c a b(L) d a b - By′ (L) = 0
a
2 4EI
2
2 EI
3
c
- By′ =
PL2
PaL
16EI
6EI
Here, it is required that M′C = ∆ C = 0. Referring to Fig. d,
a + ΣMC = 0;
c
L
1 L
1 PL
ba bdc a bd
a
2 4EI
2
3 2
-c
L
1 L
1 Pa
ba bdc a bd
a
2 2EI
2
3 2
-c
PaL
PL2
L
da b = 0
16EI
6EI
2
PL3
PaL2
PL3
PaL2
+
= 0
96EI
48EI
32EI
12EI
L
a
L
a
+
= 0
96
48
32
12
a =
L
3
Ans.
Ans.
a =
336
L
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–18. Determine the slope at D and the displacement at the
end A of the beam. EI is constant. Use the moment-area
theorems.
24 kN
A
C
B
3m
D
4m
4m
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uC>D =
tB>C =
tA>C =
1
36 kN # m
72 kN # m2
72 kN # m2
b(4 m) = a=
A
2
EI
EI
EI
1
72 kN # m
768 kN # m3
768 kN # m3
1
ab(8 m) c (8 m) d = =
T
2
EI
3
EI
EI
1
72 kN # m
1
1
72 kN # m
2
ab(8 m) c 3 m + (8 m) d + a b(3 m) c (3 m) d
2
EI
3
2
EI
3
= -
1848 kN # m3
1848 kN # m3
=
T
EI
EI
Then
uC =
∆′ =
Therefore,
tB>C
LBC
=
768 kN # m3 >EI
8m
=
11
11 768 kN # m3
1056 kN # m3
b =
(tB>C) =
a
8
8
EI
EI
+ AuC = uD + uC>D;
96 kN # m2
72 kN # m2
= uD +
EI
EI
uD =
∆ A = tA>C - ∆ ′ =
=
96 kN # m2
A
EI
24 kN # m2
EI
Ans.
1848 kN # m3 1056 kN # m2
EI
EI
792 kN # m3
T
EI
Ans.
Ans.
uD =
∆A =
337
24 kN # m2
EI
792 kN # m3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–19.
Solve Prob. 7–18 using the conjugate-beam method.
24 kN
A
C
B
3m
D
4m
4m
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + ΣMB = 0; C′y (8 m) - c
a + ΣMC = 0;
c
C′y =
96 kN # m2
EI
192 kN # m2
1 72 kN # m
2
b(8 m) d c (8 m) d - B′y (8) = 0 B′y =
a
2
EI
3
EI
Referring to Fig. d,
a + ΣMA = 0;
1 72 kN # m
1
b(8 m) d c (8 m) d = 0
a
2
EI
3
- M′A - c
1 72 kN # m
2
192 kN # m2
b(3 m) d c (3 m) d - c
a
d (3 m) = 0
2
EI
3
EI
∆ A = M′A = -
Referring to Fig. e,
+ c ΣFy = 0; V′D +
792 kN # m3
792 kN # m3
=
T
EI
EI
96 kN # m2 1 36 kN # m
b(4 m) = 0
- a
EI
2
EI
uD = V′D = -
24 kN # m2
24 kN # m2
=
EI
EI
Ans.
Ans.
Ans.
792 kN # m3
T
EI
24 kN # m2
uD =
EI
∆A =
338
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–20. Use the moment-area theorems and determine the
displacement at C and the slope of the beam at A, B, and C. EI
is constant.
A
8 kN ? m
B
C
6m
3m
Solution
tB>A =
tC>A =
1 -8
- 48
b(6)(2) =
a
2 EI
EI
1 -8
-8
- 156
b(6)(3 + 2) + a
b(3)(1.5) =
a
2 EI
EI
EI
∆ C = 0 tC>A 0 uA =
uB>A =
0 tB>A 0
6
9(48)
9
156
84
0 tB>A 0 =
=
T
6
EI
6(EI)
EI
=
8
B
EI
Ans.
Ans.
1 -8
- 24
24
b(6) =
a
=
A
2 EI
EI
EI
uB = uB>A + uA
uB =
uC>A =
24
8
16
=
A
EI
EI
EI
Ans.
1 -8
-8
- 48
48
b(6) + a
b(3) =
a
=
A
2 EI
EI
EI
EI
uC = uC>A + uA
uC =
48
8
40
=
A
EI
EI
EI
Ans.
Ans.
84
T
EI
8
B
uA =
EI
16
uB =
A
EI
40
uC =
A
EI
∆C =
339
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–21. Use the moment-area theorems and determine the
slope at B and the displacement at C. The member is an A-36
steel structural Tee for which I = 76.8 in4.
5k
1.5 k>ft
A
B
C
3 ft
3 ft
Solution
Support Reactions and Elastic Curve: As shown.
M/EI Diagrams: The M/EI diagrams for the uniform distributed load and concentrated load are drawn separately as shown.
Moment-Area Theorems: Due to symmetry, the slope at
­midspan C is zero. Hence the slope at B is
uB = 0 uB>C 0 =
=
=
1 7.50
2 6.75
b(3) + a
b(3)
a
2 EI
3 EI
24.75 kip # ft2
EI
24.75(144)
29.0(103)(76.8)
= 0.00160 radB
Ans.
The displacement at C is
∆ C = 0 tA>C 0 =
=
=
1 7.50
2
2 6.75
5
b(3) a b(3) + a
b(3) a b(3)
a
2 EI
3
3 EI
8
47.8125 kip # ft3
EI
47.8125(1728)
29.0(103)(76.8)
= 0.0371 in.T Ans.
Ans.
uB = 0.00160 rad
∆ C = 0.0371 in. T
340
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–22. Determine the displacement and slope at C. EI is
constant. Use the moment-area theorems.
A
C
B
15 ft
60 k ? ft
9 ft
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uC>A =
1
60 k # ft
60 k # ft
990 k # ft2
990 k # ft2
b(15 ft) + a b(9 ft) = a=
A
2
EI
EI
EI
EI
tB>A = c
tC>A = c
60 k # ft
2250 k # ft3
2250 k # ft3
1
1
ab(15 ft) d c (15 ft) d = =
T
2
EI
3
EI
EI
1
60 k # ft
1
60 k # ft
1
b(15 ft) d c 9 ft + (15 ft) d + c a b(9 ft) d c (9 ft) d
a2
EI
3
EI
2
= -
8730 k # ft3
8730 k # ft3
=
T
EI
EI
Then
uA =
∆′ =
tB>A
LAB
=
2250 k # ft3 >EI
15 ft
=
150 k # ft2
B
EI
24
24 2250 k # ft3
3600 k # ft3
b =
(tB>A) =
a
15
15
EI
EI
Therefore,
+ AuC = uA + uC>A; uC = -
∆ C = tC>A - ∆ ′ =
150 k # ft2
990 k # ft2
840 k # ft2
+
=
EI
EI
EI
8730 k # ft3
3600 k # ft3
5130 k # ft3
=
T
EI
EI
EI
Ans.
Ans.
Ans.
840 k # ft2
EI
5130 k # ft3
∆C =
T
EI
uC =
341
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–23.
Solve Prob. 7–22 using the conjugate-beam method.
A
C
B
15 ft
60 k ? ft
9 ft
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + ΣMA = 0; B′y (15 ft) - c
1 60 k # ft
2
b(15 ft) d c (15 ft) d = 0
a
2
EI
3
B′y =
Referring to Fig. d,
+ c ΣFy = 0;
- V′C - a
300 k # ft2
60 k # ft
b(9 ft) = 0
EI
EI
840 k # ft2
840 k # ft2
=
EI
EI
Ans.
5130 k # ft3
5130 k # ft3
=
T
EI
EI
Ans.
uC = V′C = a + ΣMC = 0; M′C + c a
∆ C = M′C = -
300 k # ft2
EI
60 k # ft
1
300 k # ft2
b(9 ft) d c (9 ft) d + a
b(9 ft) = 0
EI
2
EI
Ans.
840 k # ft2
EI
5130 k # ft3
∆C =
T
EI
uC =
342
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*7–24. Determine the slope at B and the maximum
displacement of the beam. Use the moment-area theorems.
Take E = 200 GPa, I = 550(106) mm4.
30 kN
90 kN ? m
A
C
B
1.5 m
1.5 m
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uB>A =
1
45 kN # m
1.35 kN # m
ab(1.5 m) + a b(1.5 m)
2
EI
EI
= tC>A = c
1
45 kN # m
2
1
45 kN # m
2
b(1.5 m) d c (1.5 m) d + c a b(1.5 m) d c 1.5 m + (1.5 m) d
a2
EI
3
2
EI
3
= Here,
236.25 kN # m2
945 kN # m2
=
A
EI
4EI
+ ca-
135 kN # m
1
b(1.5 m) d c 1.5 m + (1.5 m) d
EI
2
573.75 kN # m3
2295 kN # m3
=
T
EI
4EI
+ AuB = uA + uB>A; uB = 0 +
=
945 kN # m2
4EI
945 kN # m2
4EI
945(103) N # m2
=
4c 200(109) N>m2 d c 550(10 - 6) m4 d
= 0.002148 rad = 0.00215 rad
∆ max = ∆ C = tC>A =
=
2295 kN # m
T
4EI
Ans.
3
2295(103) N # m3
4c 200(109) N>m2 d c 550(10 - 6) m4 d
= 5.216(10 - 3) m = 5.22 mmT Ans.
Ans.
uB = 0.00215 rad
∆ max = ∆ C = 5.22 mmT
343
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–25.
Solve Prob. 7–24 using the conjugate-beam method.
30 kN
90 kN ? m
A
C
B
1.5 m
1.5 m
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
+ c ΣFy = 0;
- V′B -
1 45 kN # m
135 kN # m
b(1.5 m) - a
b(1.5 m) = 0
a
2
EI
EI
uB = V′B = =
=
Referring to Fig. d,
a + ΣMC = 0; M′C + c
236.25 kN # m2
EI
945 kN # m2
4EI
945(103) N # m2
43200(10 ) N>m2 43550(10 - 6) m4 4
9
= 0.002148 rad = 0.00215 rad
Ans.
1 45 kN # m
2
135 kN # m
1
b(1.5 m) d c (1.5 m) d + c a
b(1.5 m) d c 1.5 m + (1.5 m) d
a
2
EI
3
EI
2
+ c
1 45 kN # m
2
b(1.5 m) d c 1.5 m + (1.5 m) d = 0
a
2
EI
3
∆ max = ∆ C = M′C = =
=
573.75 kN # m3
EI
2295 kN # m3
T
4EI
2295(103) N # m3
43200(109) N>m2 43550(10 - 6) m4 4
= 5.216(10 - 3) m = 5.22 mmT
Ans.
uB = 0.00215 rad
∆ max = ∆ C = 5.22 mmT
344
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–26. The beam is subjected to the load P as shown. Use the
moment-area theorems and determine the magnitude of force
F that must be applied at the end of the overhang C so that the
displacement at C is zero. EI is constant.
F
P
A
D
B
C
a
a
a
Solution
Support Reactions and Elastic Curve: As shown.
M/EI Diagram: As shown.
Moment-Area Theorems:
1 Pa
1
Fa
2
a3
b(2a)(a) + a b(2a) a ab =
a
(3P - 4F)
2 2EI
2
EI
3
6EI
1 Pa
1
Fa
2
b(2a)(a + a) + a - b(2a) aa + ab
tC>A = a
2 2EI
2
EI
3
1
Fa
2
+ ab(a) a ab
2
EI
3
3
a
=
(P - 2F)
EI
tB>A =
Require ∆ C = 0, then
∆ C = 0 = 0 tC>A 0 - `
0 =
F =
3
t
`
2 B>A
a3
3 a3
(P - 2F) - c
(3P - 4F) d
EI
2 6EI
P
4
aP - 2F =
Ans.
1
P 7 0, 3P - 4F = 2P 7 0b
2
Ans.
F =
345
P
4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–27. The beam is subjected to the load P as shown. If F = P,
determine the displacement at D. Use the moment-area
theorems. EI is constant.
F
P
A
D
B
C
a
a
a
Solution
Support Reactions and Elastic Curve: As shown.
M/EI Diagram: As shown.
Moment-Area Theorems:
a
1
Pa
Pa 3
ab(a) a b = 2
EI
3
6EI
tD>A = 0
tB>A =
The displacement at D is
∆D =
1
0 tB>A 0 - 0 tD>A 0
2
1 Pa 3
a
b - 0
2 6EI
Pa 3
=
c
12EI
=
Ans.
Ans.
∆D =
346
Pa 3
c
12EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–28. Determine the slope and the displacement at C. EI is
constant. Use the moment-area theorems.
6k
12 k ? ft
A
4 ft
C
B
6 ft
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uB>C =
1
36 k # ft
12 k # ft
180 k # ft2
180 k # ft2
b(6 ft) + a b(6 ft) = a=
A
2
EI
EI
EI
EI
tC>B = c
1
36 k # ft
2
12 k # ft
1
b(6 ft) d c (6 ft) d + c a b(6 ft) d c (6 ft) d
a2
EI
3
EI
2
= -
648 k # ft3
648 k # ft3
=
T
EI
EI
Here,
a + uB = uC + uB>C; 0 = uC + a 180 k # ft2
EI
648 k # ft3
∆ C = tC>B =
T
EI
uC =
180 k # ft2
b
EI
Ans.
Ans.
Ans.
uC =
∆C =
347
180 k # ft2
EI
648 k # ft3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–29.
Solve Prob. 7–28 using the conjugate-beam method.
6k
12 k ? ft
A
4 ft
C
B
6 ft
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
+ c ΣFy = 0;
V′C - a
uC =
a + ΣMC = 0;
12 k # ft
1 36 k # ft
b(6 ft) a
b(6 ft) = 0
EI
2
EI
180 k # ft2
EI
- M′C - c a
Ans.
12 k # ft
1
1 36 k # ft
2
b(6 ft) d c (6 ft) d - c a
b(6 ft) d c (6 ft) d = 0
EI
2
2
EI
3
∆ C = M′C = -
648 k # ft3
648 k # ft3
=
T
EI
EI
Ans.
Ans.
uC =
∆C =
348
180 k # ft2
EI
648 k # ft3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–30. Use the conjugate-beam method and determine the
displacement at D and the slope at C. Assume A is a fixed
support and C is a roller. EI is constant.
P
D
C
B
A
L
L
L
Solution
uC = VC′ = -
2PL2
2PL2
=
A
3EI
3EI
a + ΣMD′ = 0; MD′ +
∆ D = MD′ = -
Ans.
2PL2
PL2 2L
b = 0
(L) +
a
3EI
2EI 3
PL3
PL3
=
T
EI
EI
Ans.
Ans.
349
uC =
2PL2
3EI
∆D =
PL3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–31. Use the conjugate-beam method and determine the
slope at C and the displacement at B. EI is constant.
w
A
B
C
a
a
Solution
+ c ΣFy = 0;
- VC′ -
1 wa 3
wa 3
= 0
2 EI
2EI
VC′ = a + ΣMB′ = 0;
MB′ +
MB′ =
Thus,
wa 3
EI
wa 3 3
wa 3 3
wa 3 5
a ab +
a ab + a
b a ab = 0
6EI 4
2EI 2
2EI 3
- 41wa 4
24EI
uC = VC′ =
wa 3
A
EI
Ans.
∆ B = MB′ =
41a 4
T
24EI
Ans.
Ans.
wa 3
EI
41a 4
∆B =
T
24EI
uC =
350
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–32. Determine the slope at B and the displacement at C.
EI is constant. Use the moment-area theorems.
12 k
6k
A
Solution
C
B
9 ft
9 ft
9 ft
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uB>A =
1 27 k # ft
1
108 k # ft
729 k # ft2
729 k # ft2
a
b(18 ft) + ab(18 ft) = =
A
2
EI
2
EI
EI
EI
tB>A = c
108 k # ft
1 27 k # ft
1
1
a
b(18 ft) d (9 ft) + c a b(18 ft) d c (18 ft) d
2
EI
2
EI
3
= tC>A = c
3645 k # ft3
3645 k # ft3
=
T
EI
EI
1 27 k # ft
1
108 k # ft
1
b(18 ft) d (18 ft) + c a b(18 ft) d c 9 ft + (18 ft) d
a
2
EI
2
EI
3
+ c
= -
1
108 k # ft
2
ab(9 ft) d c (9 ft) d
2
EI
3
13 122 k # ft3
13 122 k # ft3
=
T
EI
EI
Then
uA =
∆′ =
Therefore,
tB>A
LAB
=
3645 k # ft3 >EI
18 ft
=
405 k # ft2
B
2EI
27
27 3645 k # ft3
10 935 k # ft3
(tB>A) =
a
b =
18
18
EI
2EI
+ AuB = uA + uB>A; uB = =
∆ C = tC>A - ∆ ′ =
405 k # ft2
729 k # ft2
+
2EI
EI
1053 k # ft2
2EI
Ans.
13 122 k # ft3
10 935 k # ft3
15 309 k # ft3 Ans.
=
T
EI
2EI
2EI
Ans.
uB =
∆C =
351
1053 k # ft2
2EI
15 309 k # ft3
T
2EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–33. Determine the slope at B and the displacement at C.
EI is constant. Use the conjugate-beam method.
12 k
6k
A
C
B
9 ft
9 ft
9 ft
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + ΣMA = 0;
By′ (18 ft) + c
-c
-1 108 k # ft
2
a
b(18 ft) d c (18 ft) d = 0
2
EI
3
B′y = a
Referring to Fig. d,
+ c ΣFy = 0;
1 27 k # ft
a
b(18 ft) d (9 ft)
2
EI
V′B +
1053 k # ft2
b
2EI
1053 k # ft2
= 0
2EI
uB = V′B = Referring to Fig. e,
a + ΣMC = 0;
MC′ + a
1053 k # ft2
1053 k # ft2
=
2EI
2EI
Ans.
1053 k # ft2
1 108 k # ft
2
b(9 ft) + c a
b(9 ft) d c (9 ft) d = 0
2EI
2
EI
3
∆ C = M′C = -
15 309 k # ft3
15 309 k # ft3
=
T
2EI
2EI
Ans.
Ans.
uB =
∆C =
352
1053 k # ft2
2EI
15 309 k # ft3
T
2EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–34. Determine the maximum displacement of the beam
and the slope at A. EI is constant. Use the moment-area
theorems.
30 kN ? m
A
B
6m
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
1 5x
60x - 5x 2
30 - 5x
a
b(x) + a
bx =
B
2 EI
EI
2EI
1 30 kN # m
2
360 kN # m3
tB>A = c a
b(6 m) d c (6 m) d =
c
2
EI
3
EI
uC>A =
Then
uA =
tB>A
LAB
=
360 kN # m3 >EI
6m
=
60 kN # m2
EI
Ans.
The maximum displacement occurs at point C where uC = 0.
+ A uC = uA + uC>A; u =
(60x - 5x ) kN # m
60 kN # m2
+ cd
EI
2EI
2
2
5x 2 - 60x + 120 = 0
x = 2.5359 m
Finally,
∆ max = tB>C = e
=
1 30 - 5(2.5359)
2
c
kN # m d (6 - 2.5359) m f c (6 - 2.5359) m d
2
EI
3
69.28 kN # m3
69.3 kN # m3
=
T
EI
EI
Ans.
Ans.
uA =
60 kN # m2
EI
∆ max =
353
69.3 kN # m3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–35.
Solve Prob. 7–34 using the conjugate-beam method.
30 kN ? m
A
B
6m
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + ΣMA = 0;
a + ΣMB = 0;
c
30 kN # m2
1 30 kN # m
a
b(6 m) d (2 m) - B′y(6 m) = 0 B′y =
2
EI
EI
A′y (6 m) - c
Referring to Fig. d,
+ c ΣFy = 0;
- V′A -
60 kN # m2
1 30 kN # m
b(6 m) d (4 m) = 0 A′y =
a
2
EI
EI
60 kN # m2
= 0
EI
uA = V′A = -
60 kN # m2
60 kN # m2
=
EI
EI
Ans.
The maximum displacement occurs at point C where uC = 0.
Referring to Fig. e,
+ c ΣFy = 0;
a + ΣMC = 0;
∆ max = M′C = -
1 5x kN # m
30 kN # m2
a
bx = 0 x = 112 m
2
EI
EI
c
1 5112 kN # m
112 m
30 kN # m2
b( 112 m) d a
b - a
b( 112 m) - M′C = 0
a
2
EI
3
EI
69.28 kN # m3
69.3 kN # m3
=
T
EI
EI
Ans.
Ans.
60 kN # m2
EI
69.3 kN # m3
∆ max =
T
EI
uA =
354
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*7–36. Determine the slope to the left and right of B and the
displacement at D. EI is constant. Use the moment-area
theorems.
60 kN ? m
B
A
3m
Solution
M
diagram and the elastic curve shown in Figs. a
EI
and b, respectively, theorems 1 and 2 give
Using the
uB>A =
1 60 kN # m
90 kN # m2
a
B
b(3 m) =
2
EI
EI
tB>A = c
tC>B = c
uD>B =
1 60 kN # m
2
180 kN # m3
b(3 m) d c (3 m) d =
a
c
2
EI
3
EI
1
60 kN # m
1
90 kN # m3
90 kN # m3
b(3 m) d c (3 m) d = a=
T
2
EI
3
EI
EI
1
60 kN # m
60 kN # m
270 kN # m2
270 kN # m2
b(3 m) + a b(3 m) = a=
A
2
EI
EI
EI
EI
tD>B = c
1
60 kN # m
1
60 kN # m
1
b(3 m) d c 3 m + (3 m) d + c a b(3 m) d c (3 m) d
a2
EI
3
EI
2
= -
630 kN # m3
630 kN # m3
=
T
EI
EI
Then
a + (uB)L = uA + uB>A; (uB)L = 0 +
∆ B = tB>A =
Ans.
180 kN # m3
c
EI
∆ ′′ = ∆ B - tC>B =
∆′ =
90 kN # m2
90 kN # m2
=
EI
EI
180 kN # m3
90 kN # m3
90 kN # m3
=
EI
EI
EI
6
6 90 kN # m3
180 kN # m3
b =
∆ ′′ = a
3
3
EI
EI
355
C
3m
D
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–36.
(Continued)
180 kN # m >EI
∆′
30 kN # m2
=
=
LBD
6m
EI
3
(uB)R =
Ans.
∆ D = tD>B + ∆ ′ - ∆ B
630 kN # m3
180 kN # m3 180 kN # m3
+
EI
EI
EI
630 kN # m3
=
T
EI
=
Ans.
Ans.
(uB)L =
(uB)R =
∆D =
356
90 kN # m2
EI
30 kN # m2
EI
630 kN # m3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–37. Determine the displacement at D and the slope at
D. Assume A is a fixed support, B is a pin, and C is a roller.
Use the conjugate-beam method.
6k
B
A
12 ft
D
C
12 ft
12 ft
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + a MB′ = 0;
Cy′(12 ft) - c
= 0
C′y =
576 k # ft2
EI
Referring to Fig. d,
+ c a Fy = 0;
1 72 k # ft
a
b(12 ft) d (16 ft)
2
EI
- VD ′ -
uD = VD ′ = a + ΣMD ′ = 0; MD ′ + c
1 72 k # ft
576 k # ft2
a
b(12 ft) = 0
2
EI
EI
1008 k # ft2
1008 k # ft2
=
A
EI
EI
Ans.
1 72 k # ft
576 k # ft2
a
b(12 ft) d (8 ft) + a
b(12 ft) = 0
2
EI
EI
∆ D = MD ′ = -
10 368 k # ft3
10 368 k # ft3
=
T
EI
EI
Ans.
Ans.
uD =
∆D =
357
1008 k # ft2
EI
10 368 k # ft3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–38. Determine the displacement at C and the slope at D.
Assume A is a fixed support, B is a pin, and D is a roller. Use
the conjugate-beam method.
10 k
D
B
A
10 ft
10 ft
C
10 ft
Solution
The real beam and conjugate beam are shown in Figs. a and b,
respectively. Referring to Fig. c,
a + ΣMB = 0;
c
1 50 k # ft
1 50 k # ft
2
a
b(20 ft) d (10 ft) + c a
b(10 ft) d c (10 ft) d - Dy ′(20 ft) = 0
2
EI
2
EI
3
Dy ′ =
Referring to Fig. d,
+ c ΣFy = 0;
V′D -
Referring to Fig. e,
a + ΣMC = 0;
∆ C = MC′ = -
c
1000 k # ft2
3EI
1000 k # ft2
1000 k # ft2
= 0 uD = V′D =
3EI
3EI
Ans.
1 50 k # ft
1
1000 k # ft2
a
b(10 ft) d c (10 ft) d - a
b(10 ft) - MC ′ = 0
2
EI
3
3EI
2500 k # ft3
2500 k # ft3
=
T
EI
EI
Ans.
Ans.
uD =
∆C =
358
1000 k # ft2
3EI
2500 k # ft3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–1. Determine the vertical displacement of joint A. Assume
the members are pin connected at their end points. Take
A = 3 in2 and E = 29(103) ksi for each member. Use the
method of virtual work.
6k
C
D
8 ft
A
6 ft
6 ft
B
3k
Solution
The virtual forces and real forces in each member are shown
in Figs. a and b respectively.
Member
n (k)
N (k)
L (ft)
nNL (k2 # ft)
AB
- 0.75
- 2.25
12
20.25
AD
1.25
3.75
10
46.875
BD
- 1.25
- 11.25
10
140.625
CD
1.50
9.00
6
81.00
Σ 288.75
1 k # (∆ A)y = a
(∆ A)y =
=
nNL
288.75 k2 # ft
=
AE
AE
288.75 k # ft
AE
(288.75 k # ft) a
12 in.
b
1 ft
(3 in2)[29(103) k>in2]
= 0.03983 in. = 0.0398 in.T
Ans.
Ans.
(∆ A)y = 0.0398 in.T
359
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–2.
6k
Solve Prob. 8–1 using Castigliano’s theorem.
C
D
8 ft
A
6 ft
6 ft
B
3k
Solution
The forces in terms of P in each member are shown in Fig. a.
AB
- 0.75P
0N
0P
- 0.75
- 2.25
12
0N
bL (k # ft)
0P
20.25
AD
1.25P
1.25
3.75
10
46.875
BD
- (1.25P + 7.50)
- 1.25
- 11.25
10
140.625
CD
1.50P + 4.50
1.50
9.00
6
Member
N (k)
N(P = 3 k)
L (ft)
Na
81.00
Σ 288.75
(∆ A)y = a N a
=
=
0N L
b
0P AE
288.75 k # ft
AE
288.75 k # ft a
12 in.
b
1 ft
(3 in2)[29(103) k>in2]
= 0.03983 in. = 0.0398 in. T Ans.
Ans.
(∆ A)y = 0.0398 in.T
360
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–3. Determine the vertical displacement of joint B. For
each member A = 400 mm2, E = 200 GPa. Use the method
of virtual work.
F
E
D
1.5 m
A
2m
B
45 kN
C
2m
Solution
Member
n
AF
0
AE
AB
N
L
nNL
0
1.5
0
- 0.8333
- 37.5
2.5
78.125
0.6667
30.0
2.0
40.00
EF
0
0
2.0
0
EB
0.50
22.5
1.5
16.875
ED
- 0.6667
- 30.0
2.0
40.00
BC
0
BD
0.8333
CD
- 0.5
0
2.0
0
37.5
2.5
78.125
- 22.5
1.5
16.875
Σ = 270
1 # ∆ By = a
∆ By =
nNL
AE
270(103)
400(10 - 6)(200)(109)
= 3.375(10 - 3) m = 3.38 mmT
Ans.
Ans.
∆ By = 3.38 mmT
361
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–4.
Solve Prob. 8–3 using Castigliano’s theorem.
F
E
D
1.5 m
A
2m
B
45 kN
C
2m
Solution
Member
0N
0P
N
N(P = 45)
L
0
1.5
Na
0N
bL
0P
AF
0
0
0
AE
- 0.8333P
- 0.8333
- 37.5
2.5
78.125
AB
0.6667P
0.6667
30.0
2.0
40.00
BE
0.5P
0.5
22.5
1.5
16.875
BD
0.8333P
0.8333
37.5
2.5
78.125
BC
0
0
0
2.0
0
CD
- 0.5P
- 0.5
- 22.5
1.5
16.875
DE
- 0.6667P
- 0.6667
- 30.0
2.0
40.00
EF
0
0
0
2.0
0
Σ = 270
∆ By = a N a
=
0N L
270
b
=
0P AE
AE
270(103)
400(10 - 6)(200)(109)
= 3.375(10 - 3) m = 3.38 mmT Ans.
Ans.
∆ By = 3.38 mmT
362
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–5. Determine the vertical displacement of joint E. For
each member A = 400 mm2, E = 200 GPa. Use the method
of virtual work.
F
E
D
1.5 m
A
2m
B
45 kN
C
2m
Solution
Member
n
N
L
nNL
0
1.5
0
AF
0
AE
- 0.8333
- 37.5
2.5
78.125
AB
0.6667
30.0
2.0
40.00
EF
0
0
2.0
0
EB
- 0.5
22.5
1.5
-16.875
ED
- 0.6667
- 30.0
2.0
40.00
BC
0
0
2.0
0
BD
0.8333
37.5
2.5
78.125
CD
- 0.5
- 22.5
1.5
16.875
Σ = 236.25
1 # ∆ Ey = a
∆ Ey =
nNL
AE
236.25(103)
400(10 - 6)(200)(109)
= 2.95(10 - 3) m = 2.95 mmT
Ans.
Ans.
∆ Ey = 2.95 mmT
363
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–6.
F
Solve Prob. 8–5 using Castigliano’s theorem.
E
D
1.5 m
A
B
45 kN
2m
C
2m
Solution
Member
0N
0P
N
AF
0
AE
- (0.8333P + 37.5)
AB
0.6667P + 30
BE
22.5 - 0.5P
BD
0.8333P + 37.5
BC
0
CD
- (0.5P + 22.5)
DE
- (0.6667P + 30)
EF
0
0
- 0.8333
N(P = 45)
L
0
1.5
-37.5
2.5
Na
0N
bL
0P
0
78.125
0.6667
30.0
2.0
40.00
- 0.5
22.5
1.5
-16.875
0.8333
37.5
2.5
78.125
0
0
2.0
0
- 0.5
-22.5
1.5
16.875
- 0.6667
-30.0
2.0
40.00
0
2.0
0
0
Σ = 236.25
∆ Ey = a Na
=
0N L
236.25
b
=
0P AE
AE
236.25(103)
400(10 - 6)(200)(109)
= 2.95(10 - 3) m = 2.95 mmT
Ans.
Ans.
∆ Ey = 2.95 mmT
364
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–7. Determine the vertical displacement of joint A. Each
bar is made of steel and has the cross-sectional area shown.
Take E = 29(103) ksi. Use the method of virtual work.
2 in2
E
2 in2
D
2 in2
3 in2 8 ft
2
3 in
B
A
6k
3 in2
3 in2
8 ft
8 ft
C
3k
Solution
The virtual forces and real forces in each member are shown
in Figs. a and b, respectively.
Member
n (k)
N (k)
L (ft)
nNL (k2 # ft)
AB
- 1.00
- 6.00
8
48
AE
612
812
9612
BC
12
- 1.00
- 6.00
8
48
BE
- 2.00
- 12.0
8
192
CE
12
612
812
9612
CD
- 1.00
- 3.00
8
24
DE
0
0
8
0
1 k # (∆ A)y = a
1 k # (∆ A)y =
nNL
AE
(48 + 48 + 192 + 24) k2 # ft
(3 in2)[29(103) k>in2]
(∆ A)y = (0.008268 ft) a
+
(9612 + 9612) k2 # ft
(2 in2)[29(103) k>in2]
12 in.
b = 0.0992 in.T 1 ft
Ans.
Ans.
(∆ A)y = 0.0992 in.T
365
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–8.
Solve Prob. 8–7 using Castigliano’s theorem.
2 in2
E
2 in2
D
2 in2
3 in2 8 ft
3 in2
B
A
2
3 in2
8 ft
8 ft
3 in
C
3k
6k
Solution
The forces in terms of P in each member are shown in Fig. a.
N (k)
Member
0N
0P
N(P = 6 k)
L (ft)
Na
0N
bL (k # ft)
0P
AB
-P
-1
-6
8
AE
12P
812
9612
-P
12
612
BC
-1
-6
8
48
BE
- 2P
-2
- 12
8
192
CE
48
12P
812
9612
- (P - 3)
12
612
CD
-1
-3
8
24
DE
0
0
0
8
0
(∆ A)y = a Na
=
0N L
b
0P AE
(48 + 48 + 192 + 24) k # ft
2
3
2
(3 in )[29(10 ) k>in ]
= (0.008268 ft) a
+
(9612 + 9612) k # ft
(2 in2)[29(103) k>in2]
12 in.
b = 0.0992 in.T 1 ft
Ans.
Ans.
(∆ A)y = 0.0992 in.T
366
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–9. Use the method of virtual work and determine the
vertical displacement of joint H. Each steel member has a
cross-sectional area of 4.5 in2. Take E = 29(103) ksi.
I
J
H
G
F
9 ft
A
B
12 ft
C
12 ft
6k
E
D
12 ft
8k
12 ft
6k
Solution
Member
N
AJ
0
AI
- 16.67
AB
13.33
n
0
L
nNL
108
0
- 0.3333
180
2500
0.6667
144
1230
BI
10.00
0.500
108
540
BH
- 6.67
- 0.8333
180
1000
BC
18.67
1.333
144
3584
CH
0.03
0
108
0
CD
18.67
1.333
144
3584
DH
- 6.67
- 0.8333
180
1000
DG
10.00
0.500
108
540
DE
13.33
0.6667
144
1280
EG
- 16.67
- 0.8333
180
2500
EF
0
0
108
0
FG
0
0
144
0
GH
- 13.33
- 0.6667
144
1280
HI
- 13.33
- 0.6667
144
1280
IJ
0
144
0
0
Σ 20 368
1 # (∆ H)y = a
(∆ H)y =
nNL
AE
20 368
= 0.156 in.
4.5(29(103))
Ans.
Ans.
(∆ H)v = 0.156 in.
367
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8–10.
Solve Prob. 8–9 using Castigliano’s theorem.
I
J
H
G
F
9 ft
A
B
12 ft
C
12 ft
6k
E
D
12 ft
8k
12 ft
6k
Solution
Member Forces N: Member forces due to external forces P
and external applied forces are shown on the figure.
Castigliano’s Second Theorem: Applying Eq. 9–27, we have
Member
N
0N
0P
N(P = 0)
L
Na
0N
bL
0P
AB
0.6667P + 13.33
0.6667
13.33
144
DE
0.6667P + 13.33
0.6667
13.33
144
1280.00
BC
1.333P + 18.67
1.333
18.67
144
3584.00
CD
1.333P + 18.67
1.333
18.67
144
3584.00
AJ
0
0
0
108
0
EF
0
0
0
108
0
1280.00
IJ
0
0
0
144
0
FG
0
0
0
144
0
HI
- (0.6667P + 13.33)
- 0.6667
-13.33
144
1280.00
GH
- (0.6667P + 13.33)
- 0.6667
-13.33
144
1280.00
AI
- (0.8333P + 16.67)
- 0.8333
-16.67
180
2500.00
EG
- (0.8333P + 16.67)
- 0.8333
-16.67
180
2500.00
BI
0.500P + 10.0
0.500
10.0
108
540.00
DG
0.500P + 10.0
0.500
10.0
108
540.00
BH
- (0.8333P + 6.667)
- 0.8333
-6.667
180
1000.00
DH
- (0.8333P + 6.667)
- 0.8333
-6.667
180
1000.00
CH
8.00
8.00
108
0
0
Σ 20 368 kip # in.
∆ = a Na
(∆ H)y =
=
0N L
b
0P AE
20 368 kip # in.
AE
20 368
= 0.156 in.T
4.5 [29.0(103)]
Ans.
Ans.
(∆ H)v = 0.156 in.
368
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–11. Determine the horizontal displacement of joint A
of the truss. Each member has a cross-sectional area of
A = 300 mm2, E = 200 GPa. Use the method of virtual work.
A
30 kN
3m
B
C
60 kN
3m
E
D
4m
Solution
The virtual forces and real forces in each member are shown
in Figs. a and b, respectively.
Member
n (kN)
N (kN)
L (m)
nNL (kN2 # m)
AB
- 1.25
- 37.5
5
234.375
AC
0.75
22.5
3
50.625
BC
1.00
30.0
4
120.00
BD
- 0.75
- 22.5
3
50.625
CD
- 1.25
- 112.5
5
703.125
CE
1.50
90.0
3
405.00
Σ 1563.75
1 kN # (∆ A)y = a
(∆ A)h =
nNL
1563.75 kN2 # m
=
AE
AE
1563.75(103) N # m
1563.75 kN # m
=
AE
[0.3(10 - 3) m2][200(109) N>m2]
= 0.02606 m = 26.1 mm d
Ans.
Ans.
(∆ A)h = 26.1 mm d
369
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*8–12.
A
Solve Prob. 8–11 using Castigliano’s theorem.
30 kN
3m
B
C
60 kN
3m
E
D
4m
Solution
The forces in each of the member in terms of P are shown
in Fig. a.
0N
0P
N(P = 30 kN)
L (m)
AB
- 1.25P
- 1.25
- 37.5
5
0N
bL (kN # m)
0P
AC
0.75P
0.75
22.5
3
50.625
Member
N (kN)
Na
234.375
BC
P
1
30
4
120.00
BD
- 0.75P
- 0.75
- 22.5
3
50.625
703.125
CD
- (1.25P + 75)
- 1.25
- 112.5
5
CE
1.5P + 45
1.5
90
3
405.00
Σ 1563.75
0N L
(∆ A)h = a Na
b
0P AE
=
=
1563.75 kN # m
AE
1563.75(103) N # m
[0.3(10 - 3) m2][200(109) N>m2]
= (0.02606 m) a
1000 mm
b = 26.1 mm d 1m
Ans.
Ans.
(∆ A)h = 26.1 mm d
370
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8–13. Determine the vertical displacement of point A. Assume
the members are pin connected at their ends. Take
A = 100 mm2 and E = 200 GPa for each member. Use the
method of virtual work.
E
F
3m
B
A
C
4m
4m
20 kN
40 kN
20 kN
Solution
1 # ∆ Ay =
(33.33)(1.667)(5)
(26.67)(1.333)(4)
( - 20)( -1)(3)
ΣNnL
=
+
+
AE
AE
AE
AE
+
∆ Ay =
(100)(1.667)(5)
AE
+
( -26.67)( -1.333)(4)
AE
+
( - 106.67)( -2.667)(4)
AE
2593.33(103) N # m
2593.33
=
= 0.130 m = 130 mmT AE
100(10 - 6) m2(200(109) N>m2)
+
(20)(0)(3)
AE
Ans.
Ans.
∆ Ay = 130 mmT
371
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8–14.
Solve Prob. 8–13 using Castigliano’s theorem.
E
F
3m
B
A
C
4m
4m
20 kN
40 kN
20 kN
Solution
∆ Ay =
ΣN(0N> 0P)L
AE
+
=
=
(33.33)(1.667)(5)
AE
(100)(1.667)(5)
AE
+
+
(26.67)(1.333)(4)
+
AE
( -26.67)( - 1.333)(4)
AE
+
( - 20)( -1)(3)
AE
( - 106.67)( -2.667)(4)
AE
2593.33(103) N # m
2593.33
=
= 0.130 m = 130 mmT AE
100(10 - 6) m2(200(109) N>m2)
+
(20)(0)(3)
AE
Ans.
Ans.
∆ Ay = 130 mmT
372
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8–15. Determine the vertical displacement of joint C. Assume
the members are pin connected at their end points. Take
A = 200 mm2 and E = 200 GPa for each member. Use the
method of virtual work.
24 kN
12 kN
A
4m
B
4m
C
4m
D
E
Solution
The virtual forces and real forces in each member are shown
in Figs. a and b, respectively.
nNL (kN2 # m)
Member
n (kN)
N (kN)
L (m)
AB
2.00
24.0
4
192
AD
0
1215
215
0
BC
2.00
24.0
4
192
BD
0
- 24.0
2
0
CD
- 15
- 1215
215
- 2415
215
DE
- 15
=
24015
Σ 384 + 36015
1 kN # (∆ C)y = a
(∆ C)y =
12015
(384 + 36015) kN # m
nNL
=
AE
AE
2
(384 + 36015) kN # m
AE
(384 + 36015)(103) N # m
[0.2(10 - 3) m2][200(109) N>m2]
= 0.02972 m = 29.7 mmT Ans.
Ans.
(∆ C)y = 29.7 mmT
373
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*8–16.
24 kN
Solve Prob. 8–15 using Castigliano’s theorem.
12 kN
4m
A
B
4m
C
4m
D
E
Solution
The forces in terms of P in each member are shown in Fig. a.
Member
N (kN)
0N
0P
N(P = 12 kN)
L (m)
AB
Na
0N
bL (kN # m)
0P
2P
2
24
4
AD
1215
0
1215
215
0
BC
2P
2
24
4
192
BD
- 24
0
- 24
2
0
CD
- 15P
- 15
- 1215
215
12015
- 2415
215
24015
DE
- 15(P + 12)
- 15
(∆ C)y = a Na
=
=
192
Σ 384 + 36015
0N L
b
0P AE
(384 + 36015) kN # m
AE
(384 + 36015) (103) N # m
[0.2(10 - 3) m2][200(109) N>m2]
Ans.
= 0.02972 m = 29.7 mmT Ans.
(∆ C)y = 29.7 mmT
374
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8–17. Determine the vertical displacement of joint C if
members AB and BC experience a temperature increase of
dT = 50°C. Take a = 12(10 - 6)>°C.
A
4m
B
4m
C
4m
D
E
Solution
The virtual forces in each of the members are shown in Fig. a.
1 kN # (∆ C)y = ΣnadTL
1 kN # (∆ C)y = (2.00 kN)[12(10 - 6)>°C](50°C)(4 m)
+ (2.00 kN)[12(10 - 6)>°C](50°C)(4 m)
(∆ C)y = 9.6(10 - 3) m = 9.60 mmT Ans.
Ans.
(∆ C)y = 9.60 mmT
375
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8–18. Determine the vertical displacement of joint C if
member CD is fabricated 10 mm too long.
A
4m
B
4m
C
4m
D
E
Solution
The virtual forces in each of the member are shown in Fig. a.
1 kN # (∆ C)y = ΣndL
1 kN # (∆ C)y = ( - 15 kN)(10 mm)
(∆ C)y = - 22.36 mm = 22.4 mm c Ans.
Ans.
(∆ C)y = 22.4 mm c
376
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8–19. Determine the displacement of point C and the slope at
point B. EI is constant. Use the principle of virtual work.
P
B
C
L
2
L
2
Solution
Real Moment Function M(x): As shown on figure (a).
Virtual Moment Functions m(x) and mu(x): As shown on
figures (b) and (c).
Virtual Work Equation: For the displacement at point C,
1#∆ =
L
mM
dx
L0 EI
1
1 # ∆ C = 2c
EI L0
∆C =
L
2
PL3
T
48EI
a
x1 P
b a x1 bdx1 d
2
2
Ans.
For the slope at B,
1#u =
L
muM
dx
L0 EI
L
2
L
2
x1 P
x2 P
a b a x1 bdx1 +
a1- b a x2 bdx2 d
EI L0 L
2
L
2
L0
1
1 # uB =
c
uB =
PL2
16EI
Ans.
Ans.
377
∆C =
PL3
T
48EI
uB =
PL2
16EI
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*8–20.
P
Solve Prob. 8–19 using Castigliano’s theorem.
B
C
L
2
L
2
Solution
Internal Moment Fuction M(x): The internal moment function
in terms of the load P′ and couple moment M′ and externally
applied load are shown on figures (a) and (b), respectively.
Castigliano’s Second Theorem: The displacement at C can be
determined with
∆ =
L0
∆ C = 2c
=
L
Ma
0M(x)
0P′
=
x
and set P′ = P.
2
0M dx
b
0P′ EI
L
2
1
P
x
a xb a bdx d
EI L0 2
2
PL3
T
48EI
Ans.
To determine the slope at B, with
0M(x2)
0M′
u =
= 1-
L0
L
Ma
L
uB =
=
0M(x1)
0M′
=
x1
,
L
x2
, and setting M′ = 0,
L
0M dx
b
0M′ EI
L
2
2
x1
x2
1
P
1
P
a x1 b a bdx1 +
a x2 b a1- bdx2
EI L0 2
L
EI L0 2
L
PL2
16EI
Ans.
Ans.
378
∆C =
PL3
T
48EI
uB =
PL2
16EI
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8–21. Determine the slope and displacement at point C. Use
the principle of virtual work. EI is constant.
12 kN>m
A
B
C
2m
3m
1m
Solution
Referring to the virtual moment functions shown in Figs. a
and b, and the real moment functions in Fig. c,
1 kN # m # uC =
L0
Lm M
u
EI
dx =
L0
2 m ( -0.1667x )(16x )dx
1
1
1
EI
+
1 kN # m # uC =
uC =
L0
3m
L0
+
1 m -0.1667(x + 2)( - 6x 2 + 16x + 32)dx
2
2
2
2
EI
(0.1667x3)( - 6x 23 + 32x3)dx3
EI
4.6667 kN2 # m3
EI
4.67 kN # m2
EI
Ans.
And
1 kN # ∆ C =
2 m (0.5x )(16x )dx
1 m 0.5(x + 2)( -6x 2 + 16x + 32)dx
mM
1
1
1
2
2
2
2
dx =
+
EI
EI
L0 EI
L0
L0
L
+
1 kN # ∆ C =
∆C =
L0
3m
( - 6x 23 + 32x3)(0.5x3)dx3
EI
152.5 kN2 # m3
EI
152.5 kN # m3
T
EI
Ans.
Ans.
uC =
∆C =
379
4.67 kN # m2
EI
152.5 kN # m3
T
EI
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8–22.
Solve Prob. 8–21 using Castigliano’s theorem.
12 kN>m
A
B
C
2m
3m
1m
Solution
For the slope, the moment functions are shown in Fig. a.
Here,
0M3
x3
0M1
x1 0M2
1
= .
= - ,
= - (x2 + 2) and
0M′
6 0M′
6
0M′
6
Also, set M′ = 0. Then M1 = 16x kN # m,
M2 = [16(x2 + 2) - 6x 22] kN # m and M3 = (32x3 - 6x 23) kN # m. Thus,
L
0M dx
b
Ma
=
uC =
0M′
EI
L0
L0
+
uC =
2 m (16x1) a -
EI
x1
bdx1
6
+
L0
1
2
(x2 + 2) d dx2
1 m [16(x2 + 2) - 6x 2] c 6
EI
x3
2
bdx3
3 m (32x3 - 6x 3) a
6
L0
EI
4.6667 kN # m2
4.67 kN # m2
=
EI
EI
Ans.
For the displacement, the moment functions are shown
0M1
x1
0M2
1
in Fig. b. Here,
=
m,
= (x2 + 2) m and
0P
2
0P
2
0M3
x3
=
m. Also, set P = 0. Then M1 = (16x1) kN # m,
0P
2
M2 = 316(x2 + 2) - 6x 22 4 kN # m and
M3 = (32x3 - 6x 23) kN # m. Thus,
2m
0M dx
∆C =
Ma
b
=
0P EI
L0
L0
L
x1
1
2
bdx1
1 m 316(x2 + 2) - 6x 2 4 c (x2 + 2) d dx2
2
2
+
EI
EI
L0
(16x1) a
+
∆C =
L0
152.5 kN # m
T
EI
3
x3
2
bdx3
3 m (32x3 - 6x 3) a
2
EI
Ans.
Ans.
uC =
∆C =
380
4.67 kN # m2
EI
152.5 kN # m3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–23. Determine the displacement at point D. Use the
principle of virtual work. EI is constant.
16 kN>m
B
A
3m
C
D
4m
4m
Solution
Referring to the virtual and real moment functions shown in
Figs. a and b, respectively,
1 kN # ∆ D =
2
3 m ( -0.5x )( - 64x )dx
4 m (0.5x )(64x - 8x )dx
mM
1
1
1
2
2
2
2
dx =
+ 2
EI
EI
L0 EI
L0
L0
L
1 kN # ∆ D =
∆D =
3424 kN2 # m3
3EI
3424 kN # m3
T
3EI
Ans.
Ans.
∆D =
381
3424 kN # m3
T
3EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–24. Determine the slope and displacement at the end A of
the beam. Take E = 29(103) ksi, I = 170 in4. Use the method
of virtual work.
10 k
2 k>ft
A
C
B
10 ft
10 ft
10 ft
Solution
Referring to the virtual moment functions shown in Figs. a and b,
respectively, and the real moment functions in Fig. c,
1 k # ft # uA =
L0
Lm M
u
EI
dx =
L0
10 ft ( -1)( - x 2)dx
1
EI
+
1 k # ft # uA =
uA =
1 k # ∆A =
L
mM
dx =
L0 EI
L0
L0
∆A =
20 000 k # ft3
3EI
=
10 ft a -
L0
1
(x + 10) d ( - 10x2)dx2
20 2
EI
x3
b(0)dx3
20
EI
750 k2 # ft3
EI
750 k # ft2
EI
(750 k # ft2) a
=
12 in. 2
b
1 ft
[29(103) k>in2](170 in4)
10 ft ( -x )( - x 2)dx
1
1
1
EI
+
1 k # ∆A =
+
10 ft c -
L0
+
L0
10 ft a -
10 ft c -
= 0.0219 rad
Ans.
1
(x + 10) d ( - 10x2)dx2
2 2
EI
x3
b(0)dx3
2
EI
20 000 k2 # ft3
3EI
(20 000 k # ft3) a
12 in. 3
b
1 ft
3[29(103) k>in2](170 in4)
= 2.34 in.T Ans.
Ans.
uA = 0.0219 rad
∆ A = 2.34 in.T
382
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–25.
Solve Prob. 8–24 using Castigliano’s theorem.
10 k
2 k>ft
A
C
B
10 ft
10 ft
10 ft
Solution
For the slope, the moment functions are shown in Fig. a. Here,
0M3
x3
0M1
0M2
1
= - . Also, set
= - 1,
= - (x2 + 10) and
0M′
0M′
20
0M′
20
M′ = 0, M1 = ( - x 21) k # ft, M2 = ( - 10x2) k # ft, and M3 = 0.
Thus,
uA =
L0
L
Ma
0M dx
b
=
0M′ EI
L0
10 ft
( -x 21)( - 1)dx1
EI
+
uA =
750 k # ft2
EI
L0
+
L0
10 ft ( - 10x2) c -
10 ft 0a -
(750 k # ft2) a
=
1
(x + 10) d dx2
20 2
EI
x3
bdx3
20
EI
12 in. 2
b
1 ft
[29(103) k>in2](170 in4)
= 0.0219 rad
Ans.
For the displacement, the moment functions are shown in
0M1
0M2
1
Fig. b. Here,
= ( - x1) ft,
= c - (x2 + 10) d ft, and
0P
0P
2
0M3
x3
= ( - ) ft. Also, set P = 0. Then M1 = ( -x 21) k # ft,
0P
2
M2 = ( -10x2) k # ft, and M3 = 0. Thus,
2
10 ft ( -x )( - x )dx
10 ft
0M dx
1
1
1
b
Ma
=
+
∆A =
0P EI
EI
L0
L0
L0
L
+
∆A =
L0
10 ft 0a -
( - 10x2) c -
1
(x + 10) d dx2
2 2
EI
x3
bdx3
2
EI
(20 000 k # ft3) a
12 in. 3
b
1 ft
20 000 k # ft
=
= 2.34 in.T
3EI
3[29(103) k>in2](170 in4)
3
Ans.
Ans.
uA = 0.0219 rad
∆ A = 2.34 in.T
383
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–26. Determine the displacement and slope at point C of the
cantilever beam. The moment of inertia of each segment is
indicated in the figure. Take E = 200 GPa. Use the principle of
virtual work.
12 kN>m
A
8 kN>m
IAB 5 600 (106) mm4
C
B IBC 5 150 (106) mm4
2m
1m
Solution
Referring to the virtual moment functions shown in Figs. a and b,
respectively, and the real moment functions in Fig. c,
1 kN # m # uC =
L0
Lm M
u
EI
dx =
1 kN # m # uC =
uC =
=
EIBC
+
L0
2 m ( - 1)( - 6x 2 - 8x - 4)dx
2
2
2
EIAB
4 kN2 # m3
40 kN2 # m3
+
3EIBC
EIAB
4 kN # m2
40 kN # m2
+
3EIBC
EIAB
4(103) N # m2
9
2
3[200(10 ) N>m ][150(10
-6
4
)m]
= 0.378(10 - 3) rad
and
1 kN # ∆ C =
L0
1 m ( -1)( - 4x 2)dx
1
1
+
40(103) N # m2
[200(10 ) N>m2][600(10 - 6) m4]
9
Ans.
1 m ( -x )( - 4x 2)dx
2 m [ -(x + 1)]( -6x 2 - 8x - 4)dx
mM
1
1
1
2
2
2
2
dx =
+
EIBC
EIAB
L0 EI
L0
L0
L
1 kN # ∆ C =
1 kN2 # m3
280 kN2 # m3
+
EIBC
3EIAB
∆C =
1 kN # m3
280 kN # m3
+
EIBC
3EIAB
=
1(103) N # m3
[200(109) N>m2][150(10 - 6) m4]
= 0.8111(10 - 3) m = 0.811 mmT +
280(103) N # m3
3[200(109) N>m2][600(10 - 6) m4]
Ans.
Ans.
uC = 0.378(10 - 3) rad
∆ C = 0.811 mmT
384
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–27.
Solve Prob. 8–26 using Castigliano’s theorem.
12 kN>m
A
8 kN>m
IAB 5 600 (106) mm4
C
B IBC 5 150 (106) mm4
2m
1m
Solution
For the slope, the moment functions are shown in Fig. a.
Here,
0M1
0M2
= - 1 and
= - 1 .Also, set M′ = 0. Then
0M′
0M′
M1 = ( -4x 21) kN # m and M2 = ( - 6x 22 - 8x2 - 4) kN # m. Thus,
uC =
L0
L
Ma
0M dx
b
=
0M′ EI
L0
uC =
=
1m
( -4x 21)( - 1)dx1
EIBC
L0
+
2m
( - 6x 22 - 8x2 - 4)( - 1)dx2
EIAB
4 kN # m2
40 kN # m2
+
3EIBC
EIAB
4(103) N # m2
9
2
3[200(10 ) N>m ][150(10
-6
4
)m]
= 0.378(10 - 3) rad
+
40(103) N # m2
[200(10 ) N>m2][600(10 - 6) m4]
9
Ans.
For the displacement, the moment functions are shown
0M1
0M2
in Fig. b. Here,
= ( - x1) m and
= - (x2 + 1).
0P
0P
2
Also, set P = 0. Then, M1 = ( - 4x 1) kN # m and
M2 = ( -6x 22 - 8x2 - 4) kN # m. Thus,
∆C =
L0
L
Ma
0M dx
b
=
0P EI
L0
=
=
1m
( -4x 21)( - x1)dx1
EIBC
L0
+
2m
( - 6x 22 - 8x2 - 4)[ -(x2 + 1)]dx2
EIAB
1 kN # m3
280 kN # m3
+
EIBC
3EIAB
1(103) N # m3
9
2
[200(10 ) N>m ][150(10
-6
4
)m]
= 0.8111(10 - 3) m = 0.811 mmT +
280(103) N # m3
3[200(109) N>m2][600(10 - 6) m4]
Ans.
Ans.
uC = 0.378(10 - 3) rad
∆ C = 0.811 mmT
385
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–28. Determine the slope at A. EI is constant.
w
C
B
A
L
L
Solution
1 # uA =
L
muM
dx
L0 EI
= 0 +
uA =
L0
La
x
- wx 2
- 1b a
b
L
2
dx
EI
wL4
wL3
+
uL
6
wL3
=
A
EI
24EI
Ans.
w
C
B
L
A
L
Ans.
uA =
386
wL3
A
24EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–29.
Solve Prob. 8–28 using Castigliano’s theorem.
w
C
B
A
L
L
Solution
M′ does not influence the moment within the overhang.
M =
M′
wx 2
x - M′ L
2
0M
x
=
- 1
0M′
L
Setting M′ = 0,
uA =
=
L0
L
Ma
wL3
A
24EI
L
0M dx
1
wx 2
x
-w L3 L3
b
b a - 1bdx =
d
=
a
c 0M′ EI
EI L0
2
L
2EI 4
3
Ans.
w
C
B
L
A
L
Ans.
uA =
387
wL3
A
24EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–30. Determine the slope and displacement at point C. EI is
constant. Assume A is a pin. Use the method of virtual work.
2 k>ft
C
A
B
12 ft
6 ft
Solution
Referring to the virtual moment functions shown in Figs. a
and b, respectively, and the real moment functions in Fig. c,
1 k # ft # uC =
L
muM
dx =
L0 EI
L0
1 k # ft # uC = uC = -
12 ft a -
x1
x 32
b(11x1 - x 21)dx1
bdx2
6 ft ( - 1) a 12
18
+
EI
EI
L0
78 k2 # ft3
EI
78 k # ft2
78 k # ft2
=
EI
EI
Ans.
and
1 k # ∆C =
L
mM
dx =
L0 EI
L0
1 k # ∆C = ∆C = -
12 ft a -
x1
x 32
b(11x1 - x 21)dx1
bdx2
6 ft ( - x2) a 2
18
+
EI
EI
L0
489.6 k2 # ft3
EI
489.6 k # ft3
490 k # ft3
=
c
EI
EI
Ans.
Ans.
uC =
∆C =
388
78 k # ft2
EI
490 k # ft3
c
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–31.
Solve Prob. 8–30 using Castigliano’s theorem.
2 k>ft
C
A
B
12 ft
6 ft
Solution
For the slope, the moment functions are shown in Fig. a.
0M1
x1
0M2
Here,
= - and
= - 1. Also, set M′ = 0. Then
0M′
12
0M′
x 32
M1 = (11x1 - x 21) k # ft and M2 = a - b k # ft. Thus,
18
uC =
L0
L
Ma
0M dx
b
=
0M′ EI
L0
uC = -
2
12 ft (11x1 - x 1) a -
EI
x1
bdx1
12
+
78 k # ft2
78 k # ft2
=
EI
EI
L0
6 ft a -
x 32
b( - 1)dx2
18
EI
Ans.
For the displacement, the moment functions are shown in Fig. b.
0M1
x1
0′M2
Here
= - and
= - x2. Also, set P = 0. Then
0P
2
0P
x 32
M1 = (11x1 - x 21) k # ft and M2 = a - b k # ft. Thus,
18
∆C =
L0
L
Ma
0M dx
b
=
0P EI
L0
∆C = -
x1
2
bdx1
12 ft 111x1 - x 1 2 a 2
EI
489.6 k # ft3
490 k # ft3
=
c
EI
EI
+
L0
6 ft a -
x 32
b( - x2)dx2
18
EI
Ans.
Ans.
uC =
∆C =
389
78 k # ft2
EI
490 k # ft3
c
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–32. Bar ABC has a rectangular cross section of 300 mm by
100 mm. Attached rod DB has a diameter of 20 mm. Determine
the vertical displacement of point C due to the loading.
Consider only the effect of bending in ABC and axial force in
DB. E = 200 GPa.
D
20 kN
300 mm
A
100 mm
3m
4m
B
3m
C
Solution
Real Moment Function M(x): As shown on figure (a).
Virtual Moment Functions m(x): As shown on figure (b).
Virtual Work Equation: For the displacement at point C,
combine Eq. 8–22 and Eq. 8–15.
1#∆ =
L
mM
nNL
dx +
EI
AE
L0
1 kN # ∆ C = 2c
∆C =
=
1
EI L0
3m
(1.00x)(20.0x)dx d +
2.50(50.0)(5)
AE
360 kN # m3
625 kN # m
+
EI
AE
360(1000)
1
200(109) c (0.1)(0.33) d
12
= 0.017947 m = 17.9 mmT +
625(1000)
x
c (0.022) d [200(109)]
4
Ans.
Ans.
∆ C = 17.9 mmT
390
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–33. Bar ABC has a rectangular cross section of 300 mm by
100 mm. Attached rod DB has a diameter of 20 mm. Determine
the slope at A due to the loading. Consider only the effect of
bending in ABC and axial force in DB. E = 200 GPa.
D
20 kN
300 mm
A
100 mm
3m
4m
B
3m
C
Solution
Real Moment Function M(x): As shown on figure (a).
Virtual Moment Functions mu(x): As shown on figure (b).
Virtual Work Equation: For the slope at point A, combine
Eq. 8–23 and Eq. 8–15.
1#u =
1 kN # m # uA =
uA =
=
L
muM
nNL
dx +
EI
AE
L0
1
EI L0
3m
(1 - 0.3333x)(20.0x)dx +
( - 0.41667)(50.0)(5)
AE
30.0 kN # m2
104.167 kN
EI
AE
30.0(1000)
1
200(109) c (0.1)(0.33) d
12
-
104.167(1000)
x
c (0.022) d [200(109)]
4
= - 0.991(10 - 3) rad = 0.991(10 - 3) radA
Ans.
Ans.
uA = 0.991(10 - 3) radA
391
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–34. Determine the slope and displacement at point B.
Assume the support at A is a pin and C is a roller. Take
E = 200 GPa, I = 150(106) mm4. Use the method of virtual
work.
16 kN>m
A
C
B
2m
4m
Solution
Referring to the virtual moment functions shown in Figs. a
and b, and the real moment functions in Fig. c,
2m
muM
1 kN # m # uB =
dx =
L0 EI
L0
L
a-
x1
x2
b(48x1 - 8x 21)dx1
b(48x2 - 8x 22)dx2
4ma
6
6
+
EI
EI
L0
1 kN # m # uB =
208 kN2 # m3
3EI
uB =
208 kN # m2
3EI
=
208(103) N # m2
3[200(109) N>m2][150(10 - 6) m4]
= 0.00231 rad
Ans.
and
L
mM
1 kN # ∆ B =
dx =
L0 EI
L0
2ma
x2
2
x b(48x1 - 8x 21)dx1
b(48x2 - 8x 22)dx2
4ma
3 1
3
+
EI
EI
L0
1 kN # ∆ B =
704 kN2 # m3
3EI
∆B =
704 kN # m3
3EI
=
704 (103) N # m3
3[200(109) N>m2][150(10 - 6) m4]
= 0.007822 m = 7.82 mmT Ans.
Ans.
uB = 0.00231 rad
∆ B = 7.82 mmT
392
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–35.
Solve Prob. 8–34 using Castigliano’s theorem.
16 kN>m
A
C
B
2m
4m
Solution
For the slope, the moment functions are shown in Fig. a.
0M1
x1
0M2
x2
Here,
= and
= . Also, set M′ = 0. Then
0M′
6
0M′
6
M1 = (48x1 - 8x 21) kN # m and M2 = (48x2 - 8x 22) kN # m.
Thus,
L
0M dx
b
uB =
Ma
=
0M′
EI
L0
L0
uB =
=
x1
2
bdx1
2 m (48x1 - 8x 1) a 6
EI
+
L0
x2
2
bdx2
4 m (48x2 - 8x 2) a
6
EI
208 kN # m2
3EI
208(103) N # m2
3[200(109) N>m2][150(10 - 6) m4]
Ans.
= 0.00231 rad
For the displacement, the moment functions are shown in Fig. b.
0M1
0M2
x2
2
Here,
= x1 and
=
. Also, set P = 0. Then
0P
3
0P
3
M1 = (48x1 - 8x 21) kN # m and M2 = (48x2 - 8x 22) kN # m.
Thus,
∆B =
L0
L
Ma
0M dx
b
=
0P EI
L0
∆B =
=
2
2
2 m (48x1 - 8x 1) a x1 bdx1
3
EI
+
L0
x2
2
bdx2
4 m (48x2 - 8x 2) a
3
EI
704 kN # m3
3EI
704(103) N # m3
3[200(109) N>m2][150(10 - 6) m4]
= 0.007822 m = 7.82 mmT
Ans.
uB = 0.00231 rad
∆ B = 7.82 mmT
393
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–36. Determine the slope and displacement at point B.
Assume the support at A is a pin and C is a roller. Account for
the additional strain energy due to shear if the cross section is
a wide flange. Take E = 200 GPa, I = 150(106) mm4,
G = 75 GPa, and assume AC has a cross-sectional area of
A = 2.50(103) mm2. Use the method of virtual work.
16 kN>m
A
2m
Solution
The virtual shear and moment functions are shown in Figs. a and b,
and the real shear and moment function are shown in Fig. c.
1 kN # m # uB =
=
L
L
muM
vuV
bdx
dx +
Ka
EI
GA
L0
L0
L0
+
=
uB =
2ma -
L0
1
1
a - b(48 - 16x1)dx1
x b(48x1 - 8x 21)dx1
2m
6 1
6
+
1D
T
EI
GA
L0
4ma
1
1
a - b(16x2 - 48)dx2
x b(48x2 - 8x 22)dx2
4m
6 2
6
+
1D
T
EI
GA
L0
208 kN2 # m3
+ 0
3EI
208(103) N # m2
208 kN # m2
=
= 0.00231 rad
3EI
3[200(109) N>m2][150(10 - 6) m4]
Ans.
And
1 kN # ∆ B =
=
L
L0
+
=
∆B =
=
L
mM
vV
bdx
dx +
Ka
EI
GA
L0
L0
2ma
L0
x
2
a b(48 - 16x1)
x b(48x1 - 8x 21)dx1
2m
3 1
3
+
1D
T dx1
EI
GA
L0
4ma
x2
1
a - b(16x2 - 48)
b(48x2 - 8x 22)dx2
4m
3
3
+
1D
T dx2
EI
GA
L0
704 kN2 # m3
64 kN2 # m
+
3EI
GA
704 kN # m3
64 kN # m
+
3EI
GA
704(103) N # m3
3[200(109) N>m2][150(10 - 6) m4]
+
64(103) N # m
[75(109) N>m2][2.50(10 - 3) m2]
= 0.008164 m = 8.16 mmT
Ans.
394
C
B
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–36
(Continued)
Ans.
uB = 0.00231 rad
∆ B = 8.16 mmT
395
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–37. Determine the displacement of point C. Use the
method of virtual work. EI is constant.
3 k>ft
A
B
C
12 ft
12 ft
Solution
Referring to the virtual and real moment functions shown in
Figs. a and b, respectively,
1 k # ∆C =
L
mM
dx = 2
L0 EI
L0
1 k # ∆C =
∆C =
12 ft a
1
x3
xb a18xb dx
2
24
EI
41 472 k2 # ft3
5EI
41 472 k # ft3
T
5EI
Ans.
Ans.
∆C =
396
41 472 k # ft3
T
5EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–38.
Solve Prob. 8–37 using Castigliano’s theorem.
3 k>ft
A
B
C
12 ft
12 ft
Solution
The moment function is shown in Fig. a. Here,
Also, set P = 0. Then M = a18x ∆C =
L0
L
Ma
0M dx
b
= 2
0P EI
L0
∆C =
0M
1
= x.
0P
2
x3
b k # ft. Thus,
24
12 ft a18x -
41 472 k # ft3
T
5EI
x3
1
b a xb dx
24 2
EI
Ans.
Ans.
∆C =
397
41 472 k # ft3
T
5EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–39. Use the method of virtual work and determine the
displacement of point C of the beam made from steel.
E = 29(103) ksi, I = 245 in4.
8k
8k
A
B
C
5 ft
5 ft
5 ft
5 ft
Solution
1 # ∆C =
L
mM
dx
L0 EI
∆C = 0 + 2
=
L0
60 a -
x
b( -480)
2
dx
EI
864 000
= 0.122 in.
29(103)(245)
Ans.
Ans.
∆ C = 0.122 in.
398
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–40. Use the method of virtual work and determine the
slope at point A of the beam made from steel.
E = 29(103) ksi, I = 245 in4.
8k
8k
A
B
C
5 ft
5 ft
5 ft
5 ft
Solution
1 # uA =
L0
L
uA = 0 +
=
m0M
dx
EI
L0
120 a
x
- 1b( -480)
120
dx
EI
28 800
= 4.05(10 - 3) rad
29(103)(245)
Ans.
Ans.
uA = 4.05(10 - 3) rad
399
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–41.
Solve Prob. 8–40 using Castigliano’s theorem.
8k
8k
A
B
C
5 ft
5 ft
5 ft
5 ft
Solution
uA =
L0
L
= 0 +
=
Ma
L0
0M dx
b
0M′ EI
120 - 480a
x
1 1202
-480 c a
- 1b
b - 120 d
120
2 120
dx = 0 +
EI
EI
28 800
= 4.05(10 - 3) rad
29(103)(245)
Ans.
Ans.
uA = 4.05(10 - 3) rad
400
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8–42. Determine the displacement at point D. Use the
principle of virtual work. EI is constant.
60 kN
30 kN>m
A
B
D
2m
2m
C
3m
Solution
The virtual and real moment functions shown in Figs. a and b,
respectively.
L
mM
1 kN # ∆ D =
dx = 2
L0 EI
L0
1 kN # ∆ D =
∆D =
2ma
1
1
2
x b(30x1)dx1
3 m a - x2 b[ -(15x 2 + 30x2)] dx2
2 1
2
+
EI
EI
L0
2935 kN2 # m3
8 EI
2935 kN # m3
T
8EI
Ans.
Ans.
∆D =
401
2935 kN # m3
T
8EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–43. Determine the displacement at point D. Use Castigliano’s
theorem. EI is constant.
60 kN
30 kN>m
A
B
D
2m
2m
C
3m
Solution
The moment functions are shown in Fig. a. Here,
0M1
0M2
1
1
= x1 and
= - x2. Also, set P = 60 kN. Then
0P
2
0P
2
M1 = (30x1) kN # m and M2 = [ - (15x2 + 30x2)] kN # m. Thus,
L
0M dx
b
Ma
= 2
∆D =
0P EI
L0
L0
∆D =
1
1
2
x bdx1
3 m [ -(15x 2 + 30x2)] a - x2 b dx2
2 1
2
+
EI
EI
L0
2 m (30x1) a
2935 kN # m3
T
8EI
Ans.
Ans.
∆D =
402
2935 kN # m3
T
8EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–44. Determine the horizontal displacement at A. Take
E = 200 GPa. The moment of inertia of each segment of the
frame is indicated in the figure. Assume D is a pin support. Use
the method of virtual work.
60 kN>m
C
B
IBC 5 300(106) mm4
8m
IAB 5 200(106) mm4
ICD 5 200(106) mm4
D
A
6m
Solution
Referring to the virtual and moment functions shown in Figs. a and b,
respectively,
L
mM
1 kN # (∆ A)h =
dx = 0 +
L0 EI
L0
6 m 8 a60x2 -
EIBC
1 kN # (∆ A)h =
4320 kN2 # m3
EIBC
(∆ A)h =
4320 kN # m3
EIBC
=
5 3
x bdx2
3 2
+ 0
4320(103) N # m3
[200(109) N>m2][300(10 - 6) m4]
= 0.0720 m = 72.0 mm d Ans.
Ans.
(∆ A)h = 72.0 mm d
403
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–45.
Solve Prob. 8–44 using Castigliano’s theorem.
60 kN>m
C
B
IBC 5 300(106) mm4
8m
IAB 5 200(106) mm4
ICD 5 200(106) mm4
D
A
6m
Solution
The moment functions are shown in Fig. a. Here,
0M1
= x1,
0P
0M3
0M2
= x3. Also, set P = 0. Then M1 = 0,
= 8 m and
0P
0P
5
M2 = a60x2 - x 32 b kN # m and M3 = 0. Thus,
3
L
0M dx
b
Ma
= 0 +
(∆ A)h =
0P EI
L0
L0
(∆ A)h =
=
5 3
x b(8)dx
3 2
+ 0
EIBC
6 m a60x2 -
4320 kN # m3
EIBC
4320(103) N # m3
[200(109) N>m2][300(10 - 6) m4]
= 0.0720 m = 72.0 mm d Ans.
Ans.
(∆ A)h = 72.0 mm d
404
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–46. The L-shaped frame is made from two fixed-connected
segments. Determine the vertical displacement of the end C.
Use the method of virtual work. EI is constant.
9 ft
2 k>ft
B
C
12 ft
A
Solution
Referring to the virtual and real moment functions shown in
Figs. a and b, respectively,
x 31
b dx1
L
9 ft ( -x1) a 12 ft
( - 9)( - 27)dx2
27
mM
#
1 k (∆ C)y =
dx =
+
EI
EI
EI
L0
L0
L0
1 k # (∆ C)y =
(∆ C)y =
16 767 k2 # ft3
5EI
16 767 k # ft3
T
5EI
Ans.
Ans.
(∆ C)y =
405
16 767 k # ft3
T
5EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–47. The L-shaped frame is made from two fixed-connected
segments. Determine the slope at point C. Use the method of
virtual work. EI is constant.
9 ft
2 k>ft
B
C
12 ft
A
Solution
Referring to the virtual and real moment functions shown in
Figs. a and b, respectively,
x 31
b dx1
L
9 ft ( -1) a 12 ft
( - 1)( - 27)dx2
muM
27
#
#
dx =
+
1 k ft uC =
EI
EI
EI
L0
L0
L0
1 k # ft # uC =
1539 k2 # ft3
4EI
uC =
1539 k # ft2
4 EI
Ans.
Ans.
uC =
406
1539 k # ft2
4EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–48.
Solve Prob. 8–46 using Castigliano’s theorem.
9 ft
2 k>ft
B
C
12 ft
A
Solution
0M1
= -x1
0P
3
0M2
x1
= - 9. Also, set P = 0. Then, M1 = a - b k # ft
and
0P
27
The moment functions are shown in Fig. a. Here,
and M2 = - 27 k # ft. Thus,
(∆ C)v =
L0
L
Ma
0M dx
b
=
0P EI
L0
(∆ C)v =
9 ft a -
x 31
b( -x1)dx1
12 ft
( - 27)( -9)dx2
27
+
EI
EI
L0
16 767 k # ft3
T
5EI
Ans.
Ans.
(∆ C)v =
407
16 767 k # ft3
T
5EI
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8–49 Use the method of virtual work and determine the
horizontal and vertical displacements of point C. There is a
fixed support at A and fixed joint at B. EI is constant.
400 lb>ft
B
C
8 ft
10 ft
A
Solution
L
1#∆ =
mM
dx
L0 EI
∆ Ch =
1
640 000 lb # ft3
c
(0)(200x 21)dx1 +
(10 - x2)(12 800)dx2 d =
EI L0
EI
L0
8
1
1 228 800 lb # ft3
c
(x1)(200x 21)dx1 +
(8)(12 800)dx2 d =
EI L0
EI
L0
8
∆ Cv =
10
Ans.
10
Ans.
Ans.
∆ Ch =
∆ Cv =
408
640 000 lb # ft3
EI
1 228 800 lb # ft3
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–50
Solve Prob. 8–49 using Castigliano’s theorem.
400 lb>ft
B
C
8 ft
10 ft
A
Solution
0M1
= 0
0P
0M2
= 10 - x2
0P
Setting P = 0,
M1 = 200x 21 M2 = 12 800
∆ Ch =
L0
L
Ma
0M dx
b
0P EI
1
640 000 lb # ft3
c
(200x 21)(0)dx1 +
12 800(10 - x2)dx2 d =
EI L0
EI
L0
8
=
0M1
= x1
0P
10
Ans.
0M2
= 8
0P
Setting P = 0,
M1 = 200x 21 M2 = 12 800
∆ Cv =
L0
L
Ma
0M dx
b
0P EI
1
1 228 000 lb # ft3
c
(200x 21)(x1)dx1 +
(12 800)(8)dx2 d =
EI L0
EI
L0
8
=
10
Ans.
Ans.
∆ Ch =
∆ Cv =
409
640 000 lb # ft3
EI
1 228 000 lb # ft3
EI
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8–51. Determine the horizontal displacement at C. Take
E = 29(103) ksi, I = 150 in4 for each member. Use the method
of virtual work.
8 ft
C
8k
D
10 ft
A
SOLUTION
B
The internal moments are determined using the methods of
Ch. 7. Then,
8
10
(1.25x2)(10x2)
(x3)(8x3)
mM
dx = 0 +
dx2 +
dx3
EI
EI
L0 EI
L0
L0
L
∆ Ch =
=
4800(1728)
4800
=
EI
29(103)(150)
= 1.91 in.
Ans.
Ans.
∆ Ch = 1.91 in.
410
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*8–52.
Solve Prob. 8–51 using Castigliano’s theorem.
8 ft
C
8k
D
10 ft
A
SOLUTION
B
Setting P = 8 k,
∆ Ch =
L0
L
= 0 +
=
Ma
L0
0M dx
b
0P EI
8
(10x2)(1.25x2)
EI
dx2 +
L0
10
4800(1728)
4800
=
= 1.91 in.
EI
29(103)(150)
(8x3)(x3)
EI
dx3
Ans.
Ans.
∆ Ch = 1.91 in.
411
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–53. Determine the horizontal displacement of the rocker at
B. Take E = 29(103) ksi, I = 150 in4 for each member. Use the
method of virtual work.
8 ft
C
8k
D
10 ft
A
SOLUTION
B
8
10
(10)(10x2)
(x3)(8x3)
mM
dx = 0 +
dx2 +
dx3
EI
EI
L0 EI
L0
L0
L
∆ Bh =
=
5866.67(1728)
5866.67
=
EI
(29)(103)(150)
= 2.33 in.
Ans.
Ans.
∆ Bh = 2.33 in.
412
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8–54.
Solve Prob. 8–53 using Castigliano’s theorem.
8 ft
C
8k
D
10 ft
A
B
SOLUTION
Setting P = 0,
∆ Bh =
L0
L
= 0 +
=
Ma
L0
0M dx
b
0P EI
3
10x2(10)
EI
dx2 +
L0
10
(8x3)(x3)
EI
dx3
5866.67(1728)
5866.67
=
EI
29(103)(150)
= 2.33 in.
Ans.
Ans.
∆ Bh = 2.33 in.
413
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8–55. Determine the vertical displacement of point C.
EI is constant. Use the method of virtual work.
80 kN ? m
B
20 kN
C
3m
4m
A
SOLUTION
Referring to the virtual and real moment functions shown in
Figs. a and b, respectively,
L
1 kN # (∆ C)v =
mM
dx =
L0 EI
L0
1 kN # (∆ C)v =
1800 kN2 # m3
EI
(∆ C)v =
1800 kN # m3
T
EI
3m
( -x1)( - 80)dx1
EI
+
L0
4m
( - 3)(20x2 - 160)dx2
EI
Ans.
Ans.
(∆ C)v =
414
1800 kN # m3
T
EI
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*8–56.
Solve Prob. 8–55 using Castigliano’s theorem.
80 kN ? m
B
20 kN
C
3m
4m
A
SOLUTION
The moment functions are shown in Fig. a. Here,
and
0M1
= -x1
0P
0M2
= - 3. Also, set P = 0. Then M1 = - 80 kN # m and
0P
M2 = (20x2 - 160) kN # m. Thus,
(∆ C)v =
L0
L
Ma
0M dx
b
=
0P EI
L0
(∆ C)v =
3m
- 80( - x1)dx1
EI
+
L0
4m
(20x2 - 160)( -3)dx2
EI
1800 kN # m3
T
EI
Ans.
Ans.
(∆ C)v =
415
1800 kN # m3
T
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–57. Determine the slope at A and the vertical displacement
at B. Use the method of virtual work. EI is constant.
B
4m
60 kN
C
4m
A
3m
SOLUTION
Referring to the virtual moment functions shown in Figs. a
and b, and real moment functions shown in Fig. c,
x2
x1
b(24x1)dx1
b(30x2)dx2
5ma
4 m a1muM
10
8
1 kN # m # uA =
dx =
+
EI
EI
L0 EI
L0
L0
L
1 kN # m # uA =
260 kN2 # m3
EI
uA =
260 kN # m2
EI
Ans.
and
1 kN # (∆ B)v =
L
mM
dx =
L0 EI
L0
1 kN # (∆ B)v =
(∆ B)v =
5ma
3
3
x1 b(24x1)dx1
4 m a x2 b(30x2)dx2
10
8
+
EI
EI
L0
540 kN2 # m3
EI
540 kN # m3
c
EI
Ans.
Ans.
uA =
260 kN # m2
EI
(∆ B)v =
416
540 kN # m3
c
EI
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8–58.
B
Solve Prob. 8–57 using Castigliano’s theorem.
4m
60 kN
C
4m
A
3m
Solution
For the slope, the moment diagrams are shown in Fig. a Here,
0M1
0M2
1
1
=
x1 and
= 1- x2. Also, set M′ = 0. Then
0M′
10
0M′
8
M1 = (24x1) kN # m and M2 = (30x2) kN # m. Thus,
uA =
L0
L
Ma
0M dx
b
=
0M′ EI
L0
uA =
5 m (24x1) a
1
1
x1 bdx1
x bdx2
4 m (30x2) a110
8 2
+
EI
EI
L0
260 kN # m2
EI
Ans.
For the displacement, the moment diagrams are shown in Fig. b.
Here,
0M1
0M2
3
3
=
x and
= x2. Also, set P = 0. Then,
0P
10 1
0P
8
M1 = (24x1) kN # m and M2 = (30x2) kN # m. Thus,
L
0M dx
b
(∆ B)y =
Ma
=
0P EI
L0
L0
(∆ B)y =
5 m (24x1) a
3
3
x bdx1
4 m (30x2) a x2 bdx2
10 1
8
+
EI
EI
L0
540 kN # m3
c
EI
Ans.
Ans.
uA =
260 kN # m2
EI
(∆ B)y =
417
540 kN # m3
c
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–59. The bent rod has an E = 200 GPa, G = 75 GPa, and a
radius of 30 mm. Use Castigliano’s theorem and determine the
vertical deflection at C. Include the effects of bending, shear,
and torsional strain energy.
z
A
y
1.5 m
B
x
1m
C
2 kN
Solution
Set P = 2 kN.
M 0M
V
0V
T 0T
b dx +
ba
b dx + a
a
Ka
a bL
GA
0P
GJ 0P
L0 EI 0P
L0
L
∆ CZ =
=
L0
+
∆ CZ =
1
L
( - 2x)( - x)
EI
L0
1a
dx +
L0
1.5
(2x - 3)(x - 1.5)
EI
dx
10
10
b(2)(1)
b(2)(1)
1.5 a
(2)(1)(1.5)
9
9
dx +
dx + 0 +
GA
GA
GJ
L0
2.917(103)
x
200(109) a b(0.03)4
4
+
5.556(103)
9
75(10 )(x)(0.03)
= 0.022924 + 0.0000262 + 0.0314380
∆ CZ = 0.0544 m = 54.4 mmT 2
+
3(103)
x
75(109) a b(0.03)4
2
Ans.
Ans.
∆ CZ = 54.4 mmT
418
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–60. Determine the slope at A. Take E = 29(103) ksi. The
moment of inertia of each segment of the frame is i­ndicated in
the figure. Assume D is a pin support. Use the method of
virtual work.
12 k
5 ft
5 ft
B
C
IBC 5 900 in4
IAB 5 600 in4
12 ft
ICD 5 600 in4
A
Solution
1 # uA =
uA =
D
5
5
(1 - 0.1x)(6x)dx
(0.1x)(6x)dx
muM
dx =
+
+ 0 + 0
EIBC
EIBC
L0 EI
L0
L0
L
(75 - 25 + 25)
EIBC
=
75(144)
29(103)(900)
= 0.414(10 - 3) radA
Ans.
Ans.
uA = 0.414(10 - 3) radA
419
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–61.
Solve Prob. 8–60 using Castigliano’s theorem.
12 k
5 ft
5 ft
B
C
IBC 5 900 in4
IAB 5 600 in4
12 ft
ICD 5 600 in4
A
Solution
D
Set M′ = 0.
5
L
uA =
5
(6x)(1 - 0.1x)dx
(6x)(0.1x)dx
M 0M
bdx =
a
+
+ 0 + 0
EIBC
EIBC
L0 EI 0M′
L0
L0
=
(75 - 25 + 25)
EIBC
=
75(144)
29(103)(900)
= 0.414(10 - 3) radA
Ans.
Ans.
uA = 0.414(10 - 3) radA
420
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–1. Determine the reactions at the supports, then draw the
shear and moment diagrams. Assume the support at A is fixed
and B is a roller. EI is constant.
500 lb/ft
B
A
12 ft
Solution
Support Reactions: FBD (a).
Ans.
+ ΣFx = 0;
S
Ax = 0 + c ΣFy = 0;
Ay + By -
a + ΣMA = 0;
ByL + MA -
w0L
= 0(1)
2
w0L L
a b = 0(2)
2
3
Method of Superposition: Using the table in Appendix C, the
required displacements are
vB ′ =
w0L4
T 30EI
vB ′′ =
ByL3
3EI
c
The compatibility condition requires
0 = vB ′ + vB ′′
( + T)
0 =
By =
ByL3
w0L4
b
+ a
30EI
3EI
w0L
10
Substituting By into Eqs. (1) and (2) yields,
Ay =
By =
Ay =
MA =
2w0L
5
500 lb>ft (12 ft)
10
= 600 lb
2(500 lb>ft)(12 ft)
5
500 lb>ft (12 ft2)
15
MA =
w0L2
15
Ans.
= 2400 lb
Ans.
= 4800 lb # ft
Ans.
Ans.
By = 600 lb;
Ax = 0;
Ay = 2400 lb;
MA = 4800 lb # ft
421
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–2. Determine the reactions at the supports A, B, and C,
then draw the shear and moment diagrams. EI is constant.
3 k>ft
2 k>ft
A
C
B
18 ft
18 ft
Solution
Compatibility Equation. Referring to Fig. a, the required
displacements are
5(3)(364)
5wL4AC
32 805 k # ft3
=
=
T
768EI
768EI
EI
4
4
5(2)(36 )
5wLAC
21 870 k # ft3
∆ B″ =
=
=
T
768EI
768EI
EI
L3AC
363
972 ft3
=
=
c
fBB =
48EI
48EI
EI
∆ ′B =
Using the principle of superposition,
∆ B = ∆ B′ + ∆ ″B + NBfBB
32 805
21 870
972
b
+
- NB a
EI
EI
EI
NB = 56.25 k
( + T) 0 =
Ans.
Support Reactions. Referring to the FBD of the beam Fig. b,
+
S
Ax = 0 ΣFx = 0;
a + ΣMA = 0;
Ans.
NC(36) + 56.25(18) - 3(18)(9) - 2
(18)(27) = 0
Ans.
NC = 12.375 k = 12.4 k
+ c ΣFy = 0;
Ay + 56.25 + 12.375 - 3(18) - 2(18) = 0
Ans.
Ay = 21.375 k = 21.4 k
Using these results, the shear and moment diagram shown in
Figs. c and d, respectively, can be plotted.
Ans.
NB = 56.25 k
Ax = 0
NC = 12.4 k
Ay = 21.4 k
422
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–3. Determine the reactions at the supports, then draw the
moment diagram. Assume the support at B is a roller. EI is
constant.
w
B
A
L
Solution
Compatibility Equation:
( + T)
∆ B - ByfBB = 0
(1)
Use virtual work method:
L
wx 2
1x21 - 2 2
mM
wL4
b dx = dx =
a
∆B =
EI
8EI
L EI
L0
L
f BB =
mm
1 x2 1 x2
L3
dx =
dx =
EI
3 EI
L EI
L0
wL4
L3
- By
= 0
8EI
3EI
3wL
By =
8
wL2
MA =
8
5wL
Ay =
8
From Eq. (1) Ax = 0
Ans.
Ans.
Ans.
Ans.
Ans.
3wL
;
8
2
wL
MA =
;
8
5wL
;
Ay =
8
Ax = 0
By =
423
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–4. Determine the support reactions. Assume B is a pin
and A and C are rollers. EI is constant.
4 kN>m
A
C
B
15 m
20 m
Solution
1
mM
∆B =
dx =
L0 EI
L0
=
15 a
4
330
3
90
bx
xb a
x - 2x 2 b
20 a xb a
7
7
7
7
dx +
dx
EI
EI
L0
30 535.714
EI
L
2
fBB =
L0 EI
=
857.143
EI
m
dx =
L0
15 a
4 2
3 2
xb
20 a xb
7
7
dx +
dx
EI
EI
L0
∆ B + ByfBB = 0
30 535.714
857.143
b = 0
+ By a
EI
EI
By = -35.625 kN = - 35.6 kN = 35.6 kN c b + ΣMA = 0;
Ans.
60(7.5) - 35.625(15) + Cy(35) = 0
Ans.
Cy = 2.41 kNT + c ΣFy = 0;
+ ΣFx = 0;
d
Ay - 60 + 35.625 - 2.41 = 0
Ay = 26.8 kN c Ans.
Bx = 0
Ans.
Ans.
By = 35.6 kN c
Cy = 2.41 kNT
Ay = 26.8 kN c
Bx = 0
424
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–5. Determine the reactions at the supports, then draw the
moment diagram. The moment of inertia for each segment is
shown in the figure. Assume A and C are rollers and B is a pin.
Take E = 200 GPa.
20 kN
A
IAB 5 120(106) mm4
3m
40 kN
20 kN
3m
3m
B IBC 5 90(106) mm4 C
4m
4m
Solution
Compatibility Equation:
( + T)
∆ B - ByfBB = 0
a + ΣMB′ = 0;
- MB′ +
891.0
439.2
1310.7
(1.990) +
(5.333) (8) = 0
EIAB
EIAB
EIAB
∆ B = MB′ = a + ΣMB′ = 0;
- MB′ -
6369.6
6369.6
=
T
EIAB
EIAB
22.588
22.228
(2.667) +
(8) = 0
EIAB
EIAB
fBB = MB′ =
From Eq. 1, (1)
117.59
c
EIAB
6369.6
117.59
B = 0
EIAB
EIAB y
By = 54.2 kN c Ans.
Bx = 0
Ans.
Cy = 12.5 kN c Ans.
Ay = 13.3 kN c Ans.
Ans.
By = 54.2 kN c
Bx = 0
Cy = 12.5 kN c
Ay = 13.3 kN c
425
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–6. Determine the reactions at the supports, then draw the
moment diagram. Assume B and C are rollers and A is pinned.
The support at B settles downward 30 mm. Take E = 200 GPa,
I = 150(106) mm4.
30 kN>m
A
C
B
6m
6m
Solution
Compatibility Equation. Referring to Fig. a, the required
displacements are
∆ B′ =
[5(30)(124)](103) N # m3
5wL4AC
=
= 0.27 mT
384EI
384[200(109) N>m2][150(10 - 6) m4]
fBB =
L3AC
123 m3
=
= 1.2(10 - 6) m>N
9
48EI
48[200(10 ) N>m2][150(10 - 6) m4]
Using the principle of superposition,
∆ B = ∆ B′ + NBfBB
( + T)
0.03 = 0.27 + { -NB[1.2(10-6)]}
NB = 200(103) N = 200 kN
Ans.
Support Reactions. Referring to the FBD of the beam, Fig. b,
+
Ans.
S
Ax = 0 ΣFx = 0;
a + ΣMA = 0;
NC(12) + 200(6) - 30(12)(6) = 0
+ c ΣFy = 0;
Ay + 200 + 80 - 30(12) = 0
NC = 80 kN
Ay = 80 kN
Ans.
Ans.
Using these results, the shear and moment diagrams shown in
Figs. c and d, respectively, can be plotted.
Ans.
NB = 200 kN
Ax = 0
NC = 80 kN
Ay = 80 kN
426
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–7. Determine the value of a so that the maximum positive
moment has the same magnitude as the maximum negative
moment. EI is constant.
P
a
L
Solution
2
(tA>B)1 =
(tA>B)2 =
2(L - a)
-P(L - a) (2L + a)
1 - P(L - a)
b(L - a) aa +
b =
a
2
EI
3
6EI
AyL3
2L
1 AyL
a
b(L) a
b =
2 EI
3
3EI
tA>B = 0 = (tA>B)1 + (tA>B)2
0 =
Ay =
-P(L - a)2(2L + a)
6EI
P(L - a)2(2L + a)
+
AyL3
3EI
2L3
Require:
` M1 ` = ` M2 `
Pa(L - a)2(2L + a)
2L
3
=
Pa(L - a)(L + a)
2L2
a 2 + 2La - L2 = 0
a = 0.414 L
Ans.
Ans.
a = 0.414 L
427
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–8. Determine the reactions at the supports, then draw the
shear and moment diagrams. Assume A and C are rollers and B
is pinned. EI is constant.
60 kN ? m
A
C
B
4m
4m
Solution
Compatibility Equation. Referring to Fig. a, the required
displacements are
M0x
∆B′ =
(L2 - x 2)
6EILAC AC
= c
fBB =
60(4)
6EI(8)
d (82 - 42) =
240 kN # m3
T
EI
L3
83 m3
32 m3
=
=
c
48EI
48EI
3EI
Using the principle of superposition,
∆ B = ∆B′ + ByfBB
240 kN # m3
32 m3
bd
+ c - By a
EI
3EI
By = 22.5 kN
( + T) 0 =
Ans.
Support Reactions. Referring to the FBD of the beam, Fig. b,
+
Ans.
S
Bx = 0 ΣFx = 0;
a + ΣMA = 0;
22.5(4) + 60 - NC(8) = 0
NC = 18.75 kN
+ c ΣFy = 0;
Ans.
22.5 - 18.75 - NA = 0
NA = 3.75 kN
Ans.
Ans.
By = 22.5 kN
Bx = 0
NC = 18.75 kN
NA = 3.75 kN
428
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–9. Determine the force in the spring. Assume the support at
A is fixed. The beam has a width of 5 mm and E = 200 GPa.
50 N
200 mm
B
A
10 mm
k 5 2 N>mm
Solution
I =
1
(0.005)(0.01)3 = 0.4166(10 - 9) m4
12
(∆ B)1 =
50(0.23)
PL3
=
= 0.0016 m
3EI
3(200)(109)(0.4166)(10 - 9)
(∆ B)2 =
2000∆ B(0.23)
PL3
=
= 0.064 ∆ B
3EI
3(200)(109)(0.4166)(10 - 9)
Compatibility Condition:
+ T ∆ B = (∆ B)1 - (∆ B)2
∆ B = 0.0016 - 0.064∆ B
∆ B = 0.001503 m = 1.50 mm
By = k∆ B = 2(1.5) = 3.00 N
Ans.
Ans.
By = 3.00 N
429
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–10. Determine the reactions at the supports. The moment
of inertia for each segment is shown in the figure. Assume the
support at B is a roller. Take E = 29(103) ksi.
A
IAB
600 in4
18 ft
B IBC
300 in4
C
15 k ft
12 ft
Solution
Compatibility Equation. Referring to the FBD of the left
conjugate beam segment sectioned through B, Fig. a
a + ΣMB = 0;
MB ′ + c a
15 k # ft
b (18 ft) d (9 ft) = 0
EIAB
∆ B = MB ′ = -
2430 k # ft3
2430 k # ft3
=
T
EIAB
EIAB
And Fig. b
a + ΣMB = 0;
1 18 ft
2
MB ″ - c a
b(18 ft) d c (18 ft) d = 0
2 EIAB
3
fBB = MB ″ =
1944 ft3
c
EIAB
Using the principle of superposition, Fig. c
∆ A + ByfBB = 0
( + T)
2430 k # ft3
1944 ft3
+ By a b = 0
EIAB
EIAB
Ans.
By = 1.25 k
Equilibrium. Referring to the FBD of the beam, Fig. d
+
S
ΣFx = 0;
Ax = 0 + c ΣFy = 0;
1.25 - Ay = 0
a + ΣMA = 0;
1.25(18) - 15 - MA = 0
Ans.
Ans.
Ay = 1.25 k
MA = 7.50 k # ft
Ans.
Ans.
By = 1.25 k;
Ax = 0;
Ay = 1.25 k;
MA = 7.50 k # ft
430
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–11. Determine the reactions at the supports. Assume the
support at A is fixed and B is a roller. Take E = 29(103) ksi.
The moment of inertia for each segment is shown in the figure.
5k
2 k>ft
A
IAB 5 600 in4
18 ft
B
IBC 5 300 in4 C
12 ft
Solution
18
(x - 18)(41x - 474 - x 2)
mM
45 684
dx =
dx + 0 =
EIAB
EIAB
L0 EI
L0
L
∆B =
18
(x - 18)2
m2
1944
dx =
dx + 0 =
EIAB
EIAB
L0 EI
L0
L
fBB =
∆ B + ByfBB = 0
45 684
1944
+ By
= 0
EIAB
EIAB
By = -23.5 k = 23.5 k c Ans.
+
S
ΣFx = 0;
Ax = 0 Ans.
+ c ΣFy = 0;
Ay = 36 - 23.5 + 5 = 17.5 k c Ans.
b + ΣMA = 0;
- MA + 36(9) - 23.5(18) + 5(30) = 0
MA = 51 k # ftB
Ans.
Ans.
By = 23.5 k c
Ax = 0
Ay = 17.5 k c
MA = 51 k # ftB
431
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–12. Determine the reactions at the fixed supports, A and
B. EI is constant.
P
A
B
L
—
2
L
—
2
Solution
u′ - u″ = 0
PL2
16EI
ML
u″ = A″y =
2EI
PL2
ML
= 0
16EI
2EI
PL
M =
8
u′ = A′y =
From equilibrium and symmetry:
MA =
PL
B
8
MB =
PL
A
8
Ans.
From equilibrium and symmetry:
Ay = By =
P
c
2
Ans.
Ans.
PL
B
8
PL
MB =
A
8
P
Ay =
c
2
MA =
432
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–13. Determine the reactions at the supports. Assume A
and C are pins and the joint at B is fixed connected. EI is
constant.
12 kN>m
B
A
4m
6m
C
Solution
Compatibility Equation. Referring to Fig. a, the required displacements can be determined using the virtual work method.
Using the virtual and real moment functions shown in Figs. b
and c,
1 kN # (∆ C′ )n =
mM
dx =
L0 EI
L
L
4m 3
a x1 b(6x1 - 6x 21)dx1
2
+
L0
EI
0
(∆ C′ )n = 1 kN # fCC =
6m
x 32
bdx2
2
EI
(x2) a -
902.4
902.4
d
=
EI
EI
mm
dx =
L0 EI
L
L
4m 3
6m
3
a x1 b a x1 bdx1
(x2)(x2)dx2
2
2
+
L
0
EI
EI
0
fCC =
120
S
EI
Using the principle of superposition, Fig. a,
(∆ C)n = (∆ C′ )n + CxfCC
+ )0 = (S
902.4
120
b
+ Cx a
EI
EI
Ans.
Cx = 7.52 kN
Support Reactions. Referring to the FBD of the frame, Fig. d,
+
S
ΣFx = 0;
a + ΣMA = 0;
Ax + 7.52 -
1
(12)(6) = 0 Ax = 28.48 kN = 28.5 kN 2
1
Cy(4) + 7.52(6) - 12(4)(2) - c (12)(6) d (2) = 0
2
Ans.
Ay + 30.72 - 12(4) = 0
Ans.
Cy = 30.72 kN = 30.7 kN
+ c ΣFy = 0;
Ans.
Ay = 17.28 kN = 17.3 kN
433
12 kN>m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–13.
(Continued)
Ans.
Cx = 7.52 kN
Ax = 28.5 kN
Cy = 30.7 kN
Ay = 17.3 kN
434
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–14. Determine the reactions at the supports, then draw the
moment diagrams for each member. EI is constant.
6 kN>m
B
A
3m
2m
12 kN
2m
C
Solution
Compatibility Equation: Referring to the real and virtual
moment functions indicated in Figs. a and b, respectively,
1 # (∆ C)y =
L
mM
dx =
L0 EI
L0
2m
(0)(0)dx1
EI
+
+
1 # (∆ c)y = 0 + 0 +
(∆ c)y =
1 # fCC =
L
L0
0( -12x2)dx2
EI
2
3 m ( - x3) c -(3x 3 + 24) d dx3
EI
168.75 kN # m
EI
3
168.75 kN # m3
T
EI
mm
dx =
L0 EI
L0
2m
(0)(0)dx1
1 # fCC = 0 + 0
fCC =
L0
2m
EI
+
L0
2m
(0)(0)dx2
EI
+
L0
3m
9 m3
EI
9 m3
T
EI
435
( - x3)( - x3)dx3
EI
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–14.
(Continued)
Using the method of superposition, Fig. c,
(∆ c)y + Cy fCC = 0
( + T)
168.75 kN # m3
9 m3
+ Cy a
b = 0
EI
EI
Cy = - 18.75 kN = 18.75 kN c Ans.
Equilibrium. Referring to the FBD of the frame, Fig. d,
+ ΣFx = 0;
S
+ c ΣFy = 0;
a + ΣMA = 0;
Ax - 12 = 0
Ax = 12.0 kN Ans.
18.75 - 6(3) - Ay = 0
Ay = 0.750 kN
Ans.
18.75(3) - 6(3)(1.5) - 12(2) - MA = 0
MA = 5.25 kN # m
Ans.
The plot of the moment diagram is shown in Fig. e.
Ans.
Cy = 18.75 kN c ;
Ax = 12.0 kN;
Ay = 0.750 kN;
MA = 5.25 kN # m
436
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–15. Determine the reactions at the supports, then draw the
moment diagram for each member. EI is constant.
6k
8 ft
8 ft
B
C
12 ft
A
Solution
30 k ? ft
Compatibility Equation. Referring to Fig. a, the required
displacements can be determined using the virtual work
method. Using the virtual and real moment functions shown
in Figs. b and c,
L
1 k # (∆ A′ )y =
mM
dx = 0 +
L0 EI
L0
(∆ ′A)y = -
1 k # fCC =
(x2)( - 30)dx2
EI
+
L0
8 ft
(x3 + 8)[ -(6x3 + 30)]dx3
6400
6400
=
T
EI
EI
L
mm
dx = 0 +
L0 EI
L0
fCC =
8 ft
8 ft
(x2)(x2)dx2
EI
+
L0
8 ft
(x3 + 8)(x3 + 8)dx3
EI
4096
c
3EI
Using the principle of superposition, Fig. a,
(∆ A)y = (∆ ′A)y + NA fCC
(+ c) 0 = -
4096
6400
+ NA a
b
EI
3EI
Ans.
NA = 4.6875 k = 4.69 k
Support Reactions. Referring to the FBD of the frame, Fig. d,
+
S
ΣFx = 0;
Cx = 0 Ans.
+ c ΣFy = 0;
Cy + 4.6875 - 6 = 0 Cy = 1.3125 k = 1.31 k
Ans.
a + ΣMC = 0;
6(8) + 30 - 4.6875(16) - MC = 0 MC = 3.00 k # ft
Using these results, the moment diagram shown in Fig. e can
be plotted.
437
Ans.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–15.
(Continued)
Ans.
NA = 4.69 k
Cx = 0
Cy = 1.31 k
MC = 3.00 k # ft
438
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–16. Determine the reactions at the supports. Assume A is
fixed connected. E is constant.
8 kN>m
B
A
IAB 5 1250 (106) mm4
9m
3m
20 kN
6
IBC 5 625 (10 ) mm
4
3m
C
Solution
Compatibility Equation. Referring to Fig. a, and using the real
and virtual moment function shown in Figs. b and c, respectively,
L
∆ ′Cy =
mM
dx =
L0 EI
L0
L
fCC =
mm
dx =
L0 EI
L0
9m
( -x3)[ -(4x 23 + 60)]
EIAB
9m
( - x3)( - x3)
EIAB
dx3 =
dx3 =
8991
T
EIAB
243
T
EIAB
Using the principle of superposition,
∆ Cy = ∆ C′ y + CyfCC
( + T) 0 =
8991
243
+ Cy a
b
EIAB
EIAB
Cy = -37.0 kN = 37.0 kN c Ans.
Equilibrium. Referring to the FBD of the frame in Fig. d,
+
S
Ax - 20 = 0
Ax = 20 kN ΣFx = 0;
a + ΣMA = 0;
+ c ΣFy = 0;
Ans.
MA + 37.0(9) - 8(9)(4.5) - 20(3) = 0
MA = 51.0 kN # m
Ans.
Ay + 37.0 - 8(9) = 0 Ay = 35.0 kN
Ans.
Ans.
Cy = 37.0 kN c
Ax = 20 kN
MA = 51.0 kN # m
Ay = 35.0 kN
439
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–17. Determine the reactions at the supports, then draw the
moment diagram for each member. EI is constant.
6m
18 kN>m
B
A
4m
C
Solution
Compatibility Equation. Referring to Fig. a, the required ­
displacements can be determined using the virtual work
method. Using the virtual and real moment functions shown
in Figs. b and c,
x 31
)dx1
L
6 m (x1)( 4m
6( -108)dx2
mM
2
1 kN # (∆ ′A)y =
dx =
+
EI
EI
EI
L0
L0
L0
(∆ ′A)y = 1 kN # fCC =
fCC =
16 848
16 848
=
T
5EI
5EI
L
mm
dx =
L0 EI
L0
6m
(x1)(x1)dx1
EI
+
L0
4m
6(6)dx2
EI
216
c
EI
Using the principle of superposition, Fig. a,
(∆ A)y = (∆ ′A)y + NA fCC
(+ c) 0 = -
16 848
216
b
+ NA a
5EI
EI
Ans.
NA = 15.6 kN
Support Reactions. Referring to the FBD of the frame, Fig. d,
+
S
ΣFx = 0;
Cx = 0 + c ΣFy = 0;
Cy + 15.6 -
a + ΣMC = 0;
Ans.
1
(18)(6) = 0 Cy = 38.4 kN
2
Ans.
1
(18)(6)(2) - 15.6(6) - MC = 0 MC = 14.4 kN # m
2
Ans.
Using these results, the moment diagram shown in Fig. e can
be plotted.
440
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–17.
(Continued)
Ans.
NA = 15.6 kN
Cx = 0
Cy = 38.4 kN
MC = 14.4 kN # m
441
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–18. Determine the reactions at the supports. Assume A and
C are pin supported and B is fixed connected. EI is constant.
12 ft
60 k ? ft
B
A
10 ft
C
Solution
Compatibility Equation. Referring to Fig. a, the required displacements can be determined using the virtual work method.
Using the virtual and real moment functions shown in Figs. b
and c,
L
mM
1 k # (∆ ′A)y =
dx = 0 +
L0 EI
L0
(∆ ′A)y = -
10 ft a
6
x b( -6x2)dx2
5 2
EI
2400
2400
=
T
EI
EI
L
mm
1 k # fAA =
dx =
L0 EI
L0
12 ft
(x1)(x1)dx1
EI
+
L0
10 ft a
6
6
x b a x bdx2
5 2 5 2
EI
1056
fAA =
c
EI
Using the principle of superposition, Fig. a,
(∆ A)y = (∆ ′A)y + Ay fAA
(+ c) 0 = -
1056
2400
+ Ay a
b
EI
EI
Ans.
Ay = 2.273 k = 2.27 k
Support Reactions. Referring to the FBD of the frame, Fig. d,
+ c ΣFy = 0;
2.273 - Cy = 0
Ans.
Cy = 2.273 k = 2.27 k
a + ΣMA = 0;
60 - 2.273(12) - Cx(10) = 0 Cx = 3.273 k = 3.27 k
Ans.
+ ΣFx = 0;
S
Ax - 3.273 = 0
Ans.
Ax = 3.273 k = 3.27 k
442
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9–18.
(Continued)
Ans.
Ay = 2.27 k
Cy = 2.27 k
Cx = 3.27 k
Ax = 3.27 k
443
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–19. Determine the reactions at the fixed support A and
rocker C. The moment of inertia of each segment is listed in the
figure. Take E = 200 GPa.
8 kN>m
B
A
IAB 5 1250 (106) mm4
9m
3m
2 kN
IBC 5 625 (106) mm4
3m
C
Solution
9
L
∆C =
L
fCC =
2
(x)(6 + 4x )
mM
6804
dx =
dx + 0 + 0 =
EIAB
EIAB
L0 EI
L0
9
m2
x2
243
dx =
+ 0 + 0 =
EI
EI
EI
L0
L0
AB
AB
∆ C + CyfCC = 0
6804
243
+ Cy a
b = 0
EIAB
EIAB
Cy = - 28 kN = 28 kN c Ans.
+
S
ΣFx = 0;
Ax - 2 = 0;
Ax = 2 kN S Ans.
c + ΣFy = 0;
28 - 72 + Ay = 0;
Ay = 44 kN c Ans.
a + ΣMA = 0;
MA - 72(4.5) - 2(3) + 28(9) = 0
MA = 78 kN # mB
Ans.
Ans.
Cy = 28 kN c
Ax = 2 kN S
Ay = 44 kN c
MA = 78 kN # mB
444
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–20. Determine the reactions at the supports. Assume C is
fixed. EI is constant.
12 ft
B
A
6 ft
5k
6 ft
Solution
C
6
L
∆A =
12(30 - 5x)
mM
1080
dx =
dx + 0 + 0 =
EI
EI
L0 EI
L0
6
6
12
(12)(12)
(12)(12)
( - x)( - x)
m2
2304
dx =
dx +
dx +
dx =
EI
EI
EI
EI
L0 EI
L0
L0
L0
L
fAA =
∆ A + Ay fAA = 0
Ay =
- 1080
= - 0.469 k = 0.469 k c 2304
+
S
ΣFx = 0;
Cx = 5.0 k S + c ΣFy = 0;
- Cy + 0.469 = 0;
b + ΣMC = 0;
Ans.
Ans.
Cy = 0.469 kT MC - 5(6) + 0.469(12) = 0;
MC = 24.4 k # ftA
Ans.
Ans.
Ans.
Ay = 0.469 k c
Cx = 5.0 k S
Cy = 0.469 kT
MC = 24.4 k # ftA
445
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9–21. Determine the reactions at the supports. Assume A and
D are pins. EI is constant.
20 ft
8k
B
C
10 ft
15 ft
D
A
Solution
Compatibility Equation. Referring to Fig. a, and the real and
virtual moment functions shown in Figs. b and c, respectively,
L
∆ ′Dh =
mM
dx =
L0 EI
L0
=
mm
dx =
L0 EI
L0
+
=
(x1)(8x1)
EI
dx1 +
L0
20 ft
(0.25x2 + 10)(6x2)
EI
dx2 + 0
25 000
S
EI
L
fDD =
15 ft
15 ft
(x1)(x1)
EI
L0
10 ft
dx1 +
(x3)(x3)
EI
L0
20 ft
(0.25x2 + 10)(0.25x2 + 10)
EI
dx2
dx3
4625
S
EI
Using the principle of superposition, Fig. a,
∆ Dh = ∆ ′Dh + DxfDD
4625
25 000
+ Dx a
b
EI
EI
Dx = - 5.405 k = 5.41 k d + )0 =
(S
Ans.
Equilibrium:
+
S
ΣFx = 0;
8 - 5.405 - Ax = 0
Ax = 2.5946 k = 2.59 k Ans.
a + ΣMA = 0;
Dy(20) + 5.405(5) - 8(15) = 0 Dy = 4.649 k = 4.65 k
Ans.
+ c ΣFy = 0;
4.649 - Ay = 0
Ans.
Ay = 4.649 k = 4.65 k
446
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9–21.
(Continued)
Ans.
Dx = 5.41 k d
Ax = 2.59 k
Dy = 4.65 k
Ay = 4.65 k
447
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–22. Determine the reactions at the supports, then draw the
moment diagrams for each member. Assume A and B are pins
and the joint at C is fixed connected. EI is constant.
6 kN>m
B
C
6m
2m
12 kN
2m
A
Solution
Compatibility Equation. Referring to Fig. a, the required displacements can be determined using the virtual work method.
Using virtual and real moment functions shown in Figs. b and c,
1 kN # (∆ ′A)h =
L
mM
dx = 0 +
L0 EI
L0
(∆ ′A)h =
1 kN # fAA =
[ - (x2 + 2)]( -12x2)dx2
EI
+
L0
6ma -
x 33
2
x3 b a2x3 bdx3
3
6
EI
784
S
5EI
L
mm
dx =
L0 EI
L0
+
fAA =
2m
2m
( -x1)( - x1)dx1
EI
L0
6ma -
+
L0
2m
[ -(x2 + 2)][ -(x2 + 2)]dx2
EI
2
2
x b a - x3 bdx3
3 3
3
EI
160
S
3EI
Using the principle of superposition,
(AA)h = (AA)′h + AxfAA
+ ) 0 = 784 + Ax a 160 b
(S
5EI
3EI
Ans.
Ax = - 2.94 kN = 2.94 kN d Support Reactions. Referring to the FBD of the frame, Fig. d,
+
S
12 - 2.94 - Bx = 0 Bx = 9.06 kN ΣFx = 0;
1
a + ΣMB = 0;
(6)(6)(4) + 12(2) - 2.94(4) - Ay(6) = 0
2
Ay = 14.04 kN = 14.0 kN
c ΣFy = 0;
By + 14.04 -
Ans.
Ans.
1
(6)(6) = 0 By = 3.96 kN
2
Ans.
Using these results, the moment diagram shown in Fig. e can be plotted.
448
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–22.
(Continued)
Ans.
Ax = 2.94 kN
Bx = 9.06 kN
Ay = 14.0 kN
By = 3.96 kN
449
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–23. Determine the reactions at the supports. Assume A and
D are pins. EI is constant.
2 k>ft
9k
C
B
12 ft
8 ft
D
A
Solution
Compatibility Equation. Referring to Fig. a, the required displacements can be determined using the virtual work method.
Using the virtual and real moment functions shown in Figs. b
and c,
1 k # (∆ ′A)h =
L
mM
dx = 0 +
L0 EI
L0
12 ft
( -8)(6x2 - x 22)dx2
EI
+
L0
8 ft
2688
S
(∆ ′A)h =
EI
1 k # fAA =
L
mm
dx =
L0 EI
L0
8 ft
( - x1)( - x1)dx1
EI
3328
S
fAA =
3EI
+
L0
12 ft
1 -x3 2( -9x3)dx3
1 - 82( - 8) dx2
EI
EI
+
L0
8 ft
( - x3)( - x3)dx
EI
Using the principle of superposition, Fig. a,
(∆ A)h = (∆ A′ )h + Ax fAA
+ ) 0 = 2688 + Ax a 3328 b
(S
EI
3EI
Ax = - 2.423 k = 2.42 k d Ans.
Support Reactions.
+
S
9 - 2.423 - Dx = 0 Dx = 6.577 k = 6.58 k ΣFx = 0;
Ans.
a + ΣMA = 0;
Dy(12) - 9(8) - 2(12)(6) = 0 Dy = 18.0 k
Ans.
+ T ΣMD = 0;
2(12)(6) - (9)(8) - Ay(12) = 0 Ay = 6.00 k
Ans.
450
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9–23.
(Continued)
Ans.
Ax = 2.42 k
Dx = 6.58 k
Dy = 18.0 k
Ay = 6.00 k
451
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–24. Two boards each having the same EI and length L are
crossed perpendicular to each other as shown. Determine the
vertical reactions at the supports. Assume the boards just touch
each other before the load P is applied.
D
P
A
L
—
2
L
—
2
L
—
2
C
B
L
—
2
Solution
∆ E ′ = ME′ = = ∆ E ″ = ME′ =
(P - Ey)L2
16EI
(P - Ey)L3
(P - Ey)L L
L
b +
a b
2
16EI
6
48EI
EyL2
16EI
= -
2
a
a
EyL3
EyL2 L
L
ba b
6
16EI 2
48EI
∆E′ = ∆E″
-
(P - Ey)L3
48EI
= -
EyL3
48EI
- (P - Ey) = -Ey
Ey =
P
2
Ans.
For equilibrium:
Ay = By = Cy = Dy =
P
4
Ans.
Ans.
P
2
P
Ay =
4
Ey =
452
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9–25.
Determine the force in each member. AE is constant.
4m
D
3m
4 kN
A
C
6 kN
3m
Solution
Compatibility Equation. Referring to Fig. a and using the
real and virtual forces in each member shown in Figs. b and
c, respectively,
1 kN # ∆ ′AD = Σ
B
( -1.60)(4.00)(4)
(1.00)( -10.0)(5)
nNL
= 0 +
+
AE
AE
AE
∆ ′AD = -
1 kN # fADAD = Σ
fADAD =
75.6
AE
2
2
(1.60 )(4)
(1.00) (5)
n2L
= 2c
d +
AE
AE
AE
20.24
AE
Applying the principle of superposition, Fig. a,
∆ AD = ∆ ′AD + FAD fADAD
0 = -
75.6
20.24
b
+ FAD a
AE
AE
FAD = 3.735 kN (T) = 3.74 kN (T)
Ans.
Method of Joints. Referring to the FBD of Joint A, Fig. d,
+ c ΣFy = 0;
3
3
FAB a b + 3.735a b - 6 = 0
5
5
FAB = 6.2648 kN (C) = 6.26 kN (C) Ans.
+
S
ΣFx = 0;
4
4
4 + 3.735a b - 6.264a b - FAC = 0
5
5
FAC = 1.976 kN (C) = 1.98 kN (C)
Ans.
Ans.
FAD = 3.74 kN (T)
FAB = 6.26 kN (C)
FAC = 1.98 kN (C)
453
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–26. Determine the force in each member of the truss. AE is
constant.
D
4 ft
3 ft
C
B
800 lb
3 ft
A
Solution
Compatibility Equation.
0 = ∆ AB + FABfABAB(1)
Use the virtual work method:
∆ AB = Σ
(1.0)(1.333)(5)
( -1.6)( -1.067)(4)
nNL
13.493
=
+
=
AE
AE
AE
AE
fABAB = Σ
2(1) (5)
( -1.6) (4)
nnL
20.24
=
+
=
AE
AE
AE
AE
2
From Eq. 1, 0 =
2
13.493
20.24
+
F
AE
AE AB
FAB = -0.667 k = 0.667 k (C)
Ans.
Joint B:
+ c ΣFy = 0;
+
d
ΣFx = 0;
3
3
FBD + a b0.6667 - 0.8 = 0
5
5
FBD = 0.667 k (T)
Ans.
FBC = 0
Ans.
Ans.
FAB = 0.667 k (C)
FBD = 0.667 k (T)
FBC = 0
454
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–27. Determine the force in member AC of the truss. AE is
constant.
E
6 ft
D
A
6 ft
B
Solution
6k
Compatibility Equation. Referring to Fig. a and using the
real and virtual forces in each member shown in Figs. b and
c, respectively,
1 k # ∆ ′AC = Σ
∆ ′AC =
8 ft
C
6k
( -0.8)( -8.00)(8)
( - 0.6)(6.00)(6)
nNL
=
+ 2c
d
AE
AE
AE
8
AE
2
2
2
(1 )(10)
( - 0.8) (8)
( - 0.6) (6)
n2L
= 2c
+
+
d
AE
AE
AE
AE
34.56
fACAC =
AE
1 k # fACAC = Σ
Applying the principle of superposition, Fig. a,
∆ AC = ∆ ′AC + FACfACAC
0 =
8
34.56
b
+ FAC a
AE
AE
FAC = -0.2314 k = 0.231 k (C)
Ans.
Ans.
FAC = 0.231 k (C)
455
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–28.
Determine the force in member HG. AE is constant.
G
H
F
15 ft
E
A
B
15 ft
C
15 ft
4k
D
15 ft
4k
15 ft
4k
Solution
Compatibility equation.
0 = ∆ HG + FHGfHGHG(1)
Use the virtual work method:
(1)(12.0)(15)
(1)(14)(15)
(1)(8)(15)
nNL
=
+
+
AE
AE
AE
AE
( - 1.414)( - 11.31)(21.21)
( -1.414)( -8.485)(21.21)
1103.97
+
+
=
AE
AE
AE
∆ HG = Σ
2
fHGHG = Σ
2
4(1) (15)
2( -1.414) (21.21)
nnL
144.85
=
+
=
AE
AE
AE
AE
1103.97
144.85
+
FHG
AE
AE
FHG = -7.621 k = 7.62 k (C)
From Eq. 1, 0 =
Ans.
Ans.
FHG = 7.62 k (C)
456
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–29.
Determine the force in member BE. AE is constant.
12 kN
D
E
3m
B
A
3m
C
3m
Solution
∆ BE + FBE fBEBE = 0
∆ BE =
ΣNnL
1
=
5(6)(0)(3) + (6)(1)(3) + (0)(1)(3) + (12)(1)(3) + (0)(1)(3)
AE
AE
+ (0)( - 1.414)(312) + (8.49)(0)(312) + ( -8.49)( - 1.414)(312)6 =
fBEBE =
Σn2L
1
=
5(1)2(3) + (1)2(3) + (1)2(3) + (1)2(3) + ( - 1.414)2(312)
AE
AE
104.912
AE
+ ( - 1.414)2(312) + 0 + 06 =
28.971
AE
104.912
28.971
b = 0
+ FBE a
AE
AE
FBE = - 3.621 kN = 3.62 kN (C) Ans.
Ans.
FBE = 3.62 kN (C)
457
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–30.
Determine the force in member BD. AE is constant.
12 kN
D
C 9 kN
2m
B
A
1.5 m
Solution
Compatibility Equation. Referring to Fig. a and using the real and
virtual force in each member shown in Figs. b and c, respectively,
1 kN # ∆ ′BD = Σ
( -0.6)( -12.0)(1.5)
( -0.8)( - 28.0)(2)
(1)(35.0)(2.5)
nNL
=
+
+
AE
AE
AE
AE
143.1
∆ ′BD =
AE
1 kN # fBDBD = Σ
2
2
2
( -0.8) (2)
(1 )(2.5)
( - 0.6) (1.5)
n2L
= 2c
+
+
d
AE
AE
AE
AE
fBDBD =
8.64
AE
Applying the principle of superposition, Fig. a,
∆ BD = ∆ B′ D + FBDfBDBD
0 =
143.1
8.64
b
+ FBD a
AE
AE
FBD = - 16.5625 kN = 16.6 kN (C)
Ans.
Ans.
FBD = 16.6 kN (C)
458
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–31.
Determine the force in member AD. AE is constant.
12 kN
D
C 9 kN
2m
B
A
1.5 m
Solution
Compatibility Equation. Referring to Fig. a and using the
real and virtual forces in each member shown in Figs. b and c,
respectively,
1 kN # ∆ ′AD = Σ
(0.75)( - 12)(1.5)
1( -28.0)(2)
( - 1.25)(35.0)(2.5)
nNL
=
+
+
AE
AE
AE
AE
178.875
∆ ′AD = AE
1 kN # fADAD = Σ
2
2
2
(1 )(2)
( - 1.25) (2.5)
(0.75 )(1.5)
n2L
= 2c
+
+
d
AE
AE
AE
AE
fADAD =
13.5
AE
Applying the principle of superposition, Fig. a,
∆ AD = ∆ ′AD + FAD fADAD
0 = -
178.875
13.5
b
+ FAD a
AE
AE
FAD = 13.25 kN (T)
Ans.
Ans.
FAD = 13.25 kN (T)
459
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*9–32. Determine the force in each member of the truss. AE
is constant.
E
F
8 kN
3m
A
B
4m
D
C
4m
4m
Solution
Compatibility Equation. Referring to Fig. a and using the
real and virtual forces in each member shown in Figs. b and
c, respectively,
1 kN # ∆ ′EF = Σ
(1.25)( - 10.0)(5)
( -0.75)(6.00)(3)
nNL
=
+
AE
AE
AE
76
∆ ′EF = AE
2
2
2
(1.25 )(5)
( -0.75) (3)
(1 )(4)
n2L
= 2c
+
d +
AE
AE
AE
AE
23
fEFEF =
AE
1 kN # fEFEF = Σ
Applying the principle of superposition, Fig. a,
∆ EF = ∆ ′EF + FEF fEFEF
76
23
+ FEF a
b
AE
AE
FEF = 3.304 kN (T) = 3.30 kN (T)
0 = -
Ans.
Methods of Joint. Referring to the FBD of Joint F, Fig. d,
+
S
ΣFx = 0;
+ c ΣFy = 0;
4
3.304 - FAF a b = 0 FAF = 4.130 kN (T) = 4.13 kN (T) 5
3
FBF - 4.130a b = 0 FBF = 2.478 kN (C) = 2.48 kN (C)
5
Ans.
Ans.
And Joint E, Fig. e,
+
S
ΣFx = 0;
+ c ΣFy = 0;
4
8 - 3.304 - FDE a b = 0 FDE = 5.870 kN (C) = 5.87 kN (C) Ans.
5
3
5.870a b - FCE = 0 FCE = 3.522 kN (T) = 3.52 kN (T)
5
Ans.
Ans.
FEF = 3.30 kN (T)
FAF = 4.13 kN (T)
FBF = 2.48 kN (C)
FDE = 5.87 kN (C)
FCE = 3.52 kN (T)
460
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9–33. Determine the force in member GB of the truss. AE is
constant.
H
G
F
10 ft
E
A
B
10 ft
10 ft
10 k
D
C
10 ft
15 k
10 ft
5k
Solution
Compatibility Equation. Referring to Fig. a, and using the real and
virtual force in each member shown in Figs. b and c, respectively,
∆ GB
′ = Σ
nNL
1
=
c ( -0.7071)(10)(10) + ( - 0.7071)(16.25)(10)
AE
AE
+ 0.7071(13.75)(10) + 0.7071(15)(10) + 0.7071( - 22.5)(10)
+ ( - 0.7071)( - 22.5)(10) + 1(8.839)(14.14)
= -
+ ( - 1)(12.37)(14.14) d
103.03
AE
2
fGBGB = Σ
2
2
0.7071 (10)
( - 0.7071) (10)
( - 1) (14.14)
n2L
= 3c
d + 3c
d + 2c
d
AE
AE
AE
AE
+ 2c
=
(12)(14.14)
AE
86.57
AE
d
Applying the principle of superposition, Fig. a,
∆ GB = ∆ GB + FGBfGBGB
86.57
- 103.03
+ FGB a
b
AE
AE
FGB = 1.190 k = 1.19 k (T)
0 =
Ans.
Ans.
FGB = 1.19 k (T)
461
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9–34. Determine the reactions at the supports, then draw the
moment diagram. Each spring is originally unstretched and
has a stiffness k = 12 EI>L3. EI is constant.
w
A
k
C
k
B
L
L
Solution
Principle of superposition. Referring to Figs. a and b, the required displacements are
∆C =
∆ ′C =
∆ ″C =
FSP
FSP
L3
=
= FSP a
bT
3
k
12EI
12EI>L
wL4
T
8EI
FSPL3
PL3
=
c
3EI
3EI
The compatibility condition requires
∆ C = ∆ ′C + ∆ ″C
( + T)
FSP a
FSPL3
L
wL4
b =
b
+ a12EI
8EI
3EI
3wL
FSP =
10
3
Ans.
Equilibrium. Referring to the FBD of the beam, Fig. b,
+
S
ΣFx = 0;
Bx = 0 + c ΣFy = 0;
By + 2a
Ans.
3wL
b - w(2L) = 0
10
By =
7wL
5
Ans.
The plot of the moment diagram is shown in Fig. c.
Ans.
FSP =
3wL
;
10
Bx = 0
By =
462
7wL
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–35. Determine the force in members AB, BC, and BD,
which are used in conjunction with the beam to carry the 30-k
load. The beam has a moment of inertia of I = 600 in4, the
members AB and BC each have a cross-sectional area of 2 in2,
and BD has a cross-sectional area of 4 in2. Take E = 29(103) ksi.
Neglect the thickness of the beam and its axial compression,
and assume all members are pin connected. Also assume the
support at A is a pin and E is a roller.
B
3 ft
A
D
C
E
3 ft
4 ft
30 k
Solution
(0.57143x)(40x)
(0.42857x)(30x)
mM
nNL
dx + a
=
dx +
dx + 0 + 0 + 0
EI
AE
EI
EI
L0
L0
L0
3
L
∆ BD =
=
480 # 3
k ft
EI
(0.57143x) dx
(0.42857x) dx
m2
n2L
dx + a
=
+
AE
EI
EI
L0 EI
L0
L0
3
L
fBDBD =
4
+
=
4
2
(1)2(3)
4E
+
2
( -0.80812)2(312)
2E
+
( - 0.71429)2(5)
2E
6.8571 ft3
3.4109 ft
+
EI
E
∆ BD + FBDfBDBD = 0
3
480(12 )
E(600)
3
+ FBD a
6.8571(12 )
E(600)
+
3.4109(12)
E
b = 0
FBD = - 22.78 k = 22.8 k (C)
Ans.
Joint B:
+
S
a Fx = 0;
+ c a Fy = 0;
- FAB a
1
4
b + a bFBC = 0;
5
12
3
1
22.78 - a bFBC - FAB a
b = 0;
5
12
FAB = 18.4 k (T)
Ans.
FBC = 16.3 k (T)
Ans.
Ans.
FBD = 22.8 k (C)
FAB = 18.4 k (T)
FBC = 16.3 k (T)
463
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–36. The queen-post trussed beam is used to support a
uniform load of 4 k>ft. Determine the force developed in each
of the five struts. Neglect the thickness of the beam and assume
the truss members are pin connected to the beam. Also, neglect
the effect of axial compression and shear in the beam. The
cross-sectional area of each strut is 3 in2, and the moment of
inertia for the beam is I = 600 in4. Also, E = 29(103) ksi.
4 k>ft
B
A
4 ft
C
F
D
10 ft
E
3 ft
4 ft
Solution
Compatibility equation.
0 = ∆ CE + FCEfCECE(1)
Use virtual work method:
4
10
( -0.75x1)(36x1 - 2x 21)
( - 3)(20x2 - 2x 22 + 112)
mM
5320 # 3
dx = 2
dx1 +
dx2 = k ft
EI
EI
EI
L0 EI
L0
L0
L
∆ CE =
4
10
( -0.75x1)2
( - 3)2
2(1.25)2(5)
2( -0.75)2(3)
(1)2(10)
mM
mL
dx + Σ
= 2
dx1 +
dx2 +
+
+
EI
EI
AE
AE
AE
L0 EI
L0
L0 EI
L
fCECE =
=
114.0 3
29.0
ft +
ft
EI
AE
From Eq. 1:
0 = -
5320(1728)
E(600)
+ FCE c
114(1728)
FCE = 34.48 k = 34.5 k (T)
E(600)
+
29(12)
3E
d
Ans.
Joint E:
+
S
ΣFx = 0;
+ c ΣFy = 0;
4
F - 34.48 = 0;
FEF = 43.1 k (T) 5 EF
3
- FED + (43.10) = 0;
FED = 25.9 k (C)
5
Ans.
Ans.
Due to symmetry:
FAC = FEF = 43.1 k (T) Ans.
FCB = FED = 25.9 k (C) Ans.
Ans.
FCE = 3 4 .5 k (T)
FEF = 43.1 k (T)
FED = 25.9 k (C)
FAC = 43.1 k (T)
FCB = 25.9 k (C)
464
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–37. The king-post trussed beam supports a concentrated
force of 40 k at its center. Determine the force in each of the
three struts. The struts each have a cross-sectional area of 2 in2.
Assume they are pin connected at their end points. Neglect
both the depth of the beam and the effect of axial compression
in the beam. Take E = 29(103) ksi for both the beam and struts.
Also, IAB = 400 in4.
40 k
12 ft
12 ft
A
D
B
5 ft
C
Solution
Compatibility equation.
0 = ∆ CD + FCDfCDCD(1)
Use the virtual work method:
12
(0.5)(20x)x
mM
11 520
dx = 2
dx =
EI
EI
EI
L0
L0
L
∆ CD =
12
(0.5x)2
2(1.3)2(13)
(1)2(5)
mm
288.0
48.94
dx = 2
dx +
+
=
+
EI
AE
AE
EI
AE
L0 EI
L0
L
fCDCD =
From Eq. (1),
0 =
11 520(1728)
E(400)
+ a
288.0(1728)
E(400)
+
48.94(12)
FCD = - 32.36 k = 32.4 k (C)
2E
bFCD
Ans.
Joint C:
+
S
ΣFx = 0;
+ c ΣFy = 0;
FCA = FCB
5
2a bFCA - 32.36 = 0
13
FCA = FCB = 42.1 k (T)
Ans.
Ans.
FCD = 32.4 k (C)
FCA = FCB = 42.1 k (T)
465
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9–38.
40 k
Determine the maximum moment in the beam in
Prob. 10–37.
12 ft
12 ft
A
D
B
5 ft
C
Solution
Compatibility equation.
0 = ∆ CD + FCDfCDCD(1)
Use the virtual work method.
12
(0.5)(20x)x
mM
11 520
dx = 2
dx =
EI
EI
L0 EI
L0
L
∆ CD =
12
(0.5x)2
2(1.3)2(13)
(1)2(5)
mm
n2L
dx + Σ
= 2
dx +
+
AE
EI
AE
AE
L0 EI
L0
L
fCDCD =
=
288.0
48.94
+
EI
AE
From Eq. (1),
0 =
11 520(1728)
E(400)
+ a
288.0(1728)
E(400)
+
48.94(12)
FCD = -32.36 k = 32.4 k (C)
2E
bFCD
Joint C:
+
S
ΣFx = 0;
FCA = FCB
+ c ΣFy = 0;
2a
5
bF
- 32.36 = 0
13 CA
FCA = FCB = 42.07 k (T)
M max = 45.8 k # ft
Ans.
Ans.
M max = 45.8 k # ft
466
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30 kN
9–39. Determine the reactions at the fixed support D. EI is
constant for both beams.
D
A
C
B
3m
3m
Solution
Compatibility Equation. Referring to Fig. a, the required
displacements are
∆B =
NB (63)
PL3AC
9NB
=
=
T
48EI
48EI
2EI
∆ ′B =
30(33)
PL3BD
270
=
=
T
3EI
3EI
EI
∆ ″B =
NB(33)
PL3BD
9NB
=
=
c
3EI
3EI
EI
Using the method of superposition, Fig. a,
∆ B = ∆ ′B + ∆ ″B
( + T)
9NB
9NB
270
b
=
+ a2EI
EI
EI
NB = 20.0 kN
Support Reactions. Referring to the FBD of the cantilever
beam, Fig. b,
+
S
ΣFx = 0;
Dx = 0 + c ΣFy = 0;
Dy + 20.0 - 30 = 0
a + ΣMD = 0;
Dy = 10.0 kN
Ans.
Ans.
MD + 20.0(3) - 30(3) = 0
MD = 30.0 kN # m
Ans.
Ans.
Dx = 0
Dy = 10.0 kN
MD = 30.0 kN # m
467
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–40. The structural assembly supports the loading shown.
Draw the moment diagrams for each of the beams. Take
I = 100(106) mm4 for the beams and A = 200 mm2 for the tie
rod. All members are made of steel for which E = 200 GPa.
15 kN
2m
6m
D
E
C
4m
8 kN>m
B
A
6m
Solution
Compatibility Equation.
0 = ∆ CB + FCBfCBCB(1)
Use the virtual work method.
6
L
∆ CB =
2
6
2
(0.25x1)(3.75x1)
(0.75x2)(11.25x2)
(1x3)( - 4x 3)
mM
- 1206
dx =
dx1 +
dx2 +
dx3 =
EI
EI
EI
EI
EI
L0
L0
L0
L0
(0.25x1)
(0.75x2)
(1x3)
(1) (4)
mm
nnL
78.0
4.00
dx + a
=
dx1 +
dx2 +
dx3 +
=
+
EI
AE
EI
EI
EI
AE
EI
AE
L0
L0
L0
L0
6
L
fCBCB =
2
2
2
6
2
2
From Eq. 1,
-
1206
E100(10-6)
+ FCB c
78.0
E(100)(10-6)
+
FCB = 15.075 kN (T) = 15.1 kN (T)
4.00
200(10-6)E
d = 0
Ans.
Ans.
FCB = 15.1 kN (T)
468
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–41. The two cantilever beams are in contact using the roller
support C. Determine the reactions at the fixed supports A and B
when the load of 9 k is applied. EI is constant.
9k
A
C
B
10 ft
Solution
Compatibility Equation. Referring to Fig. a, the required
displacements are
∆C =
NC(103)
PL3BC
1000 NC
=
=
T
3EI
3EI
3EI
∆ ′C =
9(103)
PL3AC
3000
=
=
T
3EI
3EI
EI
∆ ″C =
NC(103)
PL3AC
1000 NC
=
=
c
3EI
3EI
3EI
Using the method of superposition, Fig. a,
∆ C = ∆ ′C + ∆ ″C
( + T)
1000 NC
1000 NC
3000
b
=
+ a3EI
EI
3EI
NC = 4.50 k
Support Reactions. Referring to the FBD of beam AC, Fig. b,
+
S
ΣFx = 0;
Ax = 0 + c ΣFy = 0;
Ay + 4.50 - 9 = 0
a + ΣMA = 0;
Ans.
Ay = 4.50 k
MA + 4.50 (10) - 9(10) = 0 MA = 45.0 k # ft
Ans.
Ans.
And the FBD of beam BC, Fig. c,
+
S
ΣFx = 0;
Bx = 0 + c ΣFy = 0;
By - 4.50 = 0
a + ΣMB = 0;
Ans.
Ans.
By = 4.50 k
MB - 4.50 (10) = 0 MB = 45.0 k # ft
Ans.
Ans.
Ax = 0
Ay = 4.50 k
MA = 45.0 k # ft
Bx = 0
By = 4.50 k
MB = 45.0 k # ft
469
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–42. The beam AB has a moment of inertia of I = 475 in4
and rests on the smooth supports at its ends. A 0.75-in.-diameter
rod CD is welded to the center of the beam and to the fixed
support at D. If the temperature of the rod is decreased by
150°F, determine the force developed in the rod. The beam and
rod are both made of steel for which E = 29 (10 3) ksi and
a = 6.5(10 - 6)>°F.
5 ft
5 ft
A
B
C
50 in.
D
Solution
Method of Superposition: Using the method of superposition
as discussed in Chapter 4, the required displacements are
∆C =
FCD(1203)
PL3
=
= 0.002613FCD T
48EI
48(29)(103)(475)
Using the axial force formula,
∆F =
FCD(50)
PL
=
= 0.003903FCD c
p
AE
(0.752)(29)(103)
4
The thermal contraction is
∆ T = a∆TL = 6.5(10 - 6)(150)(50) = 0.04875 in.T
The compatibility condition requires
( + T)
∆C = ∆T + ∆F
0.002613FCD = 0.04875 + ( - 0.003903FCD)
FCD = 7.48 kip (T)
Ans.
Ans.
FCD = 7.48 kip (T)
470
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–43. Draw the influence line for the reaction at C. Plot
numerical values at the peaks. Assume A is a pin and B and C
are rollers. EI is constant.
A
B
6m
Solution
The primary real beam and qualitative influence line are
shown in Fig. a and the conjugate beam is shown in Fig. b.
Referring to Fig. c,
fAC = M′A = 0, FBC = M′B = 0, fCC = M′C =
144
EI
The maximum displacement between A and B can be determined by referring to Fig. d.
+ c ΣFy = 0;
a + ΣM = 0;
1 x
6
a bx = 0 x = 112 m
2 EI
EI
M′max +
6
1 112
112
b a 112b a
b = 0
a 112b - a
EI
2 EI
3
f max = -
13.86
EI
Dividing f ’s by fCC, we obtain
X(m)
0
Cy(kN)
0
112
–0.0962
6
12
0
1
471
C
6m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–44. Draw the influence line for the moment at A. Plot
numerical values at the peaks. Assume A is fixed and the
support at B is a roller. EI is constant.
A
B
3m
Solution
The primary real beam and qualitative influence line are
shown in Fig. a and its conjugate beam is shown in Fig. b.
Referring to Fig. c,
aAA =
1
3
, fAA = M′A = 0, fBA = M′B = 0, fCA = M′C =
EI
2EI
The maximum displacement between A and B can be determined by referring to Fig. d,
+ c ΣFy = 0;
a + ΣM = 0;
1
x
1
bx a
= 0 x = 13 m
2 3EI
2EI
1 13
1
13
b a 13b a
b a
a 13b - M′max = 0
2 3EI
3
2EI
f max = M′max = -
0.5774
EI
Dividing f ’s by aAA, we obtain
x(m)
0
1.268
3
6
MA(kN # m)
0
–0.577
0
1.50
472
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–45. Draw the influence line for the vertical reaction at B.
Plot numerical values at the peaks. Assume A is fixed and the
support at B is a roller. EI is constant.
A
B
3m
Solution
The primary real beam and qualitative influence line are
shown in Fig. a and its conjugate beam is shown in Fig. b.
Referring to Fig. c,
a + ΣMB = 0;
M′B -
Referring to Fig. d,
a + ΣMC = 0;
M′C -
1 3
9
a b(3)(2) = 0 fBB = M′B =
2 EI
EI
1 3
22.5
a b(3)(5) = 0 fCB = M′C =
2 EI
EI
Also, fAB = 0. Dividing f’s by fBB, we obtain
X(m)
0
3
6
By(kN)
0
1
2.5
473
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–46. Sketch the influence line for the moment at D using
the Müller-Breslau principle. Determine the maximum
positive moment at D due to a uniform live load of 5 kN>m.
EI is constant. Assume A is a pin and B and C are rollers.
A
B
10 m
D
5m
C
5m
Solution
10
10
(0.5x)(12.5x)dx
mM
∆B =
dx =
+
EI
L0 EI
L0
L0
L
(0.5x)(37.5x -
5 2
x )dx
2
EI
=
5208.3
EI
10
(0.5x)2dx
m2
166.7
dx = 2
=
EI
EI
EI
L0
L0
L
fBB =
( + T)
∆ B + ByfBB = 0
b + ΣMA = 0;
5208.3
166.7
b = 0
+ By a
EI
EI
By = - 31.25 kN = 31.25 kN c
- 31.25(10) + 50(15) - Cy(20) = 0
Cy = 21.875 kN c
+ c ΣFy = 0;
b + ΣMD = 0;
31.25 - 50 + 21.875 - Ay = 0
Ay = 3.125 kNT
MD + 25(2.5) - 21.875(5) = 0
MD = 46.9 kN # m
Ans.
Ans.
MD = 46.9 kN # m
474
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–47 Draw the influence line for the shear at C. Plot
numerical values every 1.5 m. Assume A is fixed and the
support at B is a roller. EI is constant.
A
C
3m
Solution
The primary real beam and qualitative influence line are
shown in Fig. a, and its conjugate beam is shown in Fig. b.
Referring to Figs. c, d, e, and f,
fOC = M′0 = 0 f1.5 C = M′1.5 = +
f 3C
= M′3 + =
49.5
EI
Dividing f ’s by M′0 =
6.1875
EI
f4.5 C = M′4.5 =
f3C = M′3 - = -
22.5
EI
26.4375
EI
f6 C = M′6 = 0
3+
4.5
6
0.367
0
72
, we obtain
EI
x (m)
0
1.5
3-
VC (KN)
0
- 0.0859
- 0.3125 0.6875
475
B
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–48. Sketch the influence line for (a) the vertical reaction
at C, (b) the moment at B, and (c) the shear at E. In each case,
indicate on a sketch of the beam where a uniform distributed
live load should be placed so as to cause a maximum positive
value of these functions. Assume the beam is fixed at F. EI is
constant.
A
C
4m
Solution
476
2m
D
B
2m
2m
E
F
2m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–49. Sketch the influence line for (a) the vertical reaction at
C, (b) the moment at B, and (c) the shear at E. In each case,
indicate on a sketch of the beam where a uniform distributed
live load should be placed so as to cause a maximum positive
value of these functions. Assume the beam is fixed at F. EI is
constant.
A
Solution
477
C
B
2m
2m
D
4m
2m
E
F
2m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–50. Use the Müller-Breslau principle to sketch the general
shape of the influence line for the moment at A.
A
Solution
478
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–51. Use the Müller-Breslau principle to sketch the general
shape of the influence line for the moment at A and the shear
at B.
A
B
Solution
479
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–52. Use the Müller-Breslau principle to sketch the general
shape of the influence line for the shear at A and the moment
at B.
A
B
Solution
480
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–53. Use the Müller-Breslau principle to sketch the general
shape of the influence line for the moment at A and the shear
at B.
A
Solution
481
B
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–1. Determine the moments at the supports, then draw the
moment diagram. Assume B is a roller and A and C are fixed.
EI is constant.
15 kN 15 kN 15 kN
25 kN>m
A
3m
B
6m
2m
C
2m
2m
2m
Solution
1FEM2 AB =
1FEM2 BA =
1FEM2 BC =
111252162 2
192
51252162 2
192
- 51152182
16
= - 51.5625 kN # m
= 23.4375 kN # m
= - 37.5 kN # m
1FEM2 CB = 37.5 kN # m
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
I
MAB = 2E a b 12102 + uB - 02 - 51.5625
6
MAB =
EIuB
- 51.5625(1)
3
I
MBA = 2E a b 12uB + 0 - 02 + 23.4375
6
MBA =
2EIuB
+ 23.4375(2)
3
I
MBC = 2E a b 12uB + 0 - 02 - 37.5
8
MBC =
EIuB
- 37.5(3)
2
I
MCB = 2E a b 12102 + uB - 02 + 37.5
8
MCB =
EIuB
+ 37.5(4)
4
Equilibrium.
MBA + MBC = 0(5)
Solving:
uB =
12.054
EI
MAB = - 47.5 kN # m
MBA = 31.5 kN # m
MBC = -31.5 kN # m
MCB = 40.5 kN # m
Ans.
MAB = -47.5 kN # m
MBA = 31.5 kN # m
MBC = -31.5 kN # m
MCB = 40.5 kN # m
Ans.
Ans.
Ans.
Ans.
482
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9k
10–2. Determine the moments at A, B, C, and D, then draw
the moment diagram for the beam. Assume the supports at A
and D are fixed and B and C are rollers. EI is constant.
2 k>ft
A
B
15 ft
Solution
Fixed End Moments. Referring to the table on the inside back
cover,
2
1FEM2 AB = -
21152
wL2
= = - 37.5 k # ft
12
12
2
1FEM2 BA =
21152
wL2
= = 37.5 k # ft
12
12
1FEM2 BC = 1FEM2 CB = 0
1FEM2 CD =
21921152
- 2PL
= = - 30 k # ft
9
9
1FEM2 DC =
21921152
2PL
=
= 30 k # ft
9
9
9k
Slope-Deflection Equation. Applying Eq. 10–8,
MN = 2Ek12uN + uF - 3c2 + 1FEM2 N
For span AB,
MAB = 2E a
I
2EI
b[2102 + uB - 3102] + 1 - 37.52 = a
buB - 37.5(1)
15
15
MBA = 2E a
I
4EI
b[2uB + 0 - 3102] + 37.5 = a
buB + 37.5(2)
15
15
For span BC,
MBC = 2E a
I
4EI
2EI
b[2uB + uC - 3102] + 0 = a
buB + a
buC(3)
15
15
15
MCB = 2E a
I
4EI
2EI
b[2uC + uB - 3102] + 0 = a
buC + a
buB(4)
15
15
15
For span CD,
MCD = 2E a
I
4EI
b[2uC + 0 - 3102] + 1 - 302 = a
buC - 30(5)
15
15
MDC = 2E a
I
2EI
b[2102 + uC - 3102] + 30 = a
buC + 30(6)
15
15
483
C
15 ft
5 ft
D
5 ft
5 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–2.
(Continued)
Equilibrium. At support B,
MBA + MBC = 0
a
4EI
4EI
2EI
buB + 37.5 + a
buB + a
buC = 0
15
15
15
a
8EI
2EI
buB + a
buC = - 37.5
15
15
At support C,
MCB + MCD = 0
a
4EI
2EI
4EI
buC + a
buB + a
buC - 30 = 0
15
15
15
a
8EI
2EI
buC + a
buB = 30
15
15
Solving Eqs. (7) and (8),
uC =
78.75
EI
uB = -
90
EI
Substitute these results into Eqs. (1) to (6).
MAB = - 49.5 k # ft
MBA = 13.5 k # ft
MBC = -13.5 k # ft
MCB = 9 k # ft
MCD = - 9 k # ft
MDC = 40.5 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
The negative signs indicate that MAB, MBC, and MCD have
counterclockwise rotational sense. Using these results, the
shear at both ends of spans AB, BC, and CD are computed and
shown in Figs. a, b, and c, respectively. Subsequently, the shear
and moment diagrams can be plotted, Figs. d, and e, respectively.
Ans.
MAB = - 49.5 k # ft
MBA = 13.5 k # ft
MBC = -13.5 k # ft
MCB = 9 k # ft
MCD = - 9 k # ft
MDC = 40.5 k # ft
484
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–3. Determine the moments at A, B, C, and D, then draw
the moment diagram for the beam. Assume the supports at A
and D are fixed and B and C are rollers. EI is constant.
20 kN>m
A
5m
Solution
Fixed End Moments. Referring to the table on the inside back
cover,
1FEM2 AB = 0
1FEM2 BA = 0
1FEM2 CD = 0
1FEM2 DC = 0
2
1FEM2 BC = -
2013 2
wL2
= = - 15 kN # m
12
12
2
1FEM2 CB =
2013 2
wL2
=
= 15 kN # m
12
12
Slope-Deflection Equation. Applying Eq. 10–8,
MN = 2Ek12uN + uF - 3c2 + 1FEM2 N
For span AB,
I
2EI
buB(1)
MAB = 2E a b[2102 + uB - 3102] + 0 = a
5
5
I
4EI
buB(2)
MBA = 2E a b[2uB + 0 - 3102] + 0 = a
5
5
For span BC,
I
4EI
2EI
buB + a
buC - 15(3)
MBC = 2E a b[2uB + uC - 3102] + 1 - 152 = a
3
3
3
I
4EI
2EI
MCB = 2E a b[2uC + uB - 3102] + 15 = a
buC + a
buB + 15(4)
3
3
3
For span CD,
I
4EI
MCD = 2E a b[2uC + 0 - 3102] + 0 = a
buC(5)
5
5
I
2EI
MDC = 2E a b[2102 + uC - 3102] + 0 = a
buC(6)
5
5
Equilibrium. At support B,
MBA + MBC = 0
a
4EI
4EI
2EI
buB + a
buB + a
buC - 15 = 0
5
3
3
a
32EI
2EI
buB + a
buC = 15(7)
15
3
485
C
B
3m
D
5m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–3.
(Continued)
At support C,
MCB + MCD = 0
a
4EI
2EI
4EI
buC + a
buB + 15 + a
buC = 0
3
3
5
a
2EI
32EI
buB + a
buC = - 15(8)
3
15
Solving Eqs. (7) to (8),
uB =
225
22EI
uC =
225
22EI
Substitute these results into Eqs. (1) to (6).
MAB = 4.091 kN # m = 4.09 kN # m
MBA = 8.182 kN # m = 8.18 kN # m
MBC = -8.182 kN # m = -8.18 kN # m
MCB = 8.182 kN # m = 8.18 kN # m
MCD = - 8.182 kN # m = - 8.18 kN # m
MDC = - 4.091 kN # m = - 4.09 kN # m
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
The negative sign indicates that MBC, MCD and MDC have
counterclockwise rotational sense. Using these results, the
shear at both ends of spans AB, BC, and CD are computed and
shown in Figs. a, b, and c, respectively. Subsequently, the shear
and moment diagrams can be plotted, Figs. d and e respectively.
Ans.
MAB = 4.09 kN # m
MBA = 8.18 kN # m
MBC = -8.18 kN # m
MCB = 8.18 kN # m
MCD = -8.18 kN # m
MDC = -4.09 kN # m
486
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–4. Determine the internal moments at the supports
A, B, and C, then draw the moment diagram. Assume A is
pinned, and B and C are rollers. EI is constant.
3 k>ft
A
4 ft
B
8 ft
C
8 ft
4 ft
Solution
MN = 2E a
I
b 12uN + uF - 3c 2 + 1FEM2 N
L
3182
2EI
12uA + uB 2 8
12
2
MAB =
3182
2EI
12uB + uA 2 +
8
12
2
MBA =
3182
2EI
12uB + uC 2 8
12
2
MBC =
3182
2EI
12uC + uB 2 +
8
12
2
MCB =
Equilibrium:
MAB + 24 = 0
MBA + MBC = 0
MCB - 24 = 0
2
3182
2EI
12uA + uB 2 + 24 = 0
8
12
0.5uA + 0.25uB = 2
8
EI
2
3182
3182
2EI
2EI
12uB + uA 2 +
+
12uB + uC 2 = 0
8
12
8
12
uB + 0.25uA + 0.25uC = 0
2
3182
2EI
12uC + uB 2 +
- 24 = 0
8
12
0.5uC + 0.25uB =
8
EI
Solving,
uB = 0
uA = - uC = MAB = -24 k # ft
MBA = 12 k # ft
MBC = - 12 k # ft
MCB = 24 k # ft
16
EI
Ans.
Ans.
MAB = - 24 k # ft
MBA = 12 k # ft
MBC = - 12 k # ft
MCB = 24 k # ft
Ans.
Ans.
Ans.
487
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–5. Determine the moments at A, B, and C, then draw the
moment diagram for the beam. Assume the supports at A and
C are fixed. EI is constant.
15 kN
A
20 kN
B
8m
C
4m
4m
Solution
MAB =
2EI
EI
10 + uB - 02 + 0 =
u
8
4 B
MBA =
2EI
EI
12uB + 0 - 02 + 0 =
u
8
2 B
MBC =
2EI
EI
12uB + 0 - 02 - 20 =
u - 20
8
2 B
MCB =
2EI
EI
10 + uB - 02 + 20 =
u + 20
8
4 B
MBA + MBC = 0
Solving,
uB =
20
EI
MAB = 5 kN # m
MBA = 10 kN # m
MBC = -10 kN # m
MCB = 25 kN # m
Ans.
Ans.
Ans.
Ans.
Ans.
20
EI
MAB = 5 kN # m
MBA = 10 kN # m
MBC = -10 kN # m
MCB = 25 kN # m
uB =
488
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
40 kN
10–6. Determine the moment at B, then draw the moment
diagram for the beam. Assume the supports at A and C are
pins and B is a roller. EI is constant.
A
B
6m
Solution
20 kN
2m
C
4m
4m
Fixed End Moments. Referring to the table on the inside back
cover,
2
1FEM2 BA = a
2 162
P
40
a 2b
2
b
ab
a
+
d = 52.5 kN # m
b = a 2 b c 62 122 +
2
2
L2
8
1FEM2 BC = -
31202182
3PL
= = - 30 kN # m
16
16
Slope-Deflection Equations. Applying Eq. 10–10, since one of
the end’s support for spans AB and BC is a pin,
MN = 3Ek1uN - c2 + 1FEM2 N
For span AB,
I
3EI
MBA = 3E a b 1uB - 02 + 52.5 = a
buB + 52.5(1)
8
8
For span BC,
I
3EI
MBC = 3E a b 1uB - 02 + 1- 302 = a
buB - 30(2)
8
8
Equilibrium. At support B,
MBA + MBC = 0
a
3EI
3EI
buB + 52.5 + a
buB - 30 = 0
8
8
a
3EI
buB = - 22.5
4
uB = -
30
EI
Substitute this result into Eqs. (1) and (2).
MBA = 41.25 kN # m
MBC = - 41.25 kN # m
Ans.
Ans.
The negative sign indicates that MBC has counterclockwise
rotational sense. Using this result, the shear at both ends of
spans AB and BC are computed and shown in Figs. a and b,
respectively. Subsequently, the shear and moment diagrams
can be plotted, Figs. c and d, respectively.
Ans.
MBA = 41.25 kN # m
MBC = - 41.25 kN # m
489
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–7. Determine the moments at each support, then draw
the moment diagram. Assume A is fixed. EI is constant.
12 k
4 k>ft
B
A
20 ft
D
C
15 ft
8 ft
8 ft
Solution
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
MAB =
41202
2EI
12102 + uB - 02 20
12
MBA =
41202
2EI
12uB + 0 - 02 +
20
12
MBC =
2EI
12uB + uC - 02 + 0
15
MCB =
2EI
12uC + uB - 02 + 0
15
MN = 3E a
MCD =
2
2
I
b 1uN - c2 + 1FEM2 N
L
3112216
3EI
1uC - 02 16
16
Equilibrium.
MBA + MBC = 0
MCB + MCD = 0
Solving,
178.08
EI
336.60
uB = EI
MAB = - 167 k # ft
uC =
MBA = 66.0 k # ft
MBC = -66.0 k # ft
MCB = 2.61 k # ft
MCD = - 2.61 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
MAB = -167 k # ft
MBA = 66.0 k # ft
MBC = -66.0 k # ft
MCB = 2.61 k # ft
MCD = -2.61 k # ft
490
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6k
*10–8. Determine the moments at A, B, and C, then draw
the moment diagram. EI is constant. Assume the support at B
is a roller and A and C are fixed.
0.5 k>ft
B
A
8 ft
8 ft
C
18 ft
Solution
1FEM2 AB = 1FEM2 BA =
PL
= - 12,
8
PL
= 12,
8
1FEM2 BC = 1FEM2 CB =
wL2
= -13.5
12
wL2
= 13.5
12
uA = uC = cAB = cBC = 0
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
MAB =
2EI
1uB 2 - 12
16
MBA =
2EI
12uB 2 + 12
16
MBC =
2EI
12uB 2 - 13.5
18
MCB =
2EI
1uB 2 + 13.5
18
Moment equilibrium at B:
MBA + MBC = 0
2EI
2EI
12uB 2 + 12 +
12uB 2 - 13.5 = 0
16
18
uB =
3.1765
EI
Thus,
MAB = -11.60 = - 11.6 k # ft
MBA = 12.79 = 12.8 k # ft
MBC = -12.79 = - 12.8 k # ft
MCB = 13.853 = 13.9 k # ft
Ans.
Ans.
Ans.
Ans.
Left Segment:
b + ΣMA = 0;
- 11.60 + 6182 + 12.79 - VBL 1162 = 0
VBL = 3.0744 k
+ c ΣFy = 0;
Ay = 2.9256 k
Right Segment:
b + ΣMB = 0;
- 12.79 + 9192 - Cy 1182 + 13.85 = 0
Cy = 4.5588 k
+ c ΣFy = 0;
VBR = 4.412 k
Ans.
MAB = - 11.6 k # ft
MBA = 12.8 k # ft
MBC = - 12.8 k # ft
MCB = 13.9 k # ft
At B:
By = 3.0744 + 4.4412 = 7.52 k
491
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–9. Determine the moments at A and B, then draw the
moment diagram for the beam. EI is constant.
2400 lb
200 lb>ft
A
B
30 ft
C
10 ft
Solution
1FEM2 AB = -
1
1
1w21L2 2 = - 120021302 2 = - 15 k # ft
12
12
MAB =
2EI
10 + uB - 02 - 15
30
MBA =
2EI
12uB + 0 - 02 + 15
30
ΣMB = 0;
MBA = 2.41102
Solving,
uB =
67.5
EI
MAB = - 10.5 k # ft
MBA = 24 k # ft
MBC = - 24 k # ft
Ans.
Ans.
Ans.
Ans.
MAB = -10.5 k # ft
MBA = 24 k # ft
MBC = -24 k # ft
492
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–10. Determine the moments at A, B, and C. The support
at B settles 0.15 ft. E = 29(103) ksi and I = 8000 in4. Assume
the supports at B and C are rollers and A is fixed.
300 lb>ft
240 lb>ft
B
A
20 ft
C
30 ft
Solution
FEMAB = FEMBA =
wL2
= - 8,
12
FEMBC = -
wL2
= - 33.75
12
wL2
= 8
12
Applying Eqs. 11–8 and 11–10,
MAB =
2EI
0.15
bd - 8
c uB - 3a
20
20
MBA =
2EI
0.15
bd + 8
c 2uB - 3a
20
20
MBC =
3EI
0.15
c uB +
d - 33.75
30
30
Moment equilibrium at B:
MBA + MBC = 0;
2EI
0.15
3EI
0.15
bd + 8 +
c 2uB - 3a
c uB +
d - 33.75 = 0
20
20
30
30
uB = 5.887110 - 3 2 rad
Thus,
MAB = - 2680 k # ft
MBA = - 1720 k # ft
MBC = 1720 k # ft
MCB = 0
Ans.
Ans.
Ans.
Ans.
Ans.
MAB = - 2680 k # ft
MBA = - 1720 k # ft
MBC = 1720 k # ft
MCB = 0
493
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–11. Determine the moments at A, B, and C, then draw
the moment diagram for the beam. The moment of inertia of
each span is indicated in the figure. Assume the support at B is
a roller and A and C are fixed. E = 29(103) ksi.
C
A
IAB 5 900 in4
24 ft
Solution
Fixed End Moments. Referring to the table on the inside back
cover,
2
1FEM2 AB = -
2124 2
wL2
= = - 96 k # ft
12
12
2
1FEM2 BA =
2124 2
wL2
=
= 96 k # ft
12
12
1FEM2 BC = 1FEM2 CB =
301162
PL
= = - 60 k # ft
8
8
301162
PL
=
= 60 k # ft
8
8
Slope-Deflection Equations. Applying Eq. 10–8,
MN = 2Ek12uN + uF - 3c2 + 1FEM2 N
For span AB,
MAB = 2E c
900 in4
d [2102 + uB - 3102] + [ - 961122 k # in.]
241122 in.
MAB = 6.25EuB - 1152(1)
MBA = 2E c
900 in4
d [2uB + 0 - 3102] + 961122 k # in.
241122 in.
MBA = 12.5EuB + 1152(2)
For span BC,
MBC = 2E c
1200 in4
d [2uB + 0 - 3102] + [ - 601122 k # in.]
161122 in.
MBC = 25EuB - 720(3)
MCB = 2E c
30 k
2 k>ft
1200 in4
d [2102 + uB - 3102] + 601122 k # in.
161122in.
MCB = 12.5EuB + 720(4)
Equilibrium. At support B,
MBA + MBC = 0(5)
494
BI
4
BC 5 1200 in
8 ft
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–11.
(Continued)
Substitute Eqs. (2) and (3) into (5).
12.5EuB + 1152 + 25EuB - 720 = 0
uB = -
11.52
E
Substitute this result into Eqs. (1) to (4).
MAB = -1224 k # in. = - 102 k # ft
MBA = 1008 k # in. = 84 k # ft
MBC = -1008 k # in. = - 84 k # ft
MCB = 576 k # in. = 48 k # ft
Ans.
Ans.
Ans.
Ans.
The negative signs indicate that MAB MBC have counterclockwise rotational senses. Using these results, the shear at both
ends of spans AB and BC are computed and shown in
Figs. a and b, respectively. Subsequently, the shear and
moment diagrams can be plotted, Figs. c and d, respectively.
Ans.
MAB = - 102 k # ft
MBA = 84 k # ft
MBC = - 84 k # ft
MCB = 48 k # ft
495
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–12. Determine the moments acting at A and B. Assume
A is fixed supported, B is a roller, and C is a pin. EI is
constant.
20 kN>m
80 kN
C
B
A
9m
3m
3m
Solution
1FEM2 AB = 1FEM2 BA =
wL2
= - 54,
30
1FEM2 BC = -
3PL
= -90
16
wL2
= 81
20
Applying Eqs. 10–8 and 10–10,
MAB =
2EI
1uB 2 - 54
9
MBA =
2EI
12uB 2 + 81
9
MBC =
3EI
1uB 2 - 90
6
Moment equilibrium at B:
MBA + MBC = 0
4EI
EI
1uB 2 + 81 +
u - 90 = 0
9
2 B
uB =
9.529
EI
Thus,
MAB = - 51.9 kN # m
MBA = 85.2 kN # m
MBC = -85.2 kN # m
Ans.
Ans.
Ans.
Ans.
MAB = -51.9 kN # m
MBA = 85.2 kN # m
MBC = -85.2 kN # m
496
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–13. Determine the moments at the supports, then draw
the moment diagram. The members are fixed connected at the
supports and at joint B. The moment of inertia of each member is given in the figure. Take E = 29(103) ksi.
20 k
8 ft
A
8 ft
B
IAB 5 800 in4
6 ft
IBC 5 1200 in4
15 k
6 ft
C
Solution
- 201162
1FEM2 AB =
8
= - 40 k # ft
1FEM2 BA = 40 k # ft
MN = 2E a
MAB =
1FEM2 BC =
1FEM2 CB = 22.5 k # ft
-151122
8
= -22.5 k # ft
I
b 12uN + uF - 3c2 + 1FEM2 N
L
212921103 218002
1611442
12102 + uB - 02 - 40
MAB = 20 138.89uB - 40(1)
MBA =
212921103 218002
1611442
12uB + 0 - 02 + 40
MBA = 40 277.78uB + 40
MBC =
212921103 2112002
1211442
(2)
12uB + 0 - 02 - 22.5
MBC = 80 555.56uB - 22.5(3)
MCB =
212921103 2112002
1211442
12102 + uB - 02 + 22.5
MCB = 40 277.78uB + 22.5(4)
Equilibrium.
MBA + MBC = 0(5)
Solving Eqs. 1–5:
uB = -0.00014483
MAB = -42.9 k # ft
MBA = 34.2 k # ft
MBC = -34.2 k # ft
MCB = 16.7 k # ft
Ans.
Ans.
Ans.
MAB = - 42.9 k # ft
MCB = 16.7 k # ft
497
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–14. Determine the moment at B, then draw the moment
diagram for each member of the frame. Assume the support at
A is fixed and C is pinned. EI is constant.
2 kN>m
A
Solution
B
3m
Fixed End Moments. Referring to the table on the inside back cover,
4m
2
1FEM2 AB = -
2132
wL2
= = - 1.50 kN # m
12
12
2
1FEM2 BA =
2132
wL2
=
= 1.50 kN # m
12
12
C
1FEM2 BC = 0
Slope-Deflection Equations. Applying Eq. 10–8 for member AB,
MN = 2Ek12uN + uF - 3c2 + 1FEM2 N
I
2EI
MAB = 2E a b[2102 + uB - 3102] + 1 - 1.502 = a
buB - 1.50(1)
3
3
I
4EI
MBA = 2E a b[2uB + 0 - 3102] + 1.50 = a
buB + 1.50(2)
3
3
Applying Eq. 10–10 for member BC,
MN = 3Ek1uN - c2 + 1FEM2 N
I
3EI
MBC = 3E a b 1uB - 02 + 0 = a
buB(3)
4
4
Equilibrium. At Joint B,
MBA + MBC = 0
a
4EI
3EI
buB + 1.50 + a
buB = 0
3
4
uB =
0.72
EI
Substitute this result into Eqs. (1) to (3).
MAB = - 1.98 kN # m
MBA = 0.540 kN # mAns.
MBC = -0.540 kN # mAns.
The negative signs indicate that MAB and MBC have counterclockwise rotational sense. Using these results, the shear at both ends of member AB and
BC are computed and shown in Figs. a and b, respectively. Subsequently,
the shear and moment diagrams can be plotted, Figs. c and d, respectively.
Ans.
MBA = 0.540 kN # m
MBC = -0.540 kN # m
498
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–15. The frame is made from pipe that is fixed connected.
If it supports the loading shown, determine the moments developed at each of the joints and supports. EI is constant.
18 kN
18 kN
4m
B
4m
4m
C
4m
A
D
Solution
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
MAB =
2EI
10 + uB 2 + 0
4
MBA =
2EI
12uB + 02 + 0
4
MBC =
211821122
2EI
12uB + uC 2 12
9
MCB =
211821122
2EI
12uC + uB 2 +
12
9
MCD =
2EI
12uC + 02 + 0
4
MDC =
2EI
10 + uC 2 + 0
4
Equilibrium:
MBA + MBC = 0
2EI
2EI
12uB 2 +
12uB + uC 2 - 48 = 0
4
12
1.333uB + 0.1667uC =
48
(1)
EI
MCB + MCD = 0
2EI
2EI
12uC + uB 2 + 48 +
12uC 2 = 0
12
4
1.333uC + 0.1667uB = -
48
(2)
EI
Solving Eqs. (1) and (2),
uB = -uC =
41.143
EI
MAB = 20.6 kN # m
MBA = 41.1 kN # m
MBC = - 41.1 kN # m
MCB = 41.1 kN # m
MCD = - 41.1 kN # m
MDC = - 20.6 kN # m
Ans.
Ans.
Ans.
MAB = 20.6 kN # m
MBA = 41.1 kN # m
MBC = - 41.1 kN # m
MCB = 41.1 kN # m
MCD = - 41.1 kN # m
MDC = - 20.6 kN # m
Ans.
Ans.
Ans.
Ans.
499
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–16. Determine the moment at each joint of the gable
frame. The roof load is transmitted to each of the purlins over
simply supported sections of the roof decking. Assume the
supports at A and E are pins and the joints are fixed connected. EI is constant.
5 ft
5 ft
C
5 ft
5 ft
5 ft
5 ft
1200 lb>ft
1200 lb>ft
B
D
A
E
12 ft
12 ft
Solution
1FEM2 BC = 1FEM2 CD =
- 21621152
9
= - 20 k # ft
1FEM2 CB = 1FEM2 DC = 20 k # ft
MN = 3E a
MBA =
I
b 1uN - c2 + 1FEM2 N
L
3EI
1uB 2
12
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
MBC =
2EI
12uB + uC 2 - 20
15
MCB =
2EI
12uC + uB 2 + 20
15
MCD =
2EI
12uC + uD 2 - 20
15
MDC =
2EI
12uD + uC 2 + 20
15
MN = 3E a
MDE =
I
b 1uN - c2 + 1FEM2 N
L
3EI
1uD 2
12
Equilibrium:
MBA + MBC = 0
MCB + MCD = 0
MDC + MDE = 0
500
12 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–16.
(Continued)
or
3EI
2EI
uB +
12uB + uC 2 - 20 = 0
12
15
0.5167uB + 0.1333uC =
20
EI
2EI
2EI
12uC + uB 2 + 20 +
12uC + uD 2 - 20 = 0
15
15
4uC + uB + uD = 0
2EI
3EI
12uD + uC 2 + 20 +
u = 0
15
12 D
0.51667uD + 0.1333uC = -
20
EI
Solving these equations,
uC = 0
uB = -uD =
38.71
EI
Thus,
MBA = 9.68 k # ft
MBC = - 9.68 k # ft
MCB = 25.2 k # ft
MCD = - 25.2 k # ft
MDC = 9.68 k # ft
MDE = - 9.68 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
MBA = 9.68 k # ft
MBC = - 9.68 k # ft
MCB = 25.2 k # ft
MCD = - 25.2 k # ft
MDC = 9.68 k # ft
MDE = - 9.68 k # ft
501
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–17. Determine the moments at B and D, then draw the
moment diagram. Assume A and C are pinned and B and D
are fixed connected. EI is constant.
8k
15 ft
A
10 ft
B
10 ft
C
12 ft
D
Solution
1FEM2 BA = 0
1FEM2 BC =
- 31821202
16
MN = 3E a
I
b 1uN - c2 + 1FEM2 N
L
MBA = 3E a
I
b 1uB - 02 + 0
15
= - 30 k # ft
1FEM2 BD = 1FEM2 DB = 0
MBA = 0.2EIuB(1)
MBC = 3E a
I
b 1uB - 02 - 30
20
MBC = 0.15EIuB - 30(2)
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
MBD = 2E a
I
b 12uB + 0 - 02 + 0
12
MBD = 0.3333EIuB(3)
MDB = 2E a
I
b 12102 + uB - 02 + 0
12
MDB = 0.1667EIuB(4)
Equilibrium.
MBA + MBC + MBD = 0(5)
Solving Eqs. 1–5,
uB =
43.90
EI
MBA = 8.78 k # ft
MBC = -23.4 k # ft
MBD = 14.6 k # ft
MDB = 7.32 k # ft
Ans.
Ans.
Ans.
Ans.
MBA = 8.78 k # ft
MBC = - 23.4 k # ft
MBD = 14.6 k # ft
MDB = 7.32 k # ft
Ans.
502
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–18. Determine the moments at A, B, and C, then draw
the moment diagram for each member. Assume all joints are
fixed connected. EI is constant.
4 k>ft
B
A
18 ft
9 ft
C
Solution
1FEM2 AB =
- 41182 2
12
= - 108 k # ft
1FEM2 BA = 108 k # ft
1FEM2 BC = 1FEM2 CB = 0
MN = 2E a
I
b12uN + uF - 3c2 + 1FEM2 N
L
MAB = 2E a
I
b12102 + uB - 02 - 108
18
MAB = 0.1111EIuB - 108(1)
MBA = 2E a
I
b12uB + 0 - 02 + 108
18
MBA = 0.2222EIuB + 108(2)
I
MBC = 2E a b12uB + 0 - 02 + 0
9
MBC = 0.4444EIuB(3)
I
MCB = 2E a b12102 + uB - 02 + 0
9
MCB = 0.2222EIuB(4)
Equilibrium.
MBA + MBC = 0(5)
Solving Eqs. 1–5,
uB =
- 162.0
EI
MAB = -126 k # ft
MBA = 72 k # ft
MBC = -72 k # ft
MCB = -36 k # ft
Ans.
Ans.
Ans.
Ans.
MAB = - 126 k # ft
MBA = 72 k # ft
MBC = - 72 k # ft
MCB = - 36 k # ft
Ans.
503
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–19. Determine the moment that each member exerts on
the joint at B, then draw the moment diagram for each member of the frame. Assume the support at A is fixed and C is a
pin. EI is constant.
2 k>ft
B
C
15 ft
6 ft
10 k
6 ft
A
Solution
Fixed End Moments. Referring to the table on the inside back
cover,
1FEM2 AB = -
101122
PL
=
= - 15 k # ft
8
8
1FEM2 BC = -
2115 2
wL2
= = - 56.25 k # ft
8
8
1FEM2 BA =
101122
PL
=
= 15 k # ft
8
8
2
Slope Reflection Equations. Applying Eq. 10–8 for member AB,
MN = 2Ek12uN + uF - 3c2 + 1FEM2 N
MAB = 2E a
I
EI
b[2102 + uB - 3102] + 1 - 152 = a buB - 15(1)
12
6
MBA = 2E a
I
EI
b[2uB + 0 - 3102] + 15 = a buB + 15(2)
12
3
For member BC, applying Eq. 10–10,
MN = 3Ek1uN - c2 + 1FEM2 N
MBC = 3E a
I
EI
b 1uB - 02 + 1 - 56.252 = a buB - 56.25(3)
15
5
504
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–19.
(Continued)
Equilibrium. At joint B,
MBA + MBC = 0
a
EI
EI
buB + 15 + a buB - 56.25 = 0
3
5
uB =
77.34375
EI
Substitute this result into Eqs. (1) to (3).
MAB = -2.109 k # ft = - 2.11 k # ft
MBA = 40.78 k # ft = 40.8 k # ft
MBC = - 40.78 k # ft = -40.8 k # ft
Ans.
Ans.
The negative signs indicate that MAB and MBC have counterclockwise rotational sense. Using these results, the shear at
both ends of member AB and BC are computed and shown in
Figs. a and b, respectively. Subsequently, the shear and moment
diagrams can be plotted, Figs. c and d, respectively.
Ans.
MBA = 40.8 k # ft
MBC = - 40.8 k # ft
505
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–20. The frame at the rear of the truck is made by welding
pipe segments together. If the applied load is 1500 lb,
determine the moments at the fixed joints B, C, D, and E.
Assume the supports at A and F are pinned. EI is constant.
1.5 ft
1 ft
C
2 ft
1 ft
1.5 ft
D
B
E
1500 lb
4 ft
A
Solution
FEMCD = -
PL
= - 375;
8
FEMDC =
PL
= 375
8
cAB = cBC = cCD = cDE = cEF = 0
Applying Eqs. 11–8 and 11–10,
MBA =
3EI
1uB - 02 + 0
4
MBC =
2EI
12uB + uC - 02 + 0
2.5
MCB =
2EI
12uC + uB - 02 + 0
2.5
MCD =
2EI
12uC + uD - 02 - 375
2
MDC =
2EI
12uD + uC - 02 + 375
2
MDE =
2EI
12uD + uE - 02 + 0
2.5
MED =
2EI
12uE + uD - 02 + 0
2.5
MEF =
3EI
1uE - 02 + 0
4
Moment equilibrium at B, C, and D, E:
MBA + MBC = 0
3EI
2EI
1uB 2 +
12uB + uC 2 = 0
4
2.5
uC = - 2.9375uB(1)
MCB + MCD = 0
2EI
2EI
12uC + uB 2 +
12uC + uD 2 - 375 = 0
2.5
2
3.6uC + 0.8uB + uD =
375
(2)
EI
506
F
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–20.
(Continued)
MDC + MDE = 0
2EI
2EI
12uD + uC 2 +
12uD + uE 2 + 375 = 0
2
2.5
3.6uD + 0.8uE + uC = -
375
(3)
EI
MED + MEF = 0
2EI
3EI
12uE + uD 2 +
1uE 2 = 0
2.5
4
uD = - 2.9375uE(4)
Solving Eqs. (1) through (4),
uB = -uE =
- 54.845
EI
uC = -uD =
161.106
EI
Thus,
MBA = - 41.1 lb # ft
MEF = 41.1 lb # ft
MBC = 41.1 lb # ft
MED = - 41.1 lb # ft
MCB = 214 lb # ft
MDE = - 214 lb # ft
MCD = - 214 lb # ft
MDC = 214 lb # ft
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
MBA = - 41.1 lb # ft
MEF = 41.1 lb # ft
MBC = 41.1 lb # ft
MED = - 41.1 lb # ft
MCB = 214 lb # ft
MDE = - 214 lb # ft
MCD = - 214 lb # ft
MDC = 214 lb # ft
507
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–21. Wind loads are transmitted to the frame at joint E.
If A, B, E, D, and F are all pin connected and C is fixed connected, determine the moments at joint C and draw the moment diagram for the girder BCE. EI is constant.
E
B
Solution
12 kN
C
8m
cBC = cCE = 0
cAB = cCD = cEF = c
Applying Eq. 10–10,
A
MCB =
3EI
1uC - 02 + 0
6
MCE =
3EI
1uC - 02 + 0
4
MCD =
3EI
1uC - c2 + 0(1)
8
D
6m
F
4m
Moment equilibrium at C:
MCB + MCE + MCD = 0
3EI
3EI
3EI
1uC 2 +
1uC 2 +
1uC - c2 = 0
6
4
8
c = 4.333uC(2)
From FBDs of members AB and EF:
b + a MB = 0;
b + a ME = 0;
VA = 0
VF = 0
Since AB and FE are two-force members, then for the entire frame:
+
S
a Fx = 0;
VD - 12 = 0;
VD = 12 kN
From FBD of member CD:
b + a MC = 0;
MCD - 12182 = 0
MCD = 96 kN # m
From Eq. (1),
3
EI1uC - 4.333uC 2
8
- 76.8
uC =
EI
96 =
From Eq. (2),
c =
Thus,
- 332.8
EI
MCB = -38.4 kN # m
MCE = -57.6 kN # m
Ans.
MCB = -38.4 kN # m
MCE = -57.6 kN # m
Ans.
Ans.
508
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–22. Determine the moments at joints A, B, C, and D, then
draw the moment diagram for each member of the frame.
Assume the supports at A and B are fixed. EI is constant.
D
C
3m
A
B
30 kN>m
3m
Solution
Fixed End Moments. Referring to the table on the inside back cover,
2
1FEM2 AD = -
3013 2
wL2
= = - 13.5 kN # m
20
20
2
1FEM2 DA =
3013 2
wL2
=
= 9 kN # m
30
30
1FEM2 DC = 1FEM2 CD = 1FEM2 CB = 1FEM2 BC = 0.
Slope-Deflection Equations. Here, cAD = cDA = cBC = cCB = c and cCD = cDC = 0.
Applying Eq. 10–8,
MN = 2Ek12uN + uF - 3c2 + 1FEM2 N
For member AD,
I
2EI
buD - 2EIc - 13.5(1)
MAD = 2E a b[2102 + uD - 3c] + 1 - 13.52 = a
3
3
I
4EI
buD - 2EIc + 9(2)
MDA = 2E a b[2uD + 0 - 3c] + 9 = a
3
3
For member CD,
I
4EI
2EI
buD + a
buC(3)
MDC = 2E a b[2uD + uC - 3102] + 0 = a
3
3
3
I
4EI
2EI
MCD = 2E a b[2uC + uD - 3102] + 0 = a
buC + a
buD(4)
3
3
3
For member BC,
I
2EI
MBC = 2E a b[2102 + uC - 3c] + 0 = a
buC - 2EIc(5)
3
3
I
4EI
MCB = 2E a b[2uC + 0 - 3c] + 0 = a
buC - 2EIc(6)
3
3
509
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–22.
(Continued)
Equilibrium. At Joint D,
MDA + MDC = 0
a
4EI
4EI
2EI
buD - 2EIc + 9 + a
buD + a
buC = 0
3
3
3
a
8EI
2EI
buD + a
buC - 2EIc = - 9(7)
3
3
At joint C,
MCD + MCB = 0
a
4EI
2EI
4EI
buC + a
buD + a
buC - 2EIc = 0
3
3
3
a
2EI
8EI
buD + a
buC - 2EIc = 0(8)
3
3
Consider the horizontal force equilibrium for the entire frame.
+
S
a Fx = 0;
1
1302132 - VA - VB = 0
2
Referring to the FBD of members AD and BC in Fig. a,
a + a MD = 0;
1
1302132122 - MDA - MAD - VA 132 = 0
2
VA = 30 -
MDA
MAD
3
3
and
a + a MC = 0;
- MCB - MBC - VB 132 = 0
VB = -
MCB
MBC
3
3
Thus,
MBC
MCB
MDA
MAD
1
1302132 - a30 b - a
b = 0
2
3
3
3
3
MDA + MAD + MCB + MBC = - 45
510
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–22.
a
(Continued)
4EI
2EI
4EI
buD - 2EIc + 9a
buD - 2EIc - 13.5 + a
buC - 2EIc
3
3
3
+a
2EI
buC - 2EIc = -45
3
2EIuD + 2EIuC - 2EIc = -40.5(9)
Solving of Eqs. (7), (8) and (9),
uC =
261
56EI
uD =
9
56EI
c =
351
56EI
Substitute these results into Eqs. (1) to (6),
MAD = - 25.93 kN # m = - 25.9 kN # m
MDA = - 3.321 kN # m = - 3.32 kN # m
MDC = 3.321 kN # m = 3.32 kN # m
MCD = 6.321 kN # m = 6.32 kN # m
MBC = - 9.429 kN # m = - 9.43 kN # m
MCB = - 6.321 kN # m = - 6.32 kN # m
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
The negative signs indicate that MAD, MDA, MBC, and MCB have
counterclockwise rotational sense. Using these results, the
shear at both ends of member AD, CD, and BC are computed
and shown in Figs. b, c, and d, respectively. Subsequently, the
shear and moment diagrams can be plotted, Figs. e and d, respectively.
Ans.
MAD = - 25.9 kN # m
MDA = - 3.32 kN # m
MDC = 3.32 kN # m
MCD = 6.32 kN # m
MBC = -9.43 kN # m
MCB = -6.32 kN # m
511
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–23. Determine the moments at each joint and fixed support, then draw the moment diagram. EI is constant.
20 ft
8k
B
C
10 ft
15 ft
D
A
Solution
1FEM2 AB = 1FEM2 BA = 0
1FEM2 BC = 1FEM2 CB = 0
1FEM2 CD = 1FEM2 DC = 0
cAB =
2
c
3 DC
MN = 2E a
I
b 120N + uF - 3c2 + 1FEM2 N
L
MAB = 2E a
I
2
b a2102 + uB - 3a bcDC b + 0
15
3
MAB = 0.1333EIuB - 0.2667EIcDC(1)
MBA = 2E a
I
2
b a2uB + 0 - 3a bcDC b + 0
15
3
MBA = 0.2667EIuB - 0.2667EIcDC(2)
MBC = 2E a
I
b 12uB + uC - 31022 + 0
20
MBC = 0.2EIuB + 0.1EIuC(3)
MCB = 2E a
I
b 12uC + uB - 31022 + 0
20
MCB = 0.2EIuC + 0.1EIuB(4)
MCD = 2E a
I
b 12uC + 0 - 3cDC 2 + 0
10
MCD = 0.4EIuC - 0.6EIcDC(5)
MDC = 2E a
I
b 13102 + uC - 3cDC 2 + 0
10
MDC = 0.2EIuC - 0.6EIcDC(6)
Equilibrium:
MBA + MBC = 0(7)
MCB + MCD = 0(8)
VA + VD - 8 = 0
-
1MAB + MBA 2
15
-
1MCD + MDC 2
10
- 8 = 0
2MAB + 2MBA + 3MCD + 3MDC = - 240(9)
512
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–23.
(Continued)
Solving these equations,
uB =
33.149
EI
cDC =
89.503
EI
uC =
MAB = -19.4 k # ft
MBA = -15.0 k # ft
MBC = 15.0 k # ft
MCB = 20.1 k # ft
MCD = - 20.1 k # ft
MDC = - 36.9 k # ft
83.978
EI
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
MAB = - 19.4 k # ft
MBA = - 15.0 k # ft
MBC = 15.0 k # ft
MCB = 20.1 k # ft
MCD = - 20.1 k # ft
MDC = - 36.9 k # ft
513
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–24. Determine the moments acting at the supports A
and D of the battered-column frame. Take E = 29(103) ksi,
I = 600 in4.
4 k>ft
6k
C
B
20 ft
A
D
15 ft
Solution
1FEM2 BC = -
wL2
= - 1600 k # in.
12
uA = uD = 0 cAB = cCD =
1FEM2 CB =
∆
= c
25
cBC = -
wL2
= 1600 k # in.
12
1.2∆
= -1.5c
20
MN = 2E a
I
b 12uN + uF - 3c2 + 1FEM2 N
L
MAB = 2E a
600
b 10 + uB - 3c2 + 0 = 116 000uB - 348 000c
251122
MBA = 2E a
600
b 12uB + 0 - 3c2 + 0 = 232 000uB - 348 000c
251122
MBC = 2E a
600
b 12uB + uC - 31 - 1.5c22 - 1600
201122
= 290 00uB + 145 000uC + 652 500c - 1600
MCB = 2E a
600
b 12uC + uB - 31 - 1.5c22 + 1600
201122
= 290 000uC + 145 000uB + 652 500c + 1600
MCD = 2E a
600
b 12uC + 0 - 3c2 + 0
201122
= 232 000uC - 348 000c
MDC = 2E a
600
b 10 + uC - 3c2 + 0
251122
= 116 000uC - 348 000c
Moment equilibrium at B and C:
MBA + MBC = 0
522 000uB + 145 000uC + 304 500c = 1600(1)
MCB + MCD = 0
145 000uB + 522 000uC + 304 500c = - 1600(2)
514
20 ft
15 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*10–24.
(Continued)
Using the FBD of the frame,
b + ΣMO = 0;
MAB + MDC - a
-a
MBA + MAB
b141.66721122
251122
MDC + MCD
b 141.66721122 - 6113.33321122 = 0
251122
- 0.667MAB - 0.667MDC - 1.667MBA - 1.667MCD - 960 = 0
464 000uB + 464 000uC - 1 624 00uc = - 960(3)
Solving Eqs. (1), (2), and (3),
uB = 0.004030 rad
uC = -0.004458 rad
c = 0.0004688 rad
MAB = 25.4 k # ft Ans.
MBA = 64.3 k # ft
MBC = - 64.3 k # ft
MCB = 99.8 k # ft
MCD = - 99.8 k # ft
MDC = - 56.7 k # ft
Ans.
Ans.
MAB = 25.4 k # ft
MDC = - 56.7 k # ft
515
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–1P. The roof is supported by joists that rest on two girders.
Each joist can be considered simply supported, and the front
girder can be considered attached to the three columns by a
pin at A and rollers at B and C. Assume the roof will be made
from 3 in.-thick cinder concrete, and each joist has a weight of
550 lb. According to code the roof will be subjected to a snow
loading of 25 psf. The joists have a length of 25 ft. Draw the
shear and moment diagrams for the girder. Assume the
supporting columns are rigid.
3 ft
A
Solution
From the text,
Weight of cinder concrete = 1108 lb>ft3 2a
3
ft b = 27 psf
12
Live load = 25 psf
Total load = 52 psf
Load on joist = 152 lb>ft2 213 ft2 = 156 lb>ft
Reaction on middle joist = 156a
Reaction on end joist =
1FEM2 BA =
25
550
= 2.225 k
b +
2
2
156 25
550
a b +
= 1.25 k
2
2
2
2.225192
PL
=
= 6.675 k # ft
3
3
2
1FEM2 BC = - Σ
+ c 162 2 192 +
1122 132
P
a 2b
2.225
ab2a +
b =
c 132 2 1122 +
d
2
2
2
2
L
1152
192 2 162
2
d + c 192 2 162 +
162 2 192
2
d + c 1122 2 132 +
= -20.025 k # ft
MN =
3EI
1uN - c2 + 1FEM2 N
L
MBA =
3EI
1uB - 02 + 6.675
9
MBC =
3EI
1uB - 02 - 20.025
15
MBA - MBC = 0
3EI
3EI
u + 6.675 +
1uB 2 - 20.025 = 0
9 B
15
uB =
MBA =
25.03
EI
3EI 25.03
b + 6.675 = 15.02 k # ft
a
9
EI
516
132 2 1122
2
d
3 ft
3 ft
3 ft
B
3 ft
3 ft
3 ft
3 ft
C
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10–1P.
(Continued)
MBC =
3EI 25.03
a
b - 20.025 = - 15.02 k # ft
15
EI
a + ΣMA = 0;
- 15.02 + 2.225132 + 2.225162 + 1.250192 - Ay 192 = 0
Ay = 1.806 k
+ c ΣFy = 0;
1.806 - 1.250 - 2.225122 + VBL = 0
VBL = 3.894 k
a + ΣMB = 0;
Cx 1152 - 2.225132 - 2.225162 - 2.225192 - 2.2251122 - 1.2501152 + 15.019 = 0
Cy = 4.699 k
+ c ΣFy = 0;
VBR - 412.2252 - 1.250 + 4.699 = 0
VBR = 5.45 k
M max = 14.0 k # ft
Ans.
Ans.
M max = 14.0 k # ft
517
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–1. Determine the moments at A, B, and C, then draw the
moment diagram for the beam. The moment of inertia of each
span is indicated in the figure. Assume the support at B is a
roller and A and C are fixed. E = 29(103) ksi.
30 k
800 lb>ft
C
A
IAB 5 900 in4
B
IBC 5 1200 in4
24 ft
8 ft
8 ft
Solution
(DF)AB = 0 (DF)BA =
(DF)BC = 0.6667
(FEM)AB =
0.75IBC >24
0.75IBC >24 + IBC >16
= 0.3333
(DF)CB = 0
- 0.8(24)2
12
= - 38.4 k # ft
(FEM)BA = 38.4 k # ft
(FEM)BC = -
30(16)
8
= - 60.0 k # ft
(FEM)CB = 60.0 k # ft
Joint
A
C
B
Mem.
AB
BA
BC
CB
DF
0
0.3333
0.6667
0
FEM
- 38.4
38.4
- 60.0
60.0
7.20
14.40
aM
3.60
- 34.8
MAB = - 34.8 k # ft
MBA = 45.6 k # ft
MBC = -45.6 k # ft
MCB = 67.2 k # ft
7.20
45.6
- 45.6
67.2 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
MAB = - 34.8 k # ft
MBA = 45.6 k # ft
MBC = - 45.6 k # ft
MCB = 67.2 k # ft
518
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–2. Determine the moments at B and C. EI is constant.
Assume B and C are rollers and A and D are pinned.
3 k>ft
A
B
8 ft
C
20 ft
D
8 ft
Solution
FEMAB = FEMCD =
wL2
= - 16
12
FEMBA = FEMDC =
wL2
= 16
12
wL2
wL2
= - 100 FEMCB =
= 100
12
12
FEMBC = KAB =
3EI
,
8
KBC =
4EI
,
20
KCD =
3EI
8
DFAB = 1 = DFDC
3EI
8
= 0.652
DFBA = DFCD =
3EI
4EI
+
8
20
DFBC = DFCB = 1 - 0.652 = 0.348
Joint
A
Member
AB
B
C
D
BA
BC
CB
CD
DC
DF
1
0.652
0.348
0.348
0.652
1
FEM
- 16
16
Dist.
16
- 100
100
-16
16
54.782
29.218
- 29.218
-54.782
CO
8
- 14.609
14.609
-8
Dist.
4.310
2.299
- 2.299
-4.310
CO
Dist.
0.750
CO
Dist.
0.130
CO
Dist.
aM
0.023
0
84.0
- 1.149
1.149
0.400
- 0.400
-0.750
- 0.200
0.200
0.070
- 0.070
- 0.035
0.035
0.012
- 0.012
-0.023
84.0
-84.0
- 84.0
-16
-0.130
519
0 k # ft
Ans.
Ans.
MBA = 84.0 k # ft
MBC = -84.0 k # ft
MCB = 84.0 k # ft
MCD = -84.0 k # ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–3. Determine the reactions at the supports. Assume A is
fixed and B and C are rollers that can either push or pull on the
beam. EI is constant.
12 kN>m
A
B
5m
C
2.5 m
Solution
Member Stiffness Factor and Distribution Factor.
KAB =
4EI
4EI
=
= 0.8EI
LAB
5
(DF)AB = 0
(DF)BC =
(DF)BA =
KBC =
3EI
3EI
=
= 1.2EI
LBC
2.5
0.8EI
= 0.4
0.8EI + 1.2EI
1.2 EI
= 0.6
0.8EI + 1.2EI
(DF)CB = 1
Fixed End Moments. Referring to the table on the inside
back cover,
2
(FEM)AB =
12(15 )
wL2
=
= - 25 kN # m
12
12
(FEM)BA =
12(5 )
wL2
=
= 25 kN # m
12
12
2
(FEM)BC = (FEM)CB = 0
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
CB
DF
0
0.4
0.6
1
FEM
- 25
25
0
0
- 10
- 15
B
Dist.
CO
aM
C
-5
- 30
15
- 15 kN # m
Using these results, the shear at both ends of members AB and
BC are computed and shown in Fig. a.
From this figure,
Ax = 0 Ay = 33 kN c
MA = 30 kN # ma
By = 27 + 6 = 33 kN c Cy = 6 kNT Ans.
Ax = 0
Ans.
Ay = 33 kN c
Ans.
By = 33 kN c
MA = 30 kN # mB
Cy = 6 kNT
520
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–4. Determine the reactions at the supports and then
draw the moment diagram. Assume A is fixed. EI is constant.
500 lb
800 lb>ft
20 ft
D
C
B
A
20 ft
15 ft
500 lb
800 lb>ft
10.4 k · ft
20 ft
Solution
FEMBC =
1.555 k
A
Member
AB
BA
BC
CB
CD
DF
0
0.5
0.5
1
0
- 26.67
26.67
-7.5
13.33
-19.167
- 9.583
6.667
B
FEM
Dist.
13.33
6.667
Dist.
4.7917
2.396
Dist.
CO
1.667
0.8333
Dist.
CO
0.5990
0.2994
Dist.
CO
0.2083
0.1042
Dist.
aM
7.84 k
4EI
20
= 0.5
DFAB = 0 DFCB = 1 DFBA = DFBC =
4EI
4EI
+
20
20
Joint
CO
15 ft
wL2
wL2
= - 26.67, FEMCB =
= 26.67 MCD = -0.5(15) = - 7.5 k # ft
12
12
4EI
4EI
, KBC =
KAB =
20
20
CO
20 ft
10.215 k
D
C
B
A
0.07485
10.4
20.7
C
4.7917
-6.667
- 3.333
2.396
1.667
-2.396
- 1.1979
0.8333
0.5990
-0.8333
- 0.4167
0.2994
0.2083
-0.2994
- 0.1497
0.07485
- 20.7
MA = 10.4 k # ftA
Ans.
Ax = 0
Ans.
0.1042
Ay = 1.56 kT Ans.
-0.1042
By = 10.2 k c Ans.
Cy = 7.84 k c Ans.
7.5
-7.5 k # ft
Ans.
MA = 10.4 k # ftA
Ax = 0
Ay = 1.56 kT
By = 10.2 k c
Cy = 7.84 k c
521
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–5. Determine the moments at A, B, and C. Assume the
support at B is a roller and A and C are fixed. EI is constant.
3 k>ft
2 k>ft
A
B
36 ft
C
24 ft
Solution
(DF)AB = 0 (DF)BA =
I>36
I>36 + I>24
= 0.4
(DF)BC = 0.6 (DF)CB = 0
(FEM)AB =
- 2(36)2
12
= - 216 k # ft
(FEM)BA = 216 k # ft
(FEM)BC =
- 3(24)2
12
= - 144 k # ft
(FEM)CB = 144 k # ft
Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.4
0.6
0
FEM
- 216
216
- 144
144
- 28.8
- 43.2
Dist.
CO
aM
B
C
- 14.4
- 230
- 21.6
187
- 187
- 122 k # ft
Ans.
Ans.
MAB = -230 k # ft
MBA = 187 k # ft
MBC = -187 k # ft
MCB = - 122 k # ft
522
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–6. Determine the moments at B and C, then draw the
moment diagram for the beam. The supports at A, B, C, and D
are pins. Assume the horizontal reactions are zero. EI is
constant.
12 kN>m
C
A
4m
B
4m
D
4m
12 kN>m
Solution
Member Stiffness Factor and Distribution Factor.
KAB =
3EI
3EI
=
LAB
4
KBC =
(DF)AB = 1 (DF)BA =
6EI
6EI
3EI
=
=
LBC
4
2
3EI>4
3EI>4 + 3EI>2
=
1
3
(DF)BC =
3EI>2
3EI>4 + 3EI>2
=
2
3
Fixed End Moments. Referring to the table on the inside back cover,
2
(FEM)BA =
12(4 )
wL2
=
= 24 kN # m (FEM)BC = 0
8
8
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
DF
1
1/3
2/3
FEM
0
24
0
-8
-16
16
-16
Dist.
aM
0
B
Ans.
Using these results, the shear at both ends of members AB, BC, and
CD are computed and shown in Fig. a. Subsequently the shear and
moment diagrams can be plotted, Figs. b and c, respectively.
Ans.
MB = -16.0 kN # m
MC = 16.0 kN # m
523
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–7. Determine the moments at A, B, and C, then draw the
moment diagram. Assume the support at B is a roller and A
and C are fixed. EI is constant.
900 lb 900 lb
B
A
6 ft
400 lb
6 ft
6 ft
C
10 ft
10 ft
Solution
(DF)AB = 0 (DF)BA =
I>18
I>18 + I>20
= 0.5263
(DF)CB = 0 (DF)BC = 0.4737
(FEM)AB =
- 2(0.9)(18)
9
= - 3.60 k # ft
(FEM)BA = 3.60 k # ft
(FEM)BC =
- 0.4(20)
8
= - 1.00 k # ft
(FEM)CB = 1.00 k # ft
Joint
A
Mem.
AB
B
BA
C
BC
CB
DF
0
0.5263
0.4737
0
FEM
- 3.60
3.60
- 1.00
1.00
- 1.368
- 1.232
Dist.
CO
aM
- 0.684
- 4.28
- 0.616
2.23
- 2.23
0.384 k # ft
Ans.
Ans.
MAB = -4.28 k # ft
MBA = 2.23 k # ft
MBA = -2.23 k # ft
MCB = 0.384 k # ft
524
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–8. Determine the moments at B and C, then draw the
moment diagram for the beam. Assume the supports at B and
C are rollers and A and D are pins. EI is constant.
12 kN>m
A
12 kN>m
B
4m
C
6m
D
4m
Solution
Member Stiffness Factor and Distribution Factor.
KAB =
3EI
3EI
=
LAB
4
(DF)BA =
KBC =
3EI>4
3EI>4 + 3EI>3
=
2EI
2EI
EI
=
=
LBC
6
3
9
13
(DF)BC =
(DF)AB = 1
EI>3
3EI>4 + EI>3
=
4
13
Fixed End Moments. Referring to the table on the inside back cover,
2
(FEM)AB = (FEM)BC = 0
(FEM)BA =
12(4 )
wL2
=
= 24 kN # m
8
8
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
DF
1
9
13
4
13
FEM
0
Dist.
aM
0
B
24
0
- 6.62
- 7.38
7.38
- 7.38
kN # m
Ans.
Due to symmetry,
MCB = 7.38 kN # m MCD = - 7.38 kN # m
Ans.
Using these results, the shear at both ends of members AB, BC,
and CD are computed and shown in Fig. a. Subsequently, the shear
and moment diagram can be plotted, Figs. b and c, respectively.
Ans.
MCB = 7.38 kN # m
MCD = -7.38 kN # m
525
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–9. The bar is pin supported at points A, B, C, and D. If the
normal force in the bar can be neglected, determine the vertical
reaction at each pin. EI is constant.
16 kN
A
4m
4m
B
8m
C
4m
4m
D
16 kN
Solution
Use antisymmetric load and symmetric beam.
KBA =
3EI
8
KBC =
6EI
8
3EI
8
= 0.3333
(DF)BA =
3EI
6EI
+
8
8
6EI
8
= 0.6667
(DF)BC =
3EI
6EI
+
8
8
FEMBA =
(3)(16)(8)
16
Joint
= 24 kN # m
A
B
Member
AB
BA
DF
1
0.3333 0.6667
FEM
ΣM
BC
24
0
-8
- 16
16
- 16
kN # m
Segment AB:
a + ΣMB = 0;
- Ay(8) + 16(4) - 16 = 0 Ay = 6 kN
c + ΣFy = 0;
VBL + 6 - 16 = 0
VBL = 10 kN
a + ΣMC = 0;
- VBR(8) + 16 + 16 = 0
VBR = 4 kN
c + ΣFy = 0;
- VCL + 4 = 0
VCL = 4 kN
Ans.
Segment BC:
Segment CD:
Ans.
a + ΣMC = 0;
- Dy(8) + 16(4) - 16 = 0 Dy = 6 kN
c + ΣFy = 0;
-VCR - 6 + 16 = 0
Ans.
VCR = 10 kN
Ay = 6 kN
Dy = 6 kN
By = VBL + VBR = 10 + 4 = 14 kN
Ans.
By = 14 kN
Cy = VCL + VCR = 4 + 10 = 14 kN
Ans.
Cy = 14 kN
526
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
300 lb
11–10. Determine the moments at B and C, then draw the
moment diagram for the beam. Assume the supports at B and
C are rollers and A is a pin. EI is constant.
200 lb>ft
Solution
A
D
C
B
Member Stiffness Factor and Distribution Factor.
KAB =
3EI
3EI
=
= 0.3EI
LAB
10
(DF)BA =
0.3EI
3
=
0.3EI + 0.4EI
7
KBC =
10 ft
4EI
4EI
=
= 0.4EI
LBC
10
(DF)BC =
0.4EI
4
=
0.3EI + 0.4EI
7
(DF)CB = 1 (DF)CD = 0
Fixed End Moments. Referring to the table on the inside
back cover,
(FEM)CD = -300(8) = -2400 lb # ft (FEM)BC = (FEM)CB = 0
(FEM)BA =
200(102)
wL2AB
=
= 2500 lb # ft
8
8
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
CB
CD
DF
1
3
7
4
7
1
0
FEM
0
2500
0
0
-2400
B
- 1071.43
Dist.
- 514.29
- 685.71
714.29
357.15
-342.86
- 153.06
- 204.09
342.86
171.43
-102.05
CO
- 73.47
Dist.
CO
- 21.87
Dist.
48.98
- 10.50
- 13.99
14.58
7.29
-7.00
- 3.12
- 4.17
7.00
3.50
-2.08
- 2.00
2.08
1.04
-1.00
- 0.45
- 0.59
1.00
0.500
-0.30
- 0.21
- 0.29
0.30
0.15
-0.15
- 0.06
- 0.09
0.15
0.07
-0.04
- 0.03
- 0.04
0.04
650
- 650
2400
- 1.50
CO
Dist.
CO
Dist.
CO
Dist.
CO
Dist.
aM
0
-48.98
-14.58
CO
Dist.
102.05
51.03
24.49
CO
Dist.
- 97.96
- 29.16
CO
Dist.
2400
-714.29
CO
Dist.
- 1428.57
1200
CO
Dist.
C
-2400 lb # ft
527
Ans.
10 ft
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–10.
(Continued)
Using these results, the shear at both ends of members AB,
BC, and CD are computed and shown in Fig. a. Subsequently,
the shear and moment diagrams can be plotted, Figs. b and c,
­respectively.
Ans.
MBA = 650 lb # ft
MBC = -650 lb # ft
MCB = 2400 lb # ft
MCD = -2400 lb # ft
528
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–11. Determine the moments at A and B, then draw the
moment diagram. Assume the support at B is a roller, C is a
pin, and A is fixed.
12 kN>m
C
B
A
5m
2.5 m
Solution
(DF)AB = 0
(DF)BA =
4I>5
4I>5 + 3I>2.5
= 0.4
(DF)BC = 0.6 (DF)CB = 1
(FEM)AB =
- 12(5)2
= - 25 kN # m
12
(FEM)BA = 25 kN # m
(FEM)BC = (FEM)CB = 0
Joint
A
B
C
Mem.
AB
BA
BC
CB
DF
0
0.4
0.6
1
- 25
25
FEM
- 10
- 15
15
- 15
-5
ΣM
- 30
MAB = -30 kN # m
MBA = 15 kN # m
MBC = - 15 kN # m
0
Ans.
Ans.
Ans.
Ans.
MAB = - 30 kN # m
MBA = 15 kN # m
MBC = - 15 kN # m
529
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–12. Determine the moments at B and C, then draw the
moment diagram for the beam. Assume C is a fixed support. EI
is constant.
12 kN
8 kN>m
A
B
C
6m
Solution
4m
Moment Distribution. Tabulating the above data,
Member Stiffness Factor and Distribution Factor.
3EI
3EI
EI
KBA =
=
=
LBA
6
2
(DF)AB = 1 (DF)BA =
(DF)BC =
4m
4EI
4EI
EI
KBC =
=
=
LBC
8
2
EI>2
EI>2 + EI>2
= 0.5
Joint
A
Member
AB
BA
BC
CB
DF
1
0.5
0.5
0
FEM
0
Dist.
B
C
36
-12
-12
-12
CO
EI>2
EI>2 + EI>2
aM
= 0.5 (DF)CB = 0
12
-6
0
24
-24
6 kN # m
Ans.
Fixed End Moments.
Using these results, the shear and both ends of members AB
and BC are computed and shown in Fig. a. Subsequently, the
shear and moment diagrams can be plotted, Fig. b.
Referring to the table on the inside back cover,
2
(FEM)BA =
8(6 )
wL2
=
= 36 kN # m
8
8
(FEM)BC = (FEM)CB =
12(8)
PL
= = - 12 kN # m
8
8
12(8)
PL
=
= 12 kN # m
8
8
Ans.
MAB = 0 kN # m
MBA = 24 kN # m
MBA = -24 kN # m
MCB = -6 kN # m
530
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–13. Determine the moments at the ends of each member
of the frame. Assume the joint at B is fixed, C is pinned, and A
is fixed. The moment of inertia of each member is listed in the
figure. E = 29(103) ksi.
2 k>ft
B
IBC 5 800 in4
C
12 ft
8 ft
4k
IAB 5 550 in4
8 ft
A
Solution
(DF)AB = 0
(DF)BA =
4(0.6875IBC)>16
4(0.6875IBC)>16 + 3IBC >12
= 0.4074
(DF)BC = 0.5926 (DF)CB = 1
(FEM)AB =
(FEM)BC =
- 4(16)
8
- 2(122)
12
= - 8 k # ft (FEM)BA = 8 k # ft
= - 24 k # ft (FEM)CB = 24 k # ft
Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.4074
0.5926
1
FEM
- 8.0
Dist.
CO
aM
8.0
6.518
C
- 24.0
9.482
24.0
-24.0
- 12.0
3.259
Dist.
CO
B
4.889
7.111
2.444
- 2.30
19.4
- 19.4
0 k # ft
Ans.
Ans.
MAB = - 2.30 k # ft
MBA = 19.4 k # ft
MBC = - 19.4 k # ft
MCB = 0 k # ft
531
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–14. Determine the internal moments acting at each joint.
Assume A, D, and E are pinned and B and C are fixed joints.
The moment of inertia of each member is listed in the figure.
Take E = 29(103) ksi.
4 k>ft
A
B
6 ft
IABC 5 800 in4
(DF)AB = 1
(DF)BA =
3(IABC)>6
3(IABC)>6 + 4(IABC)>18 + 3(0.75IABC)>12
= 0.5496
(DF)BC = 0.2443
(DF)BD = 0.2061
(DF)CB =
4(IABC)>18
4(IABC)>18 + 3(1.25IABC)>12
= 0.4156
(DF)CE = 0.5844
(DF)DB = (DF)EC = 1
(FEM)AB = -
4(6)2
12
= - 12.0 k # ft
(FEM)BA = 12.0 k # ft
(FEM)BC = -
4(18)2
12
= - 108 k # ft
(FEM)CB = 108 k # ft
(FEM)CE = -
20(12)
8
= - 30.0 k # ft
(FEM)EC = 30.0 k # ft
(FEM)BD = (FEM)DB = 0
Joint
A
Mem.
AB
BA
BD
BC
CB
CE
DF
1
0.5496
0.2061
0.2443
0.4156
FEM
- 12.0
12.0
- 108
108
B
12.0 52.76
19.79
6.0
ΣM
0
C
E
D
EC
DB
0.5844
1
1
- 30
30
23.45
- 32.42 - 45.58
-30
- 15.0
1.91
- 16.21
11.73
5.61
2.10
2.49
1.36
0.68
1.25
- 0.37
- 0.14
- 0.17
- 0.52
- 0.26
- 0.08
- 0.73
0.14
0.05
0.06
0.03
0.05
76.2
21.8
- 98.0
89.4
- 89.4
0
532
0
6 ft
18 ft
IBD 5 600 in4
ICE 5 1000 in4
D
Solution
C
20 k
6 ft
E
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–14.
(Continued)
MAB = 0
MBA = 76.2 k # ft
MBD = 21.8 k # ft
MBC = - 98.0 k # ft
MCB = 89.4 k # ft
MCE = - 89.4 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
MEC = 0
Ans.
MDB = 0
Ans.
Ans.
MAB = 0
MBA = 76.2 k # ft
MBD = 21.8 k # ft
MBC = - 98.0 k # ft
MCB = 89.4 k # ft
MCE = - 89.4 k # ft
MEC = 0
MDB = 0
533
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–15. Determine the reactions at A and D. Assume the
supports at A and D are fixed and B and C are fixed connected.
EI is constant.
8 k>ft
B
C
15 ft
A
D
24 ft
Solution
(DF)AB = (DF)DC = 0
I>15
(DF)BA = (DF)CD =
I>15 + I>24
(DF)BC = (DF)CB = 0.3846
= 0.6154
(FEM)AB = (FEM)BA = 0
(FEM)BC =
- 8(24)2
12
(FEM)CB = 384 k # ft
= - 384 k # ft
(FEM)CD = (FEM)DC = 0
Joint
A
Mem.
AB
BA
BC
DF
0
0.6154
0.3846
B
- 384
FEM
236.31
118.16
45.44
22.72
8.74
4.37
1.68
0.84
0.32
0.16
146.28
CB
0.3846
D
CD
DC
0.6154
0
384
147.69
- 147.69
- 73.84
73.84
28.40
-28.40
- 14.20
14.20
5.46
- 5.46
- 2.73
2.73
1.05
- 1.05
- 0.53
0.53
0.20
- 0.20
- 0.10
0.10
-236.31
-118.16
-45.44
-22.72
-8.74
-4.37
-1.68
-0.84
-0.33
-0.17
0.04
- 0.04
- 0.02
0.02
0.01
0.01
- 0.01
-0.01
292.57
- 292.57
292.57
-292.57
0.06
0.03
aM
C
-0.06
-0.03
Dx = 29.3 k
Ay = 96.0 k
Dy = 96.0 k
MA = 146 k # ft
MD = 146 k # ft
Ay = 96.0 k
MA = 146 k # ft
Thus, from the free-body diagrams,
Ax = 29.3 k
-146.28
Ans.
Ax = 29.3 k
Ans.
Dx = 29.3 k
Ans.
Dy = 96.0 k
MD = 146 k # ft
Ans.
534
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 kN
*11–16. The frame is made from pipe that is fixed connected. If
it supports the loading shown, determine the moments developed
at each of the joints. EI is constant.
18 kN
C
B
4m
D
A
4m
4m
4m
Solution
FEMBC = -
2PL
2PL
= - 48, FEMCB =
= 48
9
9
KAB = KCD =
4EI
4EI
, KBC =
4
12
DFAB = DFDC = 0
DFBA = DFCD = 4EI
4EI
4
4
+ 4EI
12
= 0.75
DFBC = DFCB = 1 - 0.75 = 0.25
Joint
A
Member
AB
BA
BC
CB
CD
DC
DF
0
0.75
0.25
0.25
0.75
0
B
FEM
Dist.
CO
36
18
Dist.
CO
4.5
2.25
Dist.
CO
ΣM
0.281
0.0704
20.6
- 48
48
12
-12
-6
6
1.5
-1.5
- 0.75
0.5625
Dist.
C
41.1
D
-36
-18
-4.5
0.75
-2.25
0.1875
-0.1875
- 0.0938
0.0938
0.0234
-0.0234
-0.0704
41.1
-41.1
- 41.1
-0.5625
-0.281
-20.6 kN # m
Ans.
Ans.
MAB = 20.6 kN # m
MBA = 41.1 kN # m
MBC = -41.1 kN # m
MCB = 41.1 kN # m
MCD = - 41.1 kN # m
MDC = - 20.6 kN # m
535
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–17. Determine the moments at the fixed support A and
joint D and then draw the moment diagram for the frame.
Assume B is pinned.
4 k>ft
A
12 ft
12 ft
12 ft
B
Solution
Member Stiffness Factor and Distribution Factor.
KAD =
4EI
4EI
EI
=
=
L
12
3
(DF)AD = 0
EI>3
(DF)DA =
(DF)DC = (DF)DB =
3EI
3EI
EI
=
=
L
12
4
KDC = KDB =
EI>3 + EI>4 + EI>4
EI>4
EI>3 + EI>4 + EI>4
= 0.4
= 0.3 (DF)CD = (DF)BD = 1
Fixed End Moments. Referring to the table on the inside
back cover,
2
(FEM)AD = -
4(12 )
wL2
= = - 48 k # ft
12
12
2
(FEM)DA =
4(12 )
wL2
= = 48 k # ft
12
12
2
(FEM)DC = -
4(12 )
wL2
= = - 72 k # ft
8
8
(FEM)CD = (FEM)BD = (FEM)DB = 0
Moments Distribution. Tabulating the above data,
Joint
A
Member
AD
DA
DB
DF
0
0.4
0.3
FEM
- 48
48
0
9.60
7.20
7.20
57.6
7.20
- 64.8
k # ft
Dist.
CO
aM
D
C
B
DC
CD
BD
0.3
1
1
- 72
0
0
0
0
4.80
- 43.2
536
Ans.
C
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–17.
(Continued)
Using these results, the shears at both ends of members AD,
CD, and BD are computed and shown in Fig. a. Subsequently,
the shear and moment diagrams can be plotted, Figs. b and c,
respectively.
Ans.
MAD = - 43.2 k # ft
MDA = 57.6 k # ft
MDB = 7.20 k # ft
MDC = - 64.8 k # ft
MCD = 0 k # ft
MBD = 0 k # ft
537
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–18. Determine the moments at A, C, and D, then
draw the moment diagram for each member of the frame.
Support A and joints C and D are fixed connected. EI is
constant.
6 kN>m
C
D
E
6m
A
B
8m
7m
Solution
FEMAC = 0 FEMBD = 0
2
FEMCD = -
6(8 )
wL2
= = - 32 kN # m = -FEMDC
12
12
FEMDE = -
6(7 )
wL2
= = - 36.75 kN # m
8
8
2
Joint
A
B
Member
AC
BD
CA
CD
DC
DB
DF
0
1
0.5714
0.4286
0.35
0.35
- 32
32
18.286
13.714
1.6625
0.83125
6.8571
0.47500
- 0.35625 - 2.4000
C
FEM
Dist.
CO
9.1429
Dist.
CO
aM
8.905
MAC = 8.91 kN # m
E
DE
ED
0.3
1
-36.75
1.6625
1.4250
-2.4000
-2.0571
0.053438
- 1.2000
- 0.17813
0.68571
0.51429
0.062344 0.062344
18.50
- 18.50
- 0.23750
Dist.
D
38.00
-0.6752
-37.33 kN # m
Ans.
MBD = 0
MCA = 18.5 kN # m
MCD = - 18.5 kN # m
MDC = 38 kN # m
MDB = - 0.675 kN # m
MDE = - 37.3 kN # m
Ans.
Ans.
Ans.
Ans.
Ans.
MED = 0
Ans.
MAC = 8.91 kN # m
MCA = 18.5 kN # m
MCD = -18.5 kN # m
MDC = 38 kN # m
MDB = -0.675 kN # m
MDE = -37.3 kN # m
538
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–19. Determine the moment at B, then draw the moment
diagram for each member of the frame. Support A is pinned. EI
is constant.
1.5 k>ft
B
A
12 ft
C
12 ft
5 ft
Solution
FEMBC = 0
2
FEMBA =
1.5(12 )
wL2
=
= 27 k # ft
8
8
Joint
A
Member
AB
BA
BC
CB
DF
1
0.44828
0.55172
0
FEM
B
C
27
- 12.103
- 14.897
- 7.4485
ΣM
0
MB = -14.9 k # ft
14.897
- 14.897
- 7.4485
Ans.
Ans.
MB = - 14.9 k # ft
539
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10 k
*11–20. Determine the moments at B and C, then draw the
moment diagram for each member of the frame. Assume the
supports at A, E, and D are fixed. EI is constant.
2 k>ft
A
8 ft
8 ft
C
B
12 ft
16 ft
Solution
E
D
Member Stiffness Factor and Distribution Factor.
KAB =
4EI
4EI
EI
=
=
LAB
12
3
KBC = KBE = KCD =
(DF)AB = (DF)EB = (DF)DC = 0 (DF)BA =
(DF)BC = (DF)BE =
(DF)CB = (DF)CD =
EI>4
EI>3 + EI>4 + EI>4
EI>4
EI>4 + EI>4
4EI
4EI
EI
=
=
L
16
4
EI>3
EI>3 + EI>4 + EI>4
= 0.4
= 0.3
= 0.5
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = (FEM)BA =
2(122)
wL2AB
=
= 24 k # ft
12
12
(FEM)BC = (FEM)CB =
2(122)
wL2AB
= = - 24 k # ft
12
12
10(16)
PLBC
= = - 20 k # ft
8
8
10(16)
PLBC
=
= 20 k # ft
8
8
(FEM)BE = (FEM)EB = (FEM)CD = (FEM)DC = 0
Moment Distribution. Tabulating the above data,
Joint
A
B
Member
AB
BA
BE
DF
0
0.4
0.3
FEM
- 24
Dist.
CO
CO
CO
aM
EB
0
0
0
0
-5
-0.6
20
0
-10
-10
-5
-0.60
1.50
0.30
0.15
0.75
- 0.045
-0.375
0.30
- 0.06
- 0.045
- 0.1875
-0.0225
0.075
0.05625
0.05625
0.01125
0.005625
0.028125
- 0.00225
- 0.0016875
- 0.0016875
-0.01406
-0.01406
24.41
0.3096
- 24.72
10.08
-10.08 k # ft
540
E
DC
- 1.20
1.50
D
0.5
- 20
0.0375
- 23.79
CD
0
- 0.03
Dist.
0.5
- 1.20
2.00
Dist.
CB
0.3
24
1.00
Dist.
BC
- 1.60
- 0.80
Dist.
CO
C
0.15
0.75
-0.1875
-0.0225
-0.375
0.01125
0.005625
0.028125
-5.032
0.1556
Ans.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–20.
(Continued)
Using these results, the shear at both ends of members AB, BC,
BE, and CD are computed and shown in Fig. a. Subsequently,
the shear and moment diagrams can be plotted.
541
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–20.
(Continued)
Ans.
ΣM
-23.79
24.41 0.3096
542
-24.72 10.08
-10.08 k # ft
-5.032 0.1556
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–21. Determine the moments at D and C, then draw the
moment diagram for each member of the frame. Assume the
supports at A and B are pins. EI is constant.
16 kN
1m
3m
D
C
A
B
4m
Solution
KDA = KCB =
3EI
3EI
=
L
4
KCD =
4EI
4EI
=
= EI
L
4
(DF)AD = (DF)BC = 1 (DF)DA = (DF)CB =
(DF)DC = (DF)CD =
3EI>4
3EI>4 + EI
=
3
7
EI
4
=
3EI>4 + EI
7
543
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–21.
(Continued)
Moment Distribution. No sidesway, Fig. b.
Joint
A
Member
AD
DA
DC
CD
CB
BC
DF
1
3
7
4
7
4
7
3
7
1
FEM
0
0
-9
3
0
0
3.857
5.143
- 1.714
-1.286
Dist.
D
CO
Dist.
0.367
CO
Dist.
0.315
CO
Dist.
0.030
CO
Dist.
0.026
CO
Dist.
aM
0
C
- 0.857
2.572
0.490
- 1.470
- 0.735
0.245
0.420
- 0.140
- 0.070
0.210
0.040
- 0.120
- 0.060
0.020
0.034
- 0.011
B
-1.102
-0.105
-0.090
-0.009
- 0.006
0.017
0.003
0.003
- 0.010
-0.007
4.598
- 4.598
2.599
-2.599
kN # m
0
Using these results, the shears at A and B are computed and shown in Fig. d. Thus,
for the entire frame,
+
S
ΣFx = 0;
1.1495 - 0.6498 - R = 0 R = 0.4997 kN
(FEM)DC = -
16(3 )(1)
Pb2a
= = - 9 kN # m
L2
42
2
2
(FEM)CD =
16(1 )(3)
Pa 2b
=
= 3 kN # m
L2
42
544
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–21.
(Continued)
For the frame in Fig. e,
Joint
A
Member
AD
DA
DC
CD
CB
BC
DF
1
3
7
4
7
4
7
3
7
1
FEM
0
- 10
0
0
-10
0
4.286
5.714
5.714
4.286
2.857
2.857
- 1.633
- 1.633
- 0.817
- 0.817
0.350
0.467
0.467
0.234
0.234
- 0.100
- 0.134
- 0.134
- 0.067
- 0.067
0.038
0.038
0.019
0.019
- 0.008
- 0.011
- 0.011
-0.008
- 6.667
6.667
6.667
-6.667
kN # m
Dist.
D
CO
- 1.224
Dist.
CO
Dist.
CO
Dist.
CO
Dist.
0.029
CO
Dist.
aM
0
C
B
-1.224
0.350
-0.100
0.029
0
Using these results, the shears at A and B caused by the
­application of R′ are computed and shown in Fig. f. For the
entire frame,
+
S
ΣFx = 0;
R′ - 1.667 - 1.667 = 0
R′ = 3.334 kN
Thus,
MDA = 4.598 + ( - 6.667) a
0.4997
b = 3.60 kN # m
3.334
Ans.
MDC = - 4.598 + (6.667) a
0.4997
b = - 3.60 kN # m
3.334
Ans.
MCD = 2.599 + (6.667) a
0.4997
b = 3.60 kN # m
3.334
MCB = - 2.599 + ( - 6.667) a
0.4997
b = - 3.60 kN # m
3.334
Ans.
Ans.
Ans.
MDA = 3.60 kN # m
MDC = - 3.60 kN # m
MCD = 3.60 kN # m
MCB = - 3.60 kN # m
545
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–22. Determine the moments acting at the ends of each
member of the frame. EI is the constant.
1.5 k>ft
15 k
B
C
A
D
20 ft
24 ft
(DF)AB = (DF)DC = 1
(DF)BA = (DF)CD =
3I>20
3I>20 + 4I>24
= 0.4737
(DF)BC = (DF)CB = 0.5263
Consider no sidesway
Joint
A
Member
AB
BA
DF
1
0.4737
B
FEM
Dist.
34.11
CO
Dist.
8.98
CO
Dist.
2.36
CO
Dist.
0.62
CO
Dist.
0.16
CO
Dist.
0.04
aM
BC
0.5263
D
CB
CD
DC
0.5263
0.4737
1
- 72.0
72.0
37.89
- 37.89
- 18.95
18.95
9.97
- 9.97
- 4.98
4.98
2.62
- 2.62
- 1.31
1.31
0.69
- 0.69
- 0.35
0.35
0.18
- 0.18
- 0.09
0.09
0.05
- 0.05
-34.11
-8.98
-2.36
-0.62
-0.16
-0.04
- 0.02
0.02
0.01
0.01
- 0.01
-0.01
46.28
- 46.28
46.28
-46.28
k # ft
CO
Dist.
C
(FEM)AB = (FEM)BA = 0
(FEM)BC =
- 1.5(24)2
12
(FEM)CB = 72 k # ft
= - 72 k # ft
(FEM)CD = (FEM)DC = 0
+ ΣF = 0 (for the frame without sidesway):
d
x
R + 2.314 - 2.314 - 15 = 0
R = 15.0 k
546
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–22.
(Continued)
Joint
A
Mem.
AB
DF
1
B
C
BA
BC
CB
0.4737
0.5263
0.5263
- 100
FEM
Dist.
52.63
26.32
26.32
- 12.47
- 13.85
- 13.85
- 6.93
-6.93
3.64
3.64
1.82
1.82
- 0.96
-0.96
- 0.48
-0.48
0.23
0.25
0.25
0.13
0.13
- 0.06
- 0.07
-0.07
CO
Dist.
3.28
CO
Dist.
- 0.86
CO
Dist.
CO
Dist.
aM
DC
0.4737
1
52.63
47.37
-12.47
3.28
-0.86
0.23
-0.06
- 0.03
-0.03
0.02
0.02
0.02
0.02
- 62.50
62.50
62.50
-62.50
k # ft
CO
Dist.
CD
-100
47.37
CO
Dist.
D
R′ = 3.125 + 3.125 = 6.25 k
MBA = 46.28 + a
15
b( -62.5) = -104 k # ft
6.25
Ans.
15
b(62.5) = 104 k # ft
6.25
Ans.
MBC = - 46.28 + a
MCB = 46.28 + a
15
b(62.5) = 196 k # ft
6.25
MCD = - 46.28 + a
MAB = MDC = 0
15
b( -62.5) = - 196 k # ft
6.25
Ans.
Ans.
Ans.
Ans.
MBA = - 104 k # ft
MBC = 104 k # ft
MCB = 196 k # ft
MCD = - 196 k # ft
MAB = MDC = 0
547
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–23. Determine the moments at the ends of each member
of the frame. The members are fixed connected at the supports
and joints. EI is the same for each member.
4 k>ft
20 k
B
C
6 ft
Solution
15 k
PL
wL2
FEMs for AB use
, and for BA use
. Each member
8
12
4EI
has a K =
.
L
6 ft
A
Joint
A
Member
AB
BA
BC
CB
CD
DC
DF
0
0.5556
0.4444
0.4444
0.5556
0
FEM
- 22.5
22.5
- 75
75
14.583 d 29.167
23.333
- 33.333
B
4.630 d
9.259
0.720 d
1.440
0.229 d
0.036 d
- 2.303
0.457
0.071
C
- 16.667
11.667
7.407
- 5.182
- 2.593
3.704
1.152
- 1.646
- 0.823
0.576
0.366
- 0.256
- 0.128
0.183
0.057
- 0.081
D
-41.667 d -20.833
-6.482 d -3.241
-2.058 d -1.029
-0.320 d -0.160
-0.102 d -0.051
- 0.041
0.028
0.023
0.018
- 0.013
-0.016
62.917
- 62.917
50.643
-50.643
Member AB:
b + ΣMB = 0;
62.917 - 2.303 - 15(6) + Ax(12) = 0
Ax = 2.449 k
Member CD:
b + ΣMC = 0;
- 50.643 - 25.314 + Dx(12) = 0
Dx = 6.330 k
Frame:
+
S
ΣFx = 0;
15 ft
R = 20 + 15 - 2.449 - 6.330 = 26.221
548
-25.314
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–23.
(Continued)
Joint
A
Member
AB
BA
BC
CB
CD
DC
DF
0
0.5556
0.4444
0.4444
0.5556
0
- 100
- 100
-100
-100
44.444
44.444
0.5556 d
27.778
22.222
22.222
- 9.877
-9.877
- 4.938
-4.938
2.195
2.195
1.097
1.097
- 0.488
-0.488
- 0.244
-0.244
0.108
0.108
0.054
0.054
- 0.030
- 0.024
-0.024
-0.030
- 54.551
54.551
54.551
-54.551
FEM
B
27.778 d- 55.556
- 6.173 d- 12.346
1.372 d
2.744
- 0.305 d - 0.610
0.068 d
- 77.260
0.135
C
D
-12.346 d -6.173
2.744 d
1.372
-0.610 d -0.305
0.135 d
0.068
-77.260
Member AB and CD:
b + ΣMB = 0;
Ax(12) - 54.55 - 77.260 = 0
+
S
ΣFx = 0;
Ax = Dx = 10.984 k
Frame:
R′ = 10.984 + 10.984 = 21.689 k
Thus
26.221
b = -94.5 k # ft
21.968
Ans.
26.221
b = -2.19 k # ft
21.968
Ans.
26.221
b = 2.19 k # ft
21.968
Ans.
MAB = - 2.303 + ( - 77.260) a
MBA = 62.917 + ( - 54.551) a
MBC = - 62.917 + ( + 54.551) a
MCB = 50.643 + (54.551) a
26.221
b = 116 k # ft
21.968
Ans.
26.221
b = -116 k # ft
21.968
Ans.
26.221
MDC = - 25.314 + ( - 77.260) a
b = -118 k # ft
21.968
Ans.
MCD = - 50.643 + ( - 54.551) a
549
Ans.
MAB = - 94.5 k # ft
MBA = - 2.19 k # ft
MBC = 2.19 k # ft
MCB = 116 k # ft
MCD = - 116 k # ft
MDC = - 118 k # ft
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–24. Determine the moments acting at the ends of each
member. Assume the supports at A and D are fixed. The
moment of inertia of each member is indicated in the figure.
E = 29(103) ksi.
6 k>ft
B
IBC 5 1200 in4
C
ICD 5 600 in4
15 ft
D
IAB 5 800 in4
A
Solution
24 ft
(DF)AB = (DF)DC = 0
(DF)BA =
2
a IBC b >15
3
2
a IBC b >15 + IBC >24
3
= 0.5161
(DF)BC = 0.4839
(DF)CB =
IBC >24
0.5IBC >10 + IBC >24
= 0.4545
(DF)CD = 0.5455
Consider no sidesway
Joint
A
Mem.
AB
BA
DF
0
0.5161
B
C
BC
0.4839
- 288
FEM
Dist.
148.64
CO
74.32
Dist.
33.78
CO
16.89
Dist.
8.18
CO
4.09
Dist.
1.86
CO
0.93
Dist.
0.45
CO
aM
96.50
0.4545
0.5455
0
288
-130.90
- 65.45
69.68
31.67
-31.67
- 15.84
15.84
-78.55
-38.01
-19.01
7.66
-7.20
- 3.60
3.83
1.74
-1.74
- 0.87
0.87
0.42
-0.40
- 6(24)2
12
-8.64
-4.32
-2.09
-1.04
-0.47
0.21
- 0.05
0.05
0.02
0.02
-0.02
-0.03
193.02
- 193.02
206.46
-206.46
(FEM)AB = (FEM)BA = 0
(FEM)BC =
-157.10
-0.10
0.05
Dist.
DC
0.10
0.10
CO
CD
- 0.20
0.22
Dist.
139.36
D
CB
= - 288 k # ft
(FEM)CB = 288 k # ft
(FEM)CD = (FEM)DC = 0
+
S
ΣFx = 0 (for the frame without sidesway):
R + 19.301 - 30.968 = 0
R = 11.666 k
550
-0.24
-0.11
-0.06
-103.22
k # ft
10 ft
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–24.
(Continued)
Joint
A
Mem.
AB
BA
BC
CB
CD
DC
DF
0
0.5161
0.4839
0.4545
0.5455
0
FEM
59.26
Dist.
CO
- 15.29
11.73
5.87
Dist.
CO
- 1.68
-0.84
Dist.
CO
CO
aM
49.28
100
- 28.68
- 45.45
- 22.73
- 14.34
11.00
6.52
3.26
5.50
- 1.58
-2.50
100
-54.55
-27.28
7.82
3.91
-3.00
- 1.25
-0.79
0.60
0.36
0.18
0.30
- 0.09
- 0.09
-0.14
- 0.07
-0.04
0.04
0.03
0.02
0.02
39.31
- 39.31
- 50.55
50.55
-0.05
Dist.
D
0.65
0.32
Dist.
C
59.26
- 30.58
Dist.
CO
B
-1.50
0.43
0.22
-0.16
-0.08
551
75.28
k # ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–24.
(Continued)
(FEM)CD = (FEM)DC = 100 =
(FEM)AB = (FEM)BA =
6E(0.75IAB)∆ ′
6EIAB ∆ ′
152
∆′ =
102
= a
6EIAB
152
ba
100(102)
6E(0.75IAB)
100(102)
6E(0.75IAB)
b = 59.26 k # ft
R′ = 5.906 + 12.585 = 18.489 k
MAB = 96.50 + a
11.666
b(49.28) = 128 k # ft
18.489
Ans.
11.666
b(39.31) = 218 k # ft
18.489
Ans.
MBA = 193.02 + a
MBC = -193.02 + a
MCB = 206.46 + a
11.666
b( -39.31) = -218 k # ft
18.489
11.666
b( -50.55) = 175 k # ft
18.489
Ans.
Ans.
MCD = - 206.46 + a
11.666
b(50.55) = - 175 k # ft
18.489
Ans.
MDC = - 103.22 + a
11.666
b(75.28) = - 55.7 k # ft
18.489
Ans.
Ans.
MAB = 128 k # ft
MBA = 218 k # ft
MBC = - 218 k # ft
MCB = 175 k # ft
MCD = - 175 k # ft
MDC = - 55.7 k # ft
552
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–25. Determine the moments at joints B and C, then draw
the moment diagram for each member of the frame. The
supports at A and D are pinned. EI is constant.
8k
C
B
12 ft
A
5 ft
Solution
KAB = KCD =
3EI
3EI
=
L
13
KBC =
4EI
2EI
=
10
5
(DF)AB = (DF)DC = 1 (DF)BA = (DF)CD =
(DF)BC = (DF)CB =
2EI>5
3EI>13 + 2EI>5
(FEM)BA = (FEM)CD = 100 k # ft;
3EI∆ ′
= 100
L2
∆′ =
=
3EI>13
3EI>13 + 2EI>5
=
26
41
16 900
3EI
From the geometry shown in Fig. b,
∆ ′BC =
5
5
10
∆′ +
∆′ =
∆′
13
13
13
Thus,
(FEM)BC = (FEM)CB = -
= -
6EI a
6EI∆ ′BC
L2BC
10 16 900
ba
b
13
3EI
102
= - 260 k # ft
553
15
41
D
10 ft
5 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–25.
(Continued)
Moment Distribution. For the frame with P acting at C, Fig. a,
Joint
A
Member
AB
DF
1
15>41
26>41
FEM
0
100
- 260
- 260
58.54
101.46
101.46
58.54
50.73
50.73
- 18.56
- 32.17
- 32.17
- 16.09
- 16.09
10.20
10.20
5.10
5.10
- 3.23
- 3.23
- 1.62
- 1.62
0.59
1.03
1.03
0.51
0.51
- 0.19
- 0.32
- 0.32
- 0.16
- 0.16
0.10
0.10
0.05
0.05
- 0.02
- 0.03
- 0.03
-0.02
144.44
- 144.44
- 144.44
144.44
Dist.
B
BA
CO
Dist.
CO
Dist.
5.89
CO
- 1.87
Dist.
CO
Dist.
CO
Dist.
CO
Dist.
0.06
CO
Dist.
aM
0
C
BC
CB
26>41
D
CD
DC
15>41
1
100
0
-18.56
5.89
-1.87
0.59
-0.19
0.06
554
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–25.
(Continued)
Using these results, the shears at A and D are computed and
shown in Fig. c. Thus, for the entire frame,
+
S
ΣFx = 0; 24.07 + 24.07 - P = 0 P = 48.14 k
Thus, for P = 8 k,
MBA = a
8
b(144.44) = 24.0 k # ft
48.14
Ans.
MBC = a
8
b( - 144.44) = - 24.0 k # ft
48.14
Ans.
MCB = a
8
b( - 144.44) = - 24.0 k # ft
48.14
Ans.
MCD = a
8
b(144.44) = 24.0 k # ft
48.14
Ans.
Ans.
MBA = 24.0 k # ft
MBC = - 24.0 k # ft
MCB = - 24.0 k # ft
MCD = 24.0 k # ft
555
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–26. Determine the moments acting at the fixed supports
A and D of the battered-column frame. EI is constant.
4 k>ft
6k
B
C
20 ft
A
D
15 ft
Solution
(DF)AB = (DF)DC = 0
(DF)BA = (DF)CD =
I>25
I>25 + I>20
= 0.4444
(DF)BC = (DF)CB = 0.5556
Consider no sidesway.
Joint
A
Mem.
AB
DF
0
B
BA
0.4444
FEM
Dist.
CO
59.25
29.63
Dist.
CO
16.46
8.23
Dist.
CO
CO
aM
- 4(20)2
12
0
- 37.04
37.04
20.58
- 20.58
- 10.29
10.29
5.72
- 5.72
- 2.86
2.86
1.59
- 1.59
- 0.79
0.79
0.44
- 0.44
-59.25
-29.63
-16.46
-8.23
-4.57
-2.29
-1.27
-0.64
-0.35
0.22
0.12
- 0.12
- 0.06
0.06
0.03
0.03
- 0.03
-0.03
82.03
- 82.03
82.03
-82.03
0.05
(FEM)BC =
0.4444
- 74.08
0.10
41.00
0.5556
- 0.22
0.18
Dist.
DC
74.08
0.35
Dist.
CD
133.33
1.27
0.64
Dist.
CO
0.5556
D
CB
- 133.33
4.57
2.29
Dist.
CO
C
BC
-0.18
-0.10
-0.05
= - 133.33 k # ft
(FEM)CB = 133.33 k # ft
(FEM)AB = (FEM)BA = (FEM)CD = (FEM)DC = 0
R + 36.15 - 36.15 - 6 = 0
R = 6k
556
-41.00
20 ft
15 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–26.
(Continued)
(FEM)AB = (FEM)BA =
(FEM)BC = (FEM)CB =
- 6EI∆ ′
= - 100 k # ft
(25)2
(6EI)2∆ ′ cos u
(20)
2
=
(FEM)CD = (FEM)DC = - 100 k # ft
Joint
A
Mem.
AB
DF
0
FEM
- 100
Dist.
CO
CO
CO
CO
CO
aM
C
BC
0.5556
CB
0.5556
- 48.62
-38.88
- 24.31
- 24.31
13.51
13.51
6.75
6.75
- 3.75
- 3.75
- 1.88
- 1.88
5.40
-3.00
-1.50
1.04
0.52
- 0.29
- 0.29
- 0.14
- 0.14
0.06
0.08
0.08
0.04
0.04
- 0.02
- 0.02
- 0.02
-0.02
- 130.44
130.44
130.44
-130.44
R′ = 22.065 k + 22.065 k = 44.130 k
6
MAB = 41.00 + a
b( -115.21) = 25.3 k # ft
44.130
6
b( -115.21) = -56.7 k # ft
44.130
0
-100
-19.44
1.04
0.03
DC
10.80
0.52
- 0.12
MDC = - 41.00 + a
0.4444
- 48.62
- 0.23
b = 187.5 k # ft
D
CD
- 38.88
0.84
- 115.21
6EI
-100
0.42
Dist.
(100)(25)2
187.5
- 3.00
Dist.
6EI
187.5
- 1.50
Dist.
CO
0.4444
5.40
Dist.
(20)
a
2
(100)(25)2
- 100
10.80
Dist.
(6EI)(2)(0.6)
B
BA
- 19.44
Dist.
∆′ =
0.84
0.42
-0.23
-0.12
0.06
0.03
-115.21
Ans.
Ans.
Ans.
MAB = 25.3 k # ft
MDC = - 56.7 k # ft
557
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–1. Determine (approximately) the force in each member of
the truss. Assume the diagonals can support either a tensile or a
compressive force.
5k
3k
F
4k
E
D
6 ft
A
C
B
8 ft
Solution
Support Reactions Referring to Fig. a,
a + ΣMA = 0;
a + ΣMC = 0;
NC(16) - 5(8) - 4(16) = 0
NC = 6.50 k
5(8) + 3(16) - Ay(16) = 0
Ay = 5.50 k
+
S
ΣFx = 0;
Ax = 0
Method of Sections. It is required that FBF = FAE = F1.
­ eferring to Fig. b,
R
3
+ c ΣFy = 0; 5.50 - 3 - 2F1 a b = 0 F1 = 2.083 k
5
Therefore,
FBF = 2.08 k (T) FAE = 2.08 k (C)
4
a + ΣMA = 0; FEF(6) - 2.083a b162 = 0 FEF = 1.667 k (C) = 1.67 k (C)
5
4
a + ΣMF = 0; FAB(6) - 2.083a b162 = 0 FAB = 1.667 k (T) = 1.67 k (T)
5
Ans.
Ans.
Ans.
Also, FBD = FCE = F2. Referring to Fig. c,
3
+ c ΣFy = 0; 6.50 - 4 - 2F2 a b = 0 F2 = 2.083 k
5
Therefore,
FBD = 2.08 k (T) FCE = 2.08 k (C)
4
a + ΣMC = 0; 2.083a b162 - FDE(6) = 0 FDE = 1.667 k (C) = 1.67 k (C)
5
4
a + ΣMD = 0; 2.083a b162 - FBC(6) = 0 FBC = 1.667 k (T) = 1.67 k (T)
5
Ans.
Ans.
Ans.
Method of Joints.
Joint A, Fig. d.
3
+ c ΣFy = 0; 5.50 - 2.083a b - FAF = 0 FAF = 4.25 k (C)
5
558
Ans.
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–1.
(Continued)
Joint B, Fig. e.
3
+ c ΣFy = 0; 2c 2.083a b d - FBE = 0 FBE = 2.50 k (C)
5
Ans.
3
+ c ΣFy = 0; 6.50 - 2.083a b - FCD = 0 FCD = 5.25 k (C)
5
Ans.
Joint C, Fig. f.
Ans.
FBF = 2.08 k (T) FAE = 2.08 k (C)
FEF = 1.67 k (C)
FAB = 1.67 k (T)
FBD = 2.08 k (T) FCE = 2.08 k (C)
FDE = 1.67 k (C)
FBC = 1.67 k (T)
FAF = 4.25 k (C)
FBE = 2.50 k (C)
FCD = 5.25 k (C)
559
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–2. Solve Prob. 12–1 assuming that the diagonals cannot
support a compressive force.
5k
3k
F
4k
E
D
6 ft
A
C
B
8 ft
Solution
Support Reactions. Referring to Fig. a,
a + ΣMA = 0;
a + ΣMC = 0;
+
S
ΣFx = 0;
NC(16) - 5(8) - 4(16) = 0
NC = 6.50 k
5(8) + 3(16) - Ay(16) = 0
Ay = 5.50 k
Ax = 0
Method of Sections. It is required that
Ans.
FAE = FCE = 0
Referring to Fig. b,
3
+ c ΣFy = 0; 5.50 - 3 - FBF a b = 0 FBF = 4.167 k (T) = 4.17 k (T)
5
4
a + ΣMA = 0; FEF(6) - 4.167a b(6) = 0 FEF = 3.333 k (C) = 3.33 k (C)
5
a + ΣMF = 0; FAB(6) = 0 FAB = 0
Ans.
Ans.
Ans.
Referring to Fig. c,
3
+ c ΣFy = 0; 6.50 - 4 - FBD a b = 0 FBD = 4.167 k (T) = 4.17 k (T)
5
Ans.
4
a + ΣMC = 0; 4.167a b(6) - FDE(6) = 0 FDE = 3.333 k (C) = 3.33 k (C) Ans.
5
a + ΣMD = 0;
Ans.
- FBC(6) = 0 FBC = 0
Method of Joints. Joint A, Fig. d.
Ans.
+ c ΣFy = 0; 5.50 - FAF = 0 FAF = 5.50 k (C)
Joint B, Fig. e.
3
+ c ΣFy = 0; 2c 4.167a b d - FBE = 0 FBE = 5.00 k (C)
5
Ans.
+ c ΣFy = 0; 6.50 - FCD = 0 FCD = 6.50 k (C)
Ans.
Joint C, Fig. f.
560
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–2.
(Continued)
Ans.
FAE = FCE = 0
FBF = 4.17 k (T)
FEF = 3.33 k (C)
FAB = 0
FBD = 4.17 k (T)
FDE = 3.33 k (C)
FBC = 0
FAF = 5.50 k (C)
FBE = 5.00 k (C)
FCD = 6.50 k (C)
561
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–3. Determine (approximately) the force in each member of
the truss. Assume the diagonals can support either a tensile or a
compressive force.
10 kN
5 kN
H
I
J
5 kN
F
G
4m
A
B
3m
Solution
4
4
F + FAI = 5 kN
5 JB
5
FJB = 3.125 kN (C)
FAI = 3.125 kN (T)
Joint A:
3
+
S
ΣFx = 0; 3.125 kNa b - FAB = 0
5
FAB = 1.875 kN (C)
4
+ c ΣFy = 0; 3.125 kNa b - FAJ = 0
5
FAJ = 2.50 kN (C)
Joint J:
3
+
S
ΣFx = 0; 3.125 kNa b - FJI = 0
5
FJI = 1.875 kN (T)
4
4
F
+ FFC = 5 kN
5 BH
5
FBH = 3.125 kN (C)
FFC = 3.125 kN (T)
a + ΣMB = 0;
3
5 kN (3 m) - FIH(4 m) - 3.125 kN a b(4 m) = 0
5
FIH = 1.875 kN (T)
a + ΣMI = 0;
3
5 kN (3 m) - FBC(4 m) - 3.125 kN a b(4 m) = 0
5
FBC = 1.875 kN (C)
562
C
3m
D
3m
E
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–3.
(Continued)
Joint I:
+ c ΣFy = 0;
4
FIB - 2(3.125 kN) a b = 0
5
FIB = 5 kN (C)
Due to Symmetry:
Joint H:
+ c ΣFy = 0;
4
FHC - 10 kN + 2(3.125 kN) a b = 0
5
FHC = 5 kN (C)
Due to Symmetry:
FAB = FDE = 1.875 kN (C)
Ans.
FBC = FDC = 1.875 kN (C)
Ans.
FJI = FGF = 1.875 kN (T) Ans.
FIH = FHG = 1.875 kN (T)
Ans.
FJB = FFD = 3.125 kN (C) Ans.
FAI = FGE = 3.125 kN (T) Ans.
FIC = FGC = 3.125 kN (T) Ans.
FBH = FHD = 3.125 kN (C)
Ans.
FJA = FEF = 2.50 kN (C) Ans.
FIB = FDG = 5 kN (C)
Ans.
FHC = 5 kN (C)
Ans.
Ans.
FAB = FDE = 1.875 kN (C);
FBC = FDC = 1.875 kN (C);
FJI = FGF = 1.875 kN (T);
FIH = FHG = 1.875 kN (T);
FJB = FFD = 3.125 kN (C);
FAI = FGE = 3.125 kN (T);
FIC = FGC = 3.125 kN (T);
FBH = FHD = 3.125 kN (C);
FJA = FEF = 2.50 kN (C)
FIB = FDG = 5 kN (C);
FHC = 5 kN (C)
563
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10 kN
5 kN
*12–4. Determine (approximately) the force in each member
of the truss. Assume the diagonals cannot support a compressive
force.
H
I
J
5 kN
F
G
4m
A
B
3m
Solution
4
F = 5 kN
5 AI
FAI = 6.25 kN (T)
FJB = 0
Joint A:
3
+
S
ΣFx = 0; 6.25 kN a b - FAB = 0
5
FAB = 3.75 kN (C)
4
+ c ΣFy = 0; 6.25 kN a b - FAJ = 0
5
FAJ = 5 kN (C)
Joint J:
+
S
ΣFx = 0; FJI = 0
4
F = 5 kN
5 IC
FIC = 6.25 kN (T)
FBH = 0
3
a + ΣMB = 0; 5 kN (3 m) - FIH(4 m) - 6.25 kNa b(4 m) = 0
5
FIH = 0
+
S
ΣFx = 0;
3
- FBC - 6.25 kNa b = 0
5
FBC = 3.75 kN (C)
564
C
3m
E
D
3m
3m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–4.
(Continued)
Joint I:
+ c ΣFy = 0;
4
FIB - 2(6.25 kN) a b = 0
5
FIB = 10 kN (C)
Due to Symmetry:
Joint H:
+ c ΣFy = 0;
FHC - 10 kN = 0
FHC = 10 kN (C)
Due to Symmetry:
FAB = FDE = 3.75 kN (C)
Ans.
FBC = FDC = 3.75 kN (C)
Ans.
FJI = FGF = 0
Ans.
FIH = FHG = 0
Ans.
FJB = FFD = 0
Ans.
FAI = FGE = 6.25 kN (T)
Ans.
FIC = FGC = 6.25 kN (T)
Ans.
FBH = FHD = 0
Ans.
FJA = FEF = 5 kN (C)
Ans.
Ans.
FIB = FDG = 10 kN (C)
Ans.
FAB = FDE = 3.75 kN (C)
FHC = 10 kN (C)
Ans.
FBC = FDC = 3.75 kN (C)
FJI = FGF = 0
FIH = FHG = 0
FJB = FFD = 0
FAI = FGE = 6.25 kN (T)
FIC = FGC = 6.25 kN (T)
FBH = FHD = 0
FJA = FEF = 5 kN (C)
FIB = FDG = 10 kN (C)
FHC = 10 kN (C)
565
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–5. Determine (approximately) the force in each member
of the truss. Assume the diagonals can support either a tensile
or a compressive force.
10 k
10 k
10 k
10 k
H
G
F
E
B
C
5k
20 ft
A
D
20 ft
20 ft
20 ft
Solution
10 k
10 k
10 k
10 k
VPanel = 8.33 k
H
G
F
E
B
C
Assume VPanel is carried equally by FHB and FAG, so
8.33
2
FHB =
= 5.89 k (T) Ans.
sin 45°
20 ft
Ax = 5k
8.33
2
FAG =
= 5.89 k (C) Ans.
sin 45°
A
D
20 ft
Ay = 18.33 k
Joint A:
+
S
a Fx = 0;
+ c a Fy = 0;
FAB - 5 - 5.89 cos 45° = 0; FAB = 9.17 k (T)
Ans.
- FAH + 18.33 - 5.89 sin 45° = 0; FAH = 14.16 k (C) Ans.
Joint H:
+
S
a Fx = 0;
- FHG + 5.89 cos 45° = 0; FHG = 4.17 k (C)
Ans.
VPanel = 1.667 k
1.667
2
FGC =
= 1.18 k (C)
sin 45°
Ans.
1.667
2
FBF =
= 1.18 k (T)
sin 45°
Ans.
Joint G:
+
S
a Fx = 0;
+ c a Fy = 0;
4.17 + 5.89 cos 45° - 1.18 cos 45° - FGF = 0
Ans.
FGF = 7.5 k (C)
- 10 + FGB + 5.89 sin 45° + 1.18 sin 45° = 0
Ans.
FGB = 5.0 k (C)
566
20 ft
20 ft
Dy = 21.667 k
5k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–5.
(Continued)
Joint B:
+
S
a Fx = 0; FBC + 1.18 cos 45° - 9.17 - 5.89 cos 45° = 0
FBC = 12.5 k (T)
Ans.
VPanel = 21.667 - 10 = 11.667 k
11.667
2
FEC =
= 8.25 k (T)
sin 45°
11.667
2
FDF =
= 8.25 k (C)
sin 45°
Ans.
Ans.
Joint D:
+
S
a Fx = 0;
+ c a Fy = 0;
FCD = 8.25 cos 45° = 5.83 k (T)
21.667 - 8.25 sin 45° - FED = 0
FED = 15.83 k (C)
Ans.
Joint E:
+
S
a Fx = 0;
5 + FFE - 8.25 cos 45° = 0
FFE = 0.833 k (C)
Ans.
Ans.
FHB = 5.89 k (T)
Joint C:
+ c a Fy = 0;
FAG = 5.89 k (C)
- FFC + 8.25 sin 45° - 1.18 sin 45° = 0
FFC = 5.0 k (C)
FAB = 9.17 k (T)
Ans.
FAH = 14.16 k (C)
FHG = 4.17 k (C)
FGC = 1.18 k (C)
FBF = 1.18 k (T)
FGF = 7.5 k (C)
FGB = 5.0 k (C)
FBC = 12.5 k (T)
FEC = 8.25 k (T)
FDF = 8.25 k (C)
FCD = 5.83 k (T)
FED = 15.83 k (C)
FFE = 0.833 k (C)
FFC = 5.0 k (C)
567
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–6. Solve Prob. 12–5 assuming that the diagonals cannot
support a compressive force.
10 k
10 k
10 k
10 k
H
G
F
E
B
C
5k
20 ft
A
D
20 ft
20 ft
20 ft
Solution
10 k
10 k
10 k
10 k
VPanel = 8.33 k
H
G
F
E
B
C
Ans.
8.33
= 11.785 = 11.8 k
sin 45°
Ans.
FAG = 0
FHB =
Joint A:
20 ft
Ax = 5k
+
S
a Fx = 0; FAB = 5 k (T)
Ans.
+ c a Fy = 0; FAH = 18.3 k (C)
Ans.
20 ft
Ay = 18.33 k
Joint H:
+
S
a Fx = 0; 11.785 cos 45° - FHG = 0
FHG = 8.33 k (C)
Ans.
VPanel = 1.667 k
Ans.
FGC = 0
FBF =
1.667
= 2.36 k (T)
sin 45°
Ans.
Joint B:
+
S
a Fx = 0; FBC + 2.36 cos 45° - 11.785 cos 45° - 5 = 0
Ans.
FBC = 11 .7 k (T)
+ c a Fy = 0;
- FGB + 11.785 sin 45° + 2.36 sin 45° = 0
FGB = 10 k (C)
Ans.
Joint G:
+
S
a Fx = 0; FGF = 8.33 k (C)
Ans.
VPanel = 11.667 k
FDF = 0
FEC =
11.667
= 16.5 k (T)
sin 45°
A
Ans.
Ans.
568
D
20 ft
20 ft
Dy = 21.667 k
5k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–6.
(Continued)
Joint D:
+
S
ΣFx = 0; FCD = 0 Ans.
+ c ΣFy = 0; FED = 21.7 k (C)
Ans.
Joint E:
+
S
ΣFx = 0; FEF + 5 - 16.5 cos 45° = 0
FEF = 6.67 k (C)
Ans.
Joint F:
+ c ΣFy = 0; FFC - 10 - 2.36 sin 45° = 0
FFC = 11.7 k (C)
Ans.
Ans.
FAG = 0
FHB = 11.8 k
FAB = 5 k (T)
FAH = 18.3 k (C)
FHG = 8.33 k (C)
FGC = 0
FBF = 2.36 k (T)
FBC = 11 .7 k (T)
FGB = 10 k (C)
FGF = 8.33 k (C)
FDF = 0
FEC = 16.5 k (T)
FCD = 0
FED = 21.7 k (C)
FEF = 6.67 k (C)
FFC = 11.7 k (C)
569
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–7. Determine (approximately) the force in each member of
the truss. Assume the diagonals can support either a tensile or
compressive force.
4k
8k
8k
4k
H
G
F
E
B
C
3k
20 ft
A
20 ft
20 ft
D
20 ft
Solution
Assume FBH = FGA.
+ c ΣFy = 0;
- 2FBH(cos 45°) - 4 + 11 = 0
Ans.
FGA = 4.950 k = 4.95 k (C)
Ans.
Ax = 3 k
Ans.
Ans.
Joint A:
- FAH - 4.950(sin 45°) + 11 = 0
FAH = 7.50 k (C)
Ans.
Assume FBF = FGC.
+ c ΣFy = 0;
a + ΣMC = 0;
- 2FBF(cos 45°) - 8 - 4 + 13 = 0
FBF = 0.7071 k = 0.707 k (T)
Ans.
FGC = 0.7071 k = 0.707 k (C)
Ans.
- FGF(20) - 4(20) + 0.7071(sin 45°)(20) + 13(20) = 0
FGF = 9.50 k (C)
+
S
ΣFx = 0;
Ans.
- FBC + 9.50 = 0
FBC = 9.50 k (T)
Ans.
Joint B:
+ c ΣFy = 0;
- FBG + 4.950(sin 45°) + 0.7071(sin 45°) = 0
FBG = 4.00 k (C)
Ans.
Assume FCE = FFD.
+ c ΣFy = 0;
a + ΣMD = 0;
- 2FCE(sin 45°) - 4 + 13 = 0
FCE = 6.364 k = 6.36 k (T)
Ans.
FFD = 6.364 k = 6.36 k (C)
Ans.
- FFE(20) + 6.364(cos 45°)(20) = 0
FFE = 4.50 k (C)
+
S
ΣFx = 0; FCD = 4.50 k (T) 4k
H
G
F
E
B
C
A
20 ft
Ay = 11 k
+
S
ΣFx = 0; FBA - 6.5 + 3 - 3 = 0
+ c ΣFy = 0;
8k
20 ft
a + ΣMA = 0; FGH(20) - 4.950(sin 45°)(20) - 3(20) = 0
FBA = 6.50 k (T)
8k
3k
FBH = 4.950 k = 4.95 k (T)
FGH = 6.50 k (C)
4k
Ans.
Ans.
570
20 ft
D
20 ft
Dy = 13 k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–7.
(Continued)
Joint C:
+ c ΣFy = 0;
- FCF - 0.7071(sin 45°) + 6.364(sin 45°) = 0
FCF = 4.00 k (C)
Ans.
Joint D:
+ c ΣFy = 0;
- FDF - 6.364(sin 45°) + 13 = 0
FDE = 8.50 k (C)
Ans.
Ans.
FBH = 4.95 k (T)
FGA = 4.95 k (C)
FGH = 6.50 k (C)
FBA = 6.50 k (T)
FAH = 7.50 k (C)
FBF = 0.707 k (T)
FGC = 0.707 k (C)
FGF = 9.50 k (C)
FBC = 9.50 k (T)
FBG = 4.00 k (C)
FFD = 6.36 k (C)
FCE = 6.36 k (T)
FFE = 4.50 k (C)
FCD = 4.50 k (T)
FCF = 4.00 k (C)
FDE = 8.50 k (C)
571
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–8. Determine (approximately) the force in each member
of the truss. Assume the diagonals cannot support a compressive
force.
4k
8k
8k
4k
H
G
F
E
B
C
3k
20 ft
A
20 ft
Solution
Ans.
Assume FGA = 0
+ c ΣFy = 0;
- FBH(sin 45°) - 4 + 11 = 0
Ans.
FBH = 9.899 k = 9.90 k (T)
a + ΣMA = 0; FGH(20) - 9.899(cos 45°)(20) - 3(20) = 0
Ans.
+
S
ΣFx = 0; FBA + 3 + 9.899(cos 45°) - 10 - 3 = 0
FBA = 3.00 k (T)
Ans.
+ c ΣFy = 0; FAH = 11.0 k (C)
Ans.
Assume FGC = 0
Ans.
Joint A:
+ c ΣFy = 0;
- FBF(sin 45°) - 8 - 4 + 13 = 0
a + ΣMC = 0;
FBF = 1.414 k = 1.41 k (T)
Ans.
- FGF(20) + 1.414(cos 45°)(20) - 4(20) + 13(20) = 0
Ans.
FGF = 10.0 k (C)
+
S
ΣFx = 0;
- FBC - 1.414(cos 45°) + 10 = 0
Ans.
FBC = 9.00 k (T)
Joint B:
+ c ΣFy = 0;
- FBG + 9.899(sin 45°) + 1.414(sin 45°) = 0
Ans.
FBG = 8.00 k (C)
Ans.
Assume FFD = 0
+ c ΣFy = 0;
- FCE(sin 45°) - 4 + 13 = 0
a + ΣMD = 0;
FCE = 12.73 k = 12.7 k (T)
Ans.
- FFE(20) + 12.73(cos 45°)(20) = 0
Ans.
FFE = 9.00 k (C)
+
S
ΣFx = 0;
- FCD - 12.73(cos 45°) + 9.00 = 0
Ans.
FCD = 0
572
20 ft
4k
8k
8k
4k
H
G
F
E
B
C
3k
20 ft
Ax = 3 k
FGH = 10.0 k (C)
20 ft
D
A
20 ft
Ay = 11 k
20 ft
D
20 ft
Dy = 13 k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–8.
(Continued)
Joint C:
+ c ΣFy = 0;
- FCF + 12.73(sin 45°) = 0
FCF = 9.00 k (C)
Ans.
+ c ΣFy = 0; FDE = 13.0 k (C)
Ans.
Joint D:
Ans.
FGA = 0
FBH = 9.90 k (T)
FGH = 10.0 k (C)
FBA = 3.00 k (T)
FAH = 11.0 k (C)
FGC = 0
FBF = 1.41 k (T)
FGF = 10.0 k (C)
FBC = 9.00 k (T)
FBG = 8.0 k (C)
FFD = 0
FCE = 12.7 k (T)
FFE = 9.00 k (C)
FCD = 0
FCF = 9.00 k (C)
FDE = 13.0 k (C)
573
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–9. Determine (approximately) the force in each member of
the truss. Assume the cross diagonals can support both tensile
and compressive forces.
40 kN
20 kN
4m
4m
B
A
40 kN
4m
C
D
4m
G
Solution
Method of Sections. It is required that FBF = FCG = F1. Referring to Fig. a,
+ c ΣFy = 0; 2 3F1 1 1 24 - 20 - 40 = 0 F1 = 3012 kN
Therefore,
12
FCG = 3012 kN (T) = 42.4 kN (T) FBF = 3012 kN (C) = 42.4 kN (C) Ans.
a + ΣMG = 0; 20(4) + (3012)1 1 2(4) - FBC(4) = 0 FBC = 50.0 kN (T)Ans.
12
a + ΣMB = 0; 20(4) + (3012)1 1 2(4) - FGF(4) = 0 FGF = 50.0 kN (C) Ans.
12
Also, FCE = FDF = F2. Referring to Fig. b,
+ c ΣFy = 0; 2 3F2 1 1 24 - 20 - 40 - 40 = 0 F2 = 5012 kN
Therefore,
12
FCE = 5012 kN (C) = 70.7 kN (C) FDF = 5012 kN (T) = 70.7 kN (T) Ans.
a + ΣMC = 0; 20(8) + 40(4) + (5012)1 1 2(4) - FEF(4) = 0
12
FEF = 130 kN (C)
a + ΣMF = 0; 20(8) + 40(4) + (5012)1 1 2(4) - FCD(4) = 0
12
FCD = 130 kN (T)
Ans.
Ans.
Method of Joints.
Joint A, Fig c.
+ c ΣFy = 0; FAG 1 1 2 - 20 = 0 FAG = 2012 kN (C) = 28.3 kN (C)
12
Ans.
+
S
ΣFx = 0; FAB - (2012)1 1 2 = 0 FAB = 20.0 kN (T) Ans.
+ c ΣFy = 0; FBG + (3012)1 1 2 - 40 = 0 FBG = 10.0 kN (C)
Ans.
Joint B, Fig. d.
12
12
574
F
E
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–9.
(Continued)
Joint C, Fig. e.
+ c ΣFy = 0; FCF + (5012)1 1 2 - (3012)1 1 2 - 40 = 0 FCF = 20.0 kN (C)
Ans.
+ c ΣFy = 0; FDE - (5012)1 1 2 = 0 FDE = 50.0 kN (T)
Ans.
12
12
Joint E, Fig. f.
12
Ans.
FCG = 42.4 kN(T)
FBF = 42.4 kN (C)
FBC = 50.0 kN (T)
FGF = 50.0 kN (C)
FCE = 70.7 kN (C)
FDF = 70.7 kN (T)
FEF = 130 kN (C)
FCD = 130 kN (T)
FAG = 28.3 kN (C)
FAB = 20.0 kN (T)
FBG = 10.0 kN (C)
FCF = 20.0 kN (C)
FDE = 50.0 kN (T)
575
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–10. Determine (approximately) the force in each member
of the truss. Assume the cross diagonals cannot support a
compressive force.
40 kN
20 kN
4m
A
40 kN
4m
4m
C
B
D
4m
G
F
Solution
Method of Sections. It is required that
Ans.
FBF = FCE = 0
Referring to Fig. a,
+ c ΣFy = 0;
a + ΣMB = 0;
a + ΣMG = 0;
FCG a
1
22
b - 20 - 40 = 0
20(4) + (6012) a
1
22
20(4) - FBC(4) = 0
Ans.
FCG = 6012 kN (T) = 84.9 kN (T)
b(4) - FGF(4) = 0
Ans.
FGF = 80.0 kN (C)
Ans.
FDF = 10012 kN (T) = 141 kN (T)
Ans.
FBC = 20.0 kN (T)
Referring to Fig. b,
+ c ΣFy = 0;
a + ΣMC = 0;
a + ΣMF = 0;
FDF a
1
b - 20 - 40 - 40 = 0
12
20(8) + 40(4) + (10012) a
FEF = 180 kN (C)
1
b(4) - FEF(4) = 0
12
20(8) + 40(4) - FCD(4) = 0
FCD = 80.0 kN (T)
576
Ans.
Ans.
E
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–10.
(Continued)
Method of Joints.
Joint A, Fig. c.
+ c ΣFy = 0;
+
S
ΣFx = 0;
Joint B, Fig. d.
+ c ΣFy = 0;
FAG a
1
b - 20 = 0
12
FAB - (2012) a
FBG - 40 = 0
FAG = 2012 kN (C) = 28.3 kN (C)
1
b = 0
12
FAB = 20.0 kN (T)
FBG = 40.0 kN (C)
Ans.
Ans.
Ans.
Ans.
Joint C, Fig. e.
+ c ΣFy = 0;
Joint D, Fig. f.
+ c ΣFy = 0;
FCF - 40 - (6012) a
1
b = 0
12
FCF = 100 kN (C)
Ans.
FDE = 0
Ans.
FBF = FCE = 0
FCG = 84.9 kN (T)
FGF = 80.0 kN (C)
FBC = 20.0 kN (T)
FDF = 141 kN (T)
FEF = 180 kN (C)
FCD = 80.0 kN (T)
FAG = 28.3 kN (C)
FAB = 20.0 kN (T)
FBG = 40.0 kN (C)
FCF = 100 kN (C)
FDE = 0
577
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–11. Determine (approximately) the force in each member of
the truss. Assume the diagonals can support either a tensile or
compressive force.
E
8 kN
1.5 m
D
2m
10 kN
F
C
Solution
Method of Sections. It is required that FCE = FDF = F1. Referring to Fig. a,
+
S
ΣFx = 0;
Therefore,
3
8 - 2F1 a b = 0
5
FCE = 6.67 kN (C)
a + ΣME = 0;
a + ΣMD = 0;
F1 = 6.667 kN
A
Ans.
FDF = 6.67 kN (T)
4
FCD(1.5) - 6.667a b(1.5) = 0
5
4
FEF(1.5) - 6.667a b(1.5) = 0
5
FCD = 5.333 kN (C) = 5.33 kN (C)
Ans.
FEF = 5.333 kN (T) = 5.33 kN (T)
Ans.
It is required that FBF = FAC = F2. Referring to Fig. b,
+
S
ΣFx = 0;
Therefore,
3
8 + 10 - 2F2 a b = 0
5
FBF = 15.0 kN (C)
a + ΣMF = 0;
F2 = 15.0 kN
FAC = 15.0 kN (T) Ans.
4
FBC(1.5) - 15.0a b(1.5) - 8(2) = 0 5
FBC = 22.67 kN (C) = 22.7 kN (C) Ans.
a + ΣMC = 0;
2m
4
FAF(1.5) - 15.0a b(1.5) - 8(2) = 0
5
FAF = 22.67 kN (T) = 22.7 kN (T) Ans.
578
B
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–11.
(Continued)
Method of Joints.
Joint D: Referring to Fig. c,
+
S
ΣFx = 0;
3
FDE - 6.667a b = 0
5
FDE = 4.00 kN (C) Ans.
Joint C: Referring to Fig. d,
+
S
ΣFx = 0;
3
3
FCF + 6.667a b - 15.0a b = 0
5
5
FCF = 5.00 kN (C) Ans.
Joint B: Referring to Fig. e,
+
S
ΣFx = 0;
3
15.0a b - FAB = 0
5
FAB = 9.00 kN (T) Ans.
Ans.
FCE = 6.67 kN (C); FDF = 6.67 kN (T);
FCD = 5.33 kN (C); FEF = 5.33 kN (T);
FBF = 15.0 kN (C); FAC = 15.0 kN (T);
FBC = 22.7 kN (C); FAF = 22.7 kN (T);
FDE = 4.00 kN (C); FCF = 5.00 kN (C);
FAB = 9.00 kN (T)
579
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–12. Determine (approximately) the force in each member
of the truss. Assume the diagonals cannot support a compressive
force.
8 kN
E
1.5 m
D
2m
10 kN
F
C
2m
Solution
A
Method of Sections. It is required that
Ans.
FCE = FBA = 0
Referring to Fig. a,
+
S
ΣFx = 0;
a + ΣME = 0;
a + ΣMD = 0;
3
8 - FDF a b = 0
5
FDF = 13.33 kN (T) = 13.3 kN (T)
4
FCD(1.5) - 13.33a b(1.5) = 0
5
FEF(1.5) = 0
FCD = 10.67 kN (C) = 10.7 kN (C)
a + ΣMC = 0;
a + ΣMF = 0;
3
8 + 10 - FAC a b = 0
5
FAF(1.5) - 8(2) = 0
Ans.
Ans.
FEF = 0
Referring to Fig. b,
+
S
ΣFx = 0;
Ans.
FAC = 30.0 kN (T) Ans.
FAF = 10.67 kN (T) Ans.
4
FBC(1.5) - 30.0a b(1.5) - 8(2) = 0
5
FBC = 34.67 kN (C) = 34.7 kN (C) Ans.
580
B
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–12.
(Continued)
Method of Joints.
Joint E: Referring to Fig. c,
+
S
ΣFx = 0;
8 - FDE = 0
FDE = 8.00 kN (C) Ans.
Joint C: Referring to Fig. d,
+
S
ΣFx = 0;
3
FCF - 30.0a b = 0
5
FCF = 18.0 kN (C) Ans.
Joint B: Referring to Fig. e,
+
S
ΣFx = 0;
FAB = 0
Ans.
Ans.
FBF = 0
FCE = 0
FDF = 13.3 kN (T)
FCD = 10.7 kN (C)
FEF = 0
FAC = 30.0 kN (T)
FAF = 10.67 kN (T)
FBC = 34.7 kN (C)
FDE = 8.00 kN (C)
FCF = 18.0 kN (C)
FAB = 0
581
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–13. Determine (approximately) the internal moment that
member EF exerts on joint E and the internal moment that
member FG exerts on joint F.
1.5 k>ft
E
F
G
H
30 ft
A
B
20 ft
C
30 ft
D
20 ft
Solution
The frame can be simplified to that shown in Fig. a. Referring
to Fig. b,
a + ΣME = 0;
ME - 1.5(2)(1) - 12(2) = 0
Referring to Fig. c,
a + ΣMF = 0;
MF - 1.5(3)(1.5) - 18.0(3) = 0
ME = 27.0 k # ft
Ans.
MF = 60.75 k # ft
Ans.
Ans.
ME = 27.0 k # ft
MF = 60.75 k # ft
582
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–14. Determine (approximately) the internal moments at
joints A and B.
12 kN>m
F
E
A
B
D
C
8m
6m
Solution
The frame can be simplified to that shown in Fig. a. Referring
to Fig. b,
a + ΣMA = 0;
MA - 12(0.8)(0.4) - 38.4(0.8) = 0
MA = 34.56 kN # m = 34.6 kN # m
Ans.
Referring to Fig. c,
a + ΣMB = 0;
38.4(0.8) + 12(1.4)(0.1) - 28.8(0.6) - MB = 0
MB = 15.12 kN # m = 15.1 kN # m
Ans.
Ans.
MA = 34.6 kN # m
MB = 15.1 kN # m
583
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–15. Draw the approximate moment diagrams for each of
the five girders.
2 k>ft
G
F
L
K
H
I
J
E
A
B
C
D
3 k>ft
30 ft
Solution
584
40 ft
3 k>ft
30 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–16. Determine (approximately) the internal moments at
joints E and C caused by members EF and CD, respectively.
2 k>ft
E
10 ft
F
3.5 k>ft
C
D
A
B
12 ft
15 ft
Solution
a + ΣME = 0;
a + ΣMC = 0;
ME - 3(0.75) - 12(1.5) = 0
ME = 20.25 k # ft
Ans.
MC - 5.25(0.75) - 21(1.5) = 0
MC = 35.4 k # ft
Ans.
Ans.
ME = 20.25 k # ft
MC = 35.4 k # ft
585
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–17. Determine (approximately) the internal moments at
joint H from HG and at joint J from JI and JK.
5 kN>m
I
J
L
K
Solution
The frame can be simplified to that shown in Fig. a. Referring
to Figs. b and c,
a + ΣMH = 0;
5 kN>m
MHG - 6 (0.3) - 5(0.3)(0.15) = 0
MHG = 2.025 kN # m
5 kN>m
H
G
Ans.
E
F
Referring to Figs. b and d,
a + ΣMJ = 0;
- MJI + 6(0.3) + 5(0.3)(0.15) = 0
MJI = 2.025 kN # m
Ans.
Referring to Figs. b and e,
a + ΣMJ = 0;
A
3m
D
C
B
4m
4m
MJK - 8(0.4) - 5(0.4)(0.2) = 0
MJK = 3.60 kN # m
Ans.
Ans.
MHG = 2.025 kN # m;
MJI = 2.025 kN # m;
MJK = 3.60 kN # m
586
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–18. Determine (approximately) the internal moments at
joint F from FG and at joint E on the column.
1 k>ft
F
H
G
2 k>ft
D
E
A
B
30 ft
C
20 ft
Solution
The frame can be simplified to that shown in Fig. a. The vertical
reactions on the beams are shown in Fig. b. Referring to Fig. c,
a + ΣMF = 0;
MF - 1(3)(1.5) - 12.0(3) = 0
MF = 40.5 k # ft
Ans.
Referring to Fig. d,
a + ΣME = 0;
12.0(3) + 1(5)(0.5) + 24.0(3) + 2(3)(1.5) - 8.00(2) - ME = 0
ME = 103.5 k # ft
Ans.
Ans.
MF = 40.5 k # ft
ME = 103.5 k # ft
587
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–19. Determine (approximately) the internal moments at
joints D and C. Assume the supports at A and B are pins.
500 lb D
C
12 ft
A
B
10 ft
Solution
Entire Frame (1):
a + ΣMA = 0;
10By - 12(500) = 0;
+ c ΣFy = 0;
Ay = 600 lbT
By = 600 lb c
FBD (2):
a + ΣME = 0;
5(600) - 12(Ax) = 0;
+
S
ΣFx = 0;
Ex = 250 lb d
Entire Frame (1):
+
S
ΣFx = 0;
Bx = 250 lb d
FBD (3):
a + ΣMD = 0;
MD - 12(250) = 0;
MD = 3.00 k # ft
Ans.
MC - 12(250) = 0;
MC = 3.00 k # ft
Ans.
FBD (4):
a + ΣMC = 0;
Ax = 250 lb d
Ans.
MD = 3.00 k # ft
MC = 3.00 k # ft
588
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–20. Determine (approximately) the internal moment and
shear at the ends of each member of the portal frame. Assume
the supports at A and D are (a) pinned, (b) fixed and (c)
partially fixed such that the inflection point for the columns is
located h>3 = 6 ft up from A and D.
10 k B
12 ft
C
18 ft
A
D
Solution
(a) MA = MD = 0
MB = MC = 5.0(18) = 90.0 k # ft
Ans.
Ans.
or
MB = MC = 15.0(6) = 90.0 k # ft
Ans.
For members AB and CD:
VA = VB = VC = VD = 5.00 k
Ans.
For members BC:
VB = VC = - 15.0 k
(b) MA = MD = 5.0(9) = 45.0 k # ftB
MB = MC = 5.0(9) = 45.0 k # ft
Ans.
Ans.
Ans.
or
MB = MC = 7.5(6) = 45.0 k # ft
Ans.
For members AB and CD:
VA = VB = VC = VD = 5.00 k
Ans.
For member BC:
VB = VC = - 7.5 k
(c) MA = MD = 5.0(6) = 30.0 k # ftB
MB = MC = 5.0(12) = 60.0 k # ft
Ans.
Ans.
Ans.
or
MB = MC = 10(6) = 60.0 k # ft
Ans.
Ans.
MA = 0
MB = 90.0 k # ft
VA = 5.00 k
VB = - 15.0 k
MA = 45.0 k # ftB
MB = 45.0 k # ft
VA = 5.00 k
VB = - 7.5 k
MA = 30.0 k # ftB
MB = 60.0 k # ft
VA = 5.00 k
VB = - 10.0 k
For members AB and CD:
VA = VB = VC = VD = 5.00 k
Ans.
For member BC:
VB = VC = - 10.0 k
Ans.
589
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–21. Determine (approximately) the force in each truss
member of the portal frame. Assume all members of the truss
to be pin connected at their ends.
3m
3m
G
E
10 kN
F
1m
2m
18 kN
H
I
1m
C
D
A
B
4m
Solution
Support Reactions.
Assume that the horizontal reaction components on the frame at A and B are equal. Then
Ax = Bx =
10 + 18
= 14.0 kN
2
Referring to Fig. a,
a + ΣMB = 0;
Ay(6) - 10(6) - 18(4) = 0
Ay = 22.0 kN
Method of Sections. Referring to Fig. b,
a + ΣMG = 0;
a + ΣMC = 0;
22.0(3) + 18(2) - 14.0(6) - FCD(2) = 0
Ans.
FCD = 9.00 kN (C)
- 14.0(4) - (10)(2) + FEG(2) = 0
Ans.
FEG = 38.0 kN (C)
+ c ΣFy = 0;
FGH a
2
b - 22.0 = 0
113
FGH = 11113 kN (T) = 39.7 kN (T)
590
Ans.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–21.
(Continued)
Method of Joints.
Joint G, Fig. c.
+ c ΣFy = 0;
+
S
ΣFx = 0;
Joint H, Fig. d.
FGI a
2
2
b - (11113) a
b = 0
113
113
FFG + 38.0 - 2(11113) a
a + ΣFy = 0;
FEH sin u = 0
+ QΣFx = 0;
11113 - FCH = 0
3
b = 0
113
FGI = 11113 kN (C) = 39.7 kN (C)
Ans.
FFG = 28.0 kN (T)
Ans.
Ans.
FEH = 0
FCH = 11113 kN (T) = 39.7 kN (T)
Ans.
Joint I, Fig. e.
+ QΣFy = 0;
FFI sin u = 0
R + ΣFx = 0;
11113 - FDI = 0
Ans.
FFI = 0
FDI = 11113 kN (C) = 39.7 kN (C)
Ans.
Ans.
FCD = 9.00 kN (C)
FEG = 38.0 kN (C)
FGH = 39.7 kN (T)
FGI = 39.7 kN (C)
FFG = 28.0 kN (T)
FEH = 0
FCH = 39.7 kN (T)
FFI = 0
FDI = 39.7 kN (C)
591
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–22. Solve Prob. 12–21 if the supports at A and B are fixed
instead of pinned.
3m
10 kN
3m
G
E
F
1m
2m
18 kN
H
I
1m
C
D
A
B
4m
Solution
Support Reactions. Assume that zero moment occurs at the
mid-height of the column between A and C (point J) and
between B and D (point K). Also, the horizontal reaction
components at J and K are equal. Then
Jx = Kx =
10 + 18
= 14.0 kN
2
Referring to Fig. a,
a + ΣMK = 0;
Jy(6) - 18(2) - 10(4) = 0
Jy = 12.67 kN
Method of Sections. Referring to Fig. b,
a + ΣMG = 0;
a + ΣMC = 0;
12.67(3) + 18(2) - 14.0(4) - FCD(2) = 0
Ans.
FCD = 9.00 kN (C)
FEG(2) - 14.0(2) - 10(2) = 0
Ans.
FEG = 24.0 kN (C)
+ c ΣFy = 0;
FGH a
2
b - 12.67 = 0
113
Ans.
FGH = 6.333113 kN (T) = 22.8 kN (T)
592
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–22.
(Continued)
Method of Joints.
Joint G, Fig. c.
+ c ΣFy = 0;
+
S
ΣFx = 0;
FGI a
2
2
b - (6.333113) a
b = 0
113
113
FFG + 24.0 - 2c(6.333113) a
FGI = 6.333113 kN (C) = 22.8 kN (C)
3
bd = 0
113
FFG = 14.0 kN (T)
Ans.
Ans.
Joint H, Fig. d.
a + ΣFy = 0;
FEH sin u = 0
+ Q ΣFx = 0;
6.333113 - FCH = 0
Ans.
FEH = 0
FCH = 6.333113 kN (T) = 22.8 kN (T)
Ans.
Joint I, Fig. e.
+ QΣFy = 0;
FFI sin u = 0
+ R ΣFx = 0;
6.333113 - FDI = 0
Ans.
FFI = 0
FDI = 6.333113 kN (C) = 22.8 kN (C)
Ans.
Ans.
FCD = 9.00 kN (C)
FEG = 24.0 kN (C)
FGH = 22.8 kN(T)
FGI = 22.8 kN (C)
FFG = 14.0 kN (T)
FEH = 0
FCH = 22.8 kN (T)
FFI = 0
FDI = 22.8 kN (C)
593
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–23. Draw (approximately) the moment diagram for
column ACE of the portal constructed with a rigid girder and
knee braces CF and DH. Assume that all points of connection
are pins. Also determine the force in the knee brace CF.
5 ft
5 ft
8k
E
H
F
5 ft
G
5 ft
C
D
A
B
7 ft
20 ft
Solution
a + ΣME = 0;
FCF(sin 45°)(5) - 4.0(12) = 0
Ans.
FCF = 13.6 k (T)
5 ft
5 ft
8k
E
H
F
5 ft
G
5 ft
C
D
A
B
7 ft
Ax = 4 k
20 ft
Ay = 4.8 k
By = 4.8 k
Ans.
FCF = 13.6 k (T)
594
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5 ft
*12–24. Solve Prob. 12–23 if the supports at A and B are fixed
instead of pinned.
5 ft
8k
E
H
F
5 ft
G
5 ft
C
D
A
B
7 ft
20 ft
Solution
a + ΣME = 0;
FCF(sin 45°)(5) - 4.0(8.5) = 0
Ans.
FCF = 9.62 k (T)
5 ft
5 ft
8k
E
F
5 ft
C
3.5 ft
3.4 k
H
G
5 ft
D
4k
4k
3.4 k
Ans.
FCF = 9.62 k (T)
595
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–25. Draw (approximately) the moment diagram for
column BCD of the portal. Assume all truss members and the
columns to be pin connected at their ends. Also determine the
force in all the truss members.
F
8 ft
8 ft
D
E
6 ft
G
C
A
B
16 ft
Solution
Support Reactions. Assume that the horizontal reaction components
on the frame at A and B are equal. Then
6 + 8
Ax = Bx =
= 7.00 k
2
Referring to Fig. a,
a + ΣMA = 0;
6(22) + 8(16) - By(16) = 0
By = 16.25 k
Method of Sections. Referring to Fig. b,
a + ΣME = 0;
a + ΣMC = 0;
FCG(6) + 7.00(22) - 8(6) - 16.25(8) = 0
Ans.
FCG = 4.00 k (C)
7(16) + 6(6) - FDE(6) = 0
Ans.
FDE = 24.67 kN (C) = 24.7 kN (C)
+ c ΣFy = 0;
3
FCE a b - 16.25 = 0
5
FCE = 27.08 kN (T) = 27.1 kN (T)
596
6k
Ans.
8k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–25.
(Continued)
Method of Joints.
Joint E, Fig. c.
+ c ΣFy = 0;
+
S
ΣFx = 0;
3
3
FEG a b - 27.08a b = 0
5
5
FEG = 27.08 kN (C) = 27.1 kN (C)
4
2c 27.08a b d - 24.67 - FEF = 0
5
FEF = 18.67 kN = 18.7 kN (T)
Ans.
Ans.
The moment diagram for column BCD is shown in Fig. d.
Ans.
FCG = 4.00 k (C)
FDE = 24.7 kN (C)
FCE = 27.1 kN (T)
FEG = 27.1 kN (C)
FEF = 18.7 kN (T)
597
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–26. Draw (approximately) the moment diagram for
column BCD of the portal. Assume all the members of the
truss to be pin connected at their ends. The columns are fixed
at A and B. Also determine the force in all the truss members.
F
8 ft
8 ft
D
E
6 ft
G
C
A
B
16 ft
Solution
Support Reactions. Assume that zero moment occurs at the mid-height of
the column between B and C (point H) and between A and G (point I).
Also, the horizontal reaction corresponents at A and B are equal. Then
6 + 8
Ax = Bx =
= 7.00 k
2
Referring to Fig. a,
a + ΣMI = 0;
8(8) + 6(14) - Hy(16) = 0
Hy = 9.25 k
Referring to Fig. b,
+
S
ΣFx = 0;
a + ΣMB = 0;
7.00 - Hx = 0
7.00(8) - MB = 0
MB = 56.0 k # ft
+ c ΣFy = 0;
9.25 - By = 0
By = 9.25 k
Hx = 7.00 k
Method of Sections. Referring to Fig. d,
a + ΣME = 0;
a + ΣMC = 0;
FCG(6) + 7.00(14) - 9.25(8) - 8(6) = 0
Ans.
FCG = 4.00 k (C)
7.00(8) + 6(6) - FDE(6) = 0
Ans.
FDE = 15.33 k (C) = 15.3 k (C)
+ c ΣFy = 0;
3
FCE a b - 9.25 = 0
5
FCE = 15.42 k (T) = 15.4 k (T)
598
Ans.
6k
8k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–26.
(Continued)
Method of Joints.
Joint E, Fig. e.
+ c ΣFy = 0;
+
S
ΣFx = 0;
3
3
FEG a b - 15.42a b = 0
5
5
FEG = 15.42 k (C) = 15.4 k (C)
4
2c 15.42a b d - 15.33 - FEF = 0
5
FEF = 9.333 k (T) = 9.33 k (T)
Ans.
Ans.
The moment diagram for column BCD is shown in Fig. f.
Ans.
FCG = 4.00 k (C)
FDE = 15.3 k (C)
FCE = 15.4 k (T)
FEG = 15.4 k (C)
FEF = 9.33 k (T)
599
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–27. Determine (approximately) the force in each truss
member of the portal frame. Also find the reactions at the fixed
column supports A and B. Assume all members of
the truss to be pin connected at their ends.
G
8k
I
H
10 ft
E
F
4k
C
D
12 ft
A
B
10 ft
8k
G
10 ft
I
H
Solution
10 ft
E
By inspection of joints E and F,
FEG = 0
Ans.
FFI = 0
Ans.
+ c ΣFy = 0;
FCE(cos 45°) - 7.60 = 0
a + ΣMC = 0;
FCE = 10.748 k = 10.7 k (T)
+
S
ΣFx = 0;
4k
C
D
6 ft
6k
Ans.
7.6 k
FGH(10) - 8(10) - 6.0(6) = 0
FGH = 11.6 k (C)
F
6k
7.6 k
Ans.
- FCD - 6 - 11.6 + 8 + 4 + 10.748(sin 45°) = 0
FCD = 2.00 k (C)
Ans.
MA = MB = 36.0 k # ftB
Ans.
Ax = Bx = 6.00 k d Ans.
Ay = 7.6 kT
By = 7.6 k c Ans.
Joint E:
+ QΣFx = 0;
FEH = 10.7 k (T)
Ans.
Joint H:
+ c ΣFy = 0;
FHF sin 45° - 10.748 sin 45° = 0
FHF = 10.748 = 10.7 k (C)
+
S
ΣFx = 0;
Ans.
FHI + 11.6 - 2(10.748)(cos 45°) = 0
FHI = 3.60 k (T)
Ans.
FFD = 10.7 k (C)
Ans.
Joint F:
a + ΣFy = 0;
Ans.
FCE = 10.7 k (T)
FGH = 11.6 k (C)
FCD = 2.00 k (C)
MA = 36.0 k # ftB
Ax = 6.00 k d
Ay = 7.6 kT
By = 7.6 k c
FEH = 10.7 k (T)
FHF = 10.7 k (C)
FHI = 3.60 k (T)
FFD = 10.7 k (C)
600
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
G
*12–28. Determine
(approximately)
the
force
in
members GH, GJ, and JK of the portal frame. Also find the
reactions at the fixed column supports A and B. Assume all
members of the truss to be pin connected at their ends.
F
E
40 kN H
D
K
J
1.5 m
L
I
3m
C
6m
Support Reactions. Assume that the zero moment occurs at the midheight of the column between A and I (point M) and between B and
C (point N). Also, the horizontal reaction component on the frame at
A and B are equal. Then
Ax = Bx =
B
A
Solution
40
= 20.0 kN
2
4m
4m
4m
4m
Ans.
Referring to Fig. a,
a + ΣMN = 0;
a + ΣMM = 0;
My(16) - 40(4.5) = 0
My = 11.25 kN
Ny(16) - 40(4.5) = 0
Ny = 11.25 kN
Referring to Fig. b,
+
S
ΣFx = 0;
Mx - 20.0 = 0
Mx = 20.0 kN
+ c ΣFy = 0;
11.25 - Ay = 0
Ay = 11.25 kN
a + ΣMA = 0;
Ans.
MA = 60.0 kN # m Ans.
MA - 20.0(3) = 0
Referring to Fig. c,
+
S
ΣFx = 0;
Nx - 20.0 = 0
Nx = 20.0 kN
+ c ΣFy = 0;
By - 11.25 = 0
By = 11.25 kN
a + ΣMB = 0;
MB - 20.0(3) = 0
Ans.
MB = 60.0 kN # m Ans.
Method of Sections, Fig. d,
a + ΣMG = 0;
a + ΣMJ = 0;
a + ΣMH = 0;
40(3) + 11.25(4) - 20.0(7.5) - FJK(3) = 0
FJK = 5.00 kN (C)
Ans.
3
FGH a b(4) + 11.25(4) - 20.0(4.5) = 0
5
Ans.
FGH = 18.75 kN (C)
FGJ(4) - 20.0(4.5) = 0
FGJ = 22.5 kN (T)
Ans.
Ans.
Ax = Bx = 20.0 kN
Ay = 11.25 kN
MA = 60.0 kN.m
By = 11.25 kN
MB = 60.0 kN.m
FJK = 5.00 kN (C)
FGH = 18.75 kN (C)
FGJ = 22.5 kN (T)
601
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–29. Solve Prob. 12–28 if the supports at A and B are pin
connected instead of fixed.
G
F
E
40 kN H
D
K
J
1.5 m
L
I
3m
C
6m
B
A
4m
4m
4m
4m
Solution
Support Reactions. Assume that the horizontal reaction
components on the frame at A and B are equal. Then
Ax = Bx =
40
= 20.0 kN
2
Ans.
Referring to Fig. a,
a + ΣMB = 0;
a + ΣMA = 0;
Ay(16) - 40(7.5) = 0
Ay = 18.75 kN
Ans.
By(16) - 40(7.5) = 0
By = 18.75 kN
Ans.
Method of Sections. Referring to Fig. b,
a + ΣMG = 0;
a + ΣMH = 0;
a + ΣMJ = 0;
FJK(3) - 40(3) + 18.75(4) - 20.0(10.5) = 0
Ans.
FJK = 5.00 kN (T)
FGJ(4) - 20 .0(7.5) = 0
FGJ = 37.5 kN (T)
3
FGH a b(4) + 18 .75(4) - 20.0(7.5) = 0
5
FGH = 31.25 kN (C)
Ans.
Ans.
Ans.
Ax = Bx = 20.0 kN
Ay = 18.75 kN
By = 18.75 kN
FJK = 5.00 kN (T)
FGJ = 37.5 kN (T)
FGH = 31.25 kN (C)
602
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
G
12–30. Draw (approximately) the moment diagram for
column ACD of the portal. Assume all truss members and the
columns to be pin connected at their ends. Also determine the
force in members FG, FH, and EH.
F
4k
6 ft
J
D
K
H
E
I
C
6 ft
3 ft
L
12 ft
A
8 ft
Solution
Assume that the horizontal force components at pin supports
A and B are equal. Thus,
Ax = Bx =
4
= 2.00 k
2
Referring to Fig. a,
a + ΣMB = 0;
Ay(32) - 4(15) = 0
Ay = 1.875 k
Using the method of sections, Fig. b,
a + ΣMH = 0;
a + ΣMF = 0;
a + ΣMD = 0;
3
FFG a b(16) + 1.875(16) - 2.00(15) = 0
5
FFG = 0
Ans.
4(6) + 1.875(8) - 2.00(21) + FEH(6) = 0
Ans.
FEH = 0.500 k (T)
3
FFH a b(16) - 2.00(15) = 0
5
FFH = 3.125 k (C)
603
Ans.
B
8 ft
8 ft
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–30.
(Continued)
Also, referring to Fig. c,
a + ΣME = 0;
a + ΣMD = 0;
3
FDF a b(8) + 1.875(8) - 2.00(15) = 0
5
FDF = 3.125 k (C)
FCE a
3
b(8) - 2.00(15) = 0
173
FCE = 10.68 k (T)
+
S
ΣFx = 0;
4 + 10.68a
8
4
b - 3.125a b - 2.00 - FDE = 0
5
173
FDE = 9.50 k (C)
Ans.
FFG = 0
FEH = 0.500 k (T);
FFH = 3.125 k (C)
604
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–31. Solve Prob. 12–30 if the supports at A and B are fixed
instead of pinned.
G
F
4k
6 ft
J
D
K
H
E
I
C
6 ft
3 ft
L
12 ft
A
Solution
8 ft
Assume that the horizontal force components at fixed supports
A and B are equal. Thus,
Ax = Bx =
4
= 2.00 k
2
Also, the points of inflection N and O are 6 ft above A and B,
respectively. Referring to Fig. a,
a + ΣMO = 0;
Ny(32) - 4(9) = 0
Ny = 1.125 k
Referring to Fig. b,
+ ΣFx = 0;
S
Nx - 2.00 = 0
Nx = 2.00 k
a + ΣMA = 0;
MA - 2.00(6) = 0
MA = 12.0 k # ft
+ c ΣFy = 0;
1.125 - Ay = 0
Ay = 1.125 k
605
B
8 ft
8 ft
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–31.
(Continued)
Using the method of sections Fig. d,
a + ΣMH = 0;
a + ΣMF = 0;
a + ΣMD = 0;
3
FFG a b(16) + 1.125(16) - 2.00(9) = 0 FFG = 0
5
Ans.
-FEH(6) + 4(6) + 1.125(8) - 2.00(15) = 0 FEH = 0.500 k (C)
Ans.
3
FFH a b(16) - 2.00(9) = 0
5
FFH = 1.875 k (C)
Ans.
Also, referring to Fig. e,
a + ΣME = 0;
a + ΣMD = 0;
3
FDF a b(8) + 1.125(8) - 2.00(9) = 0
5
FCE a
3
b(8) - 2.00(9) = 0
173
FDF = 1.875 k (C)
FCE = 6.408 k (T)
8
4
+
S
ΣFx = 0; 4 + 6.408a
b - 1.875a b - 2.00 - FDE = 0 FDE = 6.50 k (C)
5
173
Ans.
FFG = 0; FEH = 0.500 k (C);
FFH = 1.875 k (C)
606
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–32. Draw (approximately) the moment diagram for
column AJI of the portal. Assume all truss members and the
columns to be pin connected at their ends. Also determine the
force in members HG, HL, and KL.
2 kN
H
I
4 kN
J
K
6 @ 1.5 m 5 9 m
G
F
E
L
O
M
N
D
1.5 m
1m
C
4m
A
B
Solution
Assume the horizontal force components at pin supports A and B to be equal. Thus,
Ax = Bx =
2 + 4
= 3.00 kN
2
Referring to Fig. a,
a + ΣMB = 0;
Ay(9) - 4(4) - 2(5) = 0
Ay = 2.889 kN
Using the method of sections, Fig. b,
a + ΣML = 0;
FHG cos 6.340°(1.167) + FHG sin 6.340°(1.5) + 2.889(3)
- 2(1) - 3.00(4) = 0
a + ΣMH = 0;
FHG = 4.025 kN (C) = 4.02 kN (C)
FKL(1.167) + 2(0.167) + 4(1.167) + 2.889(1.5) - 3.00(5.167) = 0
FKL = 5.286 kN (T) = 5.29 kN (T)
+ c ΣFy = 0;
Ans.
Ans.
FHL cos 52.13° - 4.025 sin 6.340° - 2.889 = 0
FHL = 5.429 kN (C) = 5.43 kN (C)
Also, referring to Fig. c,
a + ΣMH = 0;
Ans.
FJK(1.167) + 2(0.167) + 4(1.167) + 2.889(1.5) - 3.00(5.167) = 0
FJK = 5.286 kN (T)
607
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–32.
(Continued)
a + ΣMJ = 0;
FIH cos 6.340°(1) - 2(1) - 3.00(4) = 0
FIH = 14.09 kN (C)
+ c ΣFy = 0;
FJH sin 37.87° - 14.09 sin 6.340° - 2.889 = 0
FJH = 7.239 kN (T)
Ans.
FHG = 4.02 kN (C)
FKL = 5.29 kN (T)
FHL = 5.43 kN (C)
608
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–33. Solve Prob. 12–32 if the supports at A and B are fixed
instead of pinned.
2 kN
H
I
4 kN
J
K
6 @ 1.5 m 5 9 m
G
F
E
L
O
M
N
D
1.5 m
1m
C
4m
A
B
Solution
Assume that the horizontal force components at fixed supports
A and B are equal. Therefore,
Ax = Bx =
2 + 4
= 3.00 kN
2
Also, the reflection points P and R are located 2 m above A and
B, respectively. Referring to Fig. a,
a + ΣMR = 0;
Py(9) - 4(2) - 2(3) = 0 Py = 1.556 kN
Referring to Fig. b,
+
S
ΣFx = 0;
Px - 3.00 = 0 Px = 3.00 kN
a + ΣMA = 0;
MA - 3.00(2) = 0 MA = 6.00 kN # m
+ c ΣFy = 0;
1.556 - Ay = 0 Ay = 1.556 kN
Using the method of sections, Fig. d,
a + ΣML = 0;
a + ΣMH = 0;
FHG cos 6.340°(1.167) + FHG sin 6.340°(1.5) + 1.556(3) - 3.00(2) - 2(1) = 0
FHG = 2.515 kN (C) = 2.52 kN (C)
Also referring to Fig. e,
a + ΣMJ = 0;
FJK(1.167) + 4(1.167) + 2(0.167) + 1.556(1.5) - 3.00(3.167) = 0
FJK = 1.857 kN (T)
FIH cos 6.340°(1) - 2(1) - 3.00(2) = 0
FIH = 8.049 kN (C)
+ c ΣFy = 0;
Ans.
Ans.
FHL cos 52.13° - 2.515 sin 6.340° - 1.556 = 0
FHL = 2.986 kN (C) = 2.99 kN (C)
a + ΣMH = 0;
Ans.
FKL(1.167) + 2(0.167) + 4(1.167) + 1.556(1.5) - 3.00(3.167) = 0
FKL = 1.857 kN (T) = 1.86 kN (T)
+ c ΣFy = 0;
FJH sin 37.87° - 8.049 sin 6.340° - 1.556 = 0
FJH = 3.982 kN (T)
609
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–33.
(Continued)
610
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–33.
(Continued)
Ans.
FHG = 2.52 kN (C);
FKL = 1.86 kN (T);
FHL = 2.99 kN (C)
611
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10 k
12–34. Use the portal method of analysis and draw the
moment diagram for girder FED.
F
E
D
30 ft
A
B
20 ft
C
20 ft
Solution
Ans.
M max = 37.5 k # ft
612
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–35. Use the portal method and determine (approximately)
the reactions at A, B, C, and D of the frame.
36 kN
E
F
G
H
10 m
A
B
8m
C
12 m
D
8m
Solution
+
S
ΣFx = 0;
- 6V + 8 = 0
V = 1.333 k
Ay = 1.33 k
Ans.
By = 0.444 k
Ans.
Ax = 1.33 k
Ans.
Bx = 2.67 k
Ans.
MA = 13.3 k # ft
Ans.
MB = 26.7 k # ft
Ans.
Cy = 0.444 k
Ans.
Dy = 1.33 k
Ans.
C x = 2.67 k
Ans.
Dx = 1.33 k
Ans.
MC = 26.7 k # ft
MD = 13.3 k # ft
Ans.
Ans.
Ans.
Ay = 1.33 k
Ax = 1.33 k
MA = 13.3 k # ft
Cy = 0.444 k
C x = 2.67 k
MC = 26.7 k # ft
By = 0.444 k
Bx = 2.67 k
MB = 26.7 k # ft
Dy = 1.33 k
Dx = 1.33 k
MD = 13.3 k # ft
613
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
36 kN
*12–36. Draw (approximately) the moment diagram for the
girder EFGH. Use the portal method.
E
F
G
H
10 m
A
B
8m
Solution
614
C
12 m
D
8m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–36.
(Continued)
Ans.
Ax = 6.00 kN
MA = 30.0 kN # m
Ay = 7.50 kN
Bx = 12.0 kN
MB = 60.0 kN # m
By = 2.50 kN
Cx = 12.0 kN
MC = 60.0 kN # m
Cy = 2.50 kN
Dx = 6.00 kN
MD = 30.0 kN # m
Dy = 7.50 kN
M max = 30.0 kN # m
615
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–37. Use the portal method and determine (approximately)
the reactions at supports A, B, C, and D.
I
9 kN
J
L
K
4m
E
12 kN
F
G
H
4m
A
B
C
5m
Solution
9 kN
I
N
M
12 kN
E
J
P
F
5m
K
R
Q
W
G
L
S
Y
V
X
Z
A
B
C
D
5m
5m
4m
H
T
5m
616
5m
O
U
D
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–37.
(Continued)
d Ans.
B Ans.
d Ans.
Ans.
c Ans.
T Ans.
617
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–37.
(Continued)
d Ans.
B Ans.
Ans.
d Ans.
B Ans.
c Ans.
Ans.
Ax = 3.50 kN d
MA = 7.00 kN # mB
Ay = 5.20 kNT
Bx = 7.00 kN d
MB = 14.0 kN # mB
By = 0
Cx = 7.00 kN d
MC = 14.0 kN # mB
Cy = 0
Dx = 3.50 kN d
MD = 7.00 kN # mB
Dy = 5.20 kN c
618
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–38. Use the cantilever method and determine
(approximately) the reactions at supports A, B, C, and D. All
columns have the same cross-sectional area.
I
9 kN
J
L
K
4m
E
12 kN
F
G
H
4m
A
B
C
5m
Solution
9 kN
I
N
M
12 kN
E
J
P
F
5m
K
R
Q
W
G
L
S
Y
V
X
Z
A
B
C
D
5m
5m
4m
H
T
5m
619
5m
O
U
D
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–38.
(Continued)
d Ans.
d Ans.
B Ans.
B Ans.
T Ans.
T Ans.
620
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–38.
(Continued)
d Ans.
d Ans.
B Ans.
B Ans.
c Ans.
c Ans.
Ans.
Ax = 3.15 kN d
MA = 6.30 kN # mB
Ay = 4.68 kNT
Bx = 7.35 kN d
MB = 14.7 kN # mB
By = 1.56 kNT
Cx = 7.35 kN d
MC = 14.7 kN # mB
Cy = 1.56 kN c
Dx = 3.15 kN d
MD = 6.30 kN # mB
Dy = 4.68 kN c
621
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–39. Use the portal method of analysis and draw the
moment diagram for column AFE.
4m
30 kN
H
E
15 kN
G
4m
I
K
F
J
B
6m
622
C
L
A
Solution
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–40. Solve Prob. 12–39 using the cantilever method of
analysis. All the columns have the same cross-sectional area.
4m
30 kN
H
E
15 kN
G
4m
I
K
F
J
B
6m
623
C
L
A
Solution
D
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–41. Use
the
portal
method
and
determine
(approxi­
mately) the axial force, shear force, and moment
at A.
3k
G
H
I
D
E
F
A
B
12 ft
6k
18 ft
20 ft
12 in2
C
20 ft
8 in2
10 in2
Solution
+
S
ΣFx = 0;
4V - 3 = 0
V = 0.75 k
+
S
ΣFx = 0;
4V′ - 9 = 0
V′ = 2.25 k
Ay = 2.925 k c
NA = - 2.925 k
Ans.
Ax = 2.25 k S
VA = 2.25 k
Ans.
MA = 20.25 k # ftA
Ans.
Ans.
Ay = 2.925 k c NA = - 2.925 k
Ax = 2.25 k S VA = 2.25 k
MA = 20.25 k # ftA
624
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–42. Solve Prob. 12–41 using the cantilever method. Each
column has the cross-sectional area indicated.
3k
G
H
I
D
E
F
A
B
12 ft
6k
18 ft
20 ft
12 in2
Solution
a + MNA = 0;
8(20) = 10(40)
30
- Ky(1.333) - Ly(21.333) - Jy(18.667) + 3(6) = 0
1.333 Jy
a b
18.667 12
Ky
21.333
sL =
s;
18.667 j
Ly
Jy = 0.461 k
Ky = 0.02195 k
8
=
Jy
21.333
=
a b
10
18.667 12
Ky = 0.04762 Jy
12 ft
20 ft
8 in2
x =
1.333
s;
18.667 j
sK =
C
10 in2
= 18.667 ft
3k
G
H
I
J
K
L
D
E
F
M
N
O
A
B
Ly = 0.9524 Jy
Ly = 0.439 k
18 ft
6k
x
12 in2
20 ft
C
20 ft
8 in2
10 in2
a + ΣMNA = 0;
- My(18.667) - Ny(1.333) - Oy(21.333) + 6(9) + 3(21) = 0
Ny = 0.04762My
Oy = 0.9524My
My = 2.9963 k
Ny = 0.1427 k Oy = 2.8537 k
Ay = 3.00 k c
Ax = 2.30 k S
NA = - 3 k
MA = 20.7 k # ftA
VA = 2.3 k
Ans.
Ay = 3.00 k c NA = - 3 k
Ax = 2.30 k S VA = 2.3 k
MA = 20.7 k # ftA
625
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–43. Draw (approximately) the moment diagram for girder
PQRST and column AFKP of the building frame. Use the
portal method.
12 kN
24 kN
24 kN
P
Q
R
S
T
K
L
M
N
O
4m
F
G
H
I
J
4m
A
B
C
D
E
6m
Solution
626
6m
8m
8m
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–43.
(Continued)
627
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–44. Draw (approximately) the moment diagram for
girder PQRST and column AFKP of the building frame. All
columns have the same cross-sectional area. Use the cantilever
method.
12 kN
24 kN
24 kN
P
Q
R
S
T
K
L
M
N
O
4m
F
G
H
I
J
4m
A
B
C
D
E
6m
6m
8m
8m
Solution
Referring to Fig. a,
x =
0 + 6 + 12 + 20 + 28
= 13.2 m
5
a + ΣMU = 0;
a
17
37
1
6
F b(20) + a F1 b(28) - a F1 b(12) - a F1 b(6) - 12(2) = 0 F1 = 0.6429 kN
33 1
33
11
11
a
17
37
1
6
F b(20) + a F2 b(28) - a F2 b(12) - a F2 b(6) - 12(6) - 24(2) = 0 F2 = 3.2143 kN
33 2
33
11
11
Referring to Fig. b,
a + ΣMV = 0;
Referring to Fig. c,
a + ΣMW = 0;
a
17
37
1
6
F b(20) + a F3 b(28) - a F3 b(12) - a F2 b(6) - 12(10) - 24(6) - 24(2) = 0 F3 = 8.3571 kN
33 3
33
11
11
628
4m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–44.
(Continued)
629
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–45. Draw the moment diagram for girder IJKL of the
building frame. Use the portal method of analysis. Each column
has the cross-sectional area indicated.
20 kN
I
J
K
L
4m
40 kN
G
F
E
H
4m
A
B
4m
Area
Solution
+
S
ΣFx = 0;
10 - 6V = 0;
V = 1.667 kN
The equilibrium of each segment is shown on the FBDs.
630
24 (1023) m2 16 (1023) m2
C
5m
D
4m
16 (1023) m2
24 (1023) m2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–46. Draw the moment diagram for girder IJKL of the
building frame. Use the cantilever method of analysis. Each
column has the cross-sectional area indicated.
20 kN
I
J
K
L
4m
40 kN
G
F
E
H
4m
A
B
4m
Area
Solution
The centroid of column area is in the center of the framework.
Since s =
s1 = a
F
, then
A
6.5
bs ;
2.5 2
F1
6.5 F2
=
a b;
12
2.5 8
s4 = s1;
F4 = F 1
s2 = s3;
F2 = F 3
a + MM = 0;
F1 = 3.90 F2
- 2(10) - 4(F2) + 9(F2) + 13(3.90F2) = 0
F2 = 0.359 k
F1 = 1.400 k
The equilibrium of each segment is shown on the FBDs.
631
24 (1023) m2 16 (1023) m2
C
5m
D
4m
16 (1023) m2
24 (1023) m2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–1. Determine the moments at A, B, and C by the momentdistribution method. Assume the supports at A and C are fixed
and a roller support at B is on a rigid base. The girder has a
thickness of 4 ft. Use Table 13.1. E is constant. The haunches
are tapered.
8 k>ft
A
B
4 ft
6 ft
4 ft
2 ft
20 ft
C
4 ft
4 ft
4 ft
20 ft
6 ft
Solution
aA =
6
= 0.3
20
rA = rB =
aB =
4
= 0.2
20
4 - 2
= 1
2
From Table 13–1,
for span AB,
CAB = 0.622
CBA = 0.748
KAB = 10.06
KBA = 8.37
KBA =
KBAEIC
8.37EIC
=
= 0.4185EIC
L
20
1FEM2 AB = - 0.10891821202 2 = - 348.48 k # ft
1FEM2 BA = 0.09421821202 2 = 301.44 k # ft
For span BC,
CBC = 0.748
CCB = 0.622
KBC = 8.37
KCB = 10.06
KBC = 0.4185EIC
1FEM2 BC = -301.44 k # ft
1FEM2 CB = 348.48 k # ft
Joint
A
Mem.
AB
K
DF
0
B
C
BA
BC
0.4185EIC
0.4185EIC
0.5
0.5
CB
0
COF
0.622
0.748
0.748
0.622
FEM
- 348.48
301.44
- 301.44
348.48
0
0
301.44
- 301.44
ΣM
- 348.48
348.48 k # ft
Ans.
Ans.
MAB = -348.48 k # ft; MBA = 301.44 k # ft;
MBC = -301.44 k # ft; MCB = 348.48 k # ft
632
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–2.
Solve Prob. 13–1 using the slope-deflection equations.
8 k>ft
A
B
4 ft
6 ft
2 ft
20 ft
C
4 ft
4 ft
4 ft
4 ft
20 ft
6 ft
Solution
6
4
= 0.3
aB =
= 0.2
20
20
4 - 2
rA = rB =
= 1
2
aA =
For span AB,
CAB = 0.622
CBA = 0.748
KAB = 10.06
KBA = 8.37
KBAEIC
8.37EIC
KBA =
=
= 0.4185EIC
L
20
1FEM2 AB = -0.10891821202 2 = - 348.48 k # ft
1FEM2 BA = 0.09421821202 2 = 301.44 k # ft
For span BC,
CBC = 0.748
CCB = 0.622
KBC = 8.37
KCB = 10.06
KBC = 0.4185EIC
1FEM2 BC = - 301.44 k # ft
1FEM2 CB = 348.48 k # ft
MN = KN[uN + CNuF - c11 + CN 2] + 1FEM2 N
MAB = 0.503EI10 + 0.622uB - 02 - 348.48
MAB = 0.312866EIuB - 348.8
(1)
MBA = 0.4185EI1uB + 0 - 02 + 301.44
MBA = 0.4185EIuB + 301.44
(2)
MBC = 0.4185EI1uB + 0 - 02 - 301.44
MBC = 0.4185EIuB - 301.44
(3)
MCB = 0.503EI10 + 0.622uB - 02 + 348.48
MCB = 0.312866EIuB - 348.48
(4)
Equilibrium.
MBA + MBC = 0(5)
Solving Eqs. 1–5:
uB = 0
MAB = -348 k # ft
MBA = 301 k # ft
MBC = - 301 k # ft
MCB = 348 k # ft
Ans.
Ans.
Ans.
MAB = -348 k # ft; MBA = 301 k # ft;
MBC = -301 k # ft; MCB = 348 k # ft
Ans.
Ans.
633
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–3. Apply the moment-distribution method to determine
the moment at each joint of the parabolic haunched frame.
Supports A and B are fixed. Use Table 13.2. The members are
each 1 ft thick. E is constant.
2 ft
1.5 k>ft
4 ft
C
5 ft
8 ft
12 ft
2 ft
B
40 ft
40 ft
A
Solution
The necessary data for member BC can be found from Table 13–2.
Here,
aC =
8
= 0.2
40
aB =
12
= 0.3
40
CBC =
12
= 0.589
40
rC =
4 - 2
= 1.0
2
rB =
5 - 2
= 1.5
2
Thus,
CCB = 0.735
KCB = 7.02
KBC = 8.76
Then,
KCBEIC
KCB =
=
LBC
7.02E c
1
112123 2 d
12
= 0.117E
40
The fixed end moment are given by
1FEM2 CB = -0.086211.521402 2 = - 206.88 k # ft
1FEM2 BC = 0.113311.521402 2 = 271.92 k # ft
Since member AC is prismatic,
4EI
KCA =
=
LAC
4E c
1
112122 3 d
12
= 0.0667E
40
Tabulating these data,
Joint
A
Mem
AC
C
B
CA
CB
BC
0.0667E
0.117E
DF
0
0.3630
0.6370
0
COF
0
0.5
0.735
0
K
FEM
- 206.88
Dist.
CO
37.546
ΣM
37.546
75.10
131.78
271.92
75.10
- 75.10
368.78
96.86
Thus,
MAC = 37.54 k # ft = 37.5 k # ft
MCA = 75.10 k # ft = 75.1 k # ft
MCB = -75.10 k # ft = - 75.1 k # ft
MBC = 368.78 k # ft = 369 k # ft
Ans.
Ans.
Ans.
MAC = 37.5 k # ft; MCA = 75.1 k # ft;
MCB = -75.1 k # ft; MBC = 369 k # ft
Ans.
Ans.
634
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 ft
*13–4. Solve Prob. 13–3 using the slope-deflection equations.
1.5 k>ft
4 ft
C
5 ft
8 ft
12 ft
2 ft
40 ft
40 ft
A
Solution
The necessary data for member BC can be found from Table 13.2.
Here,
aC =
8
= 0.2
40
aB =
12
= 0.3
40
rC =
4 - 2
= 1.0
2
rB =
5 - 2
= 1.5
2
Thus,
CC = 0.735
CBC = 0.589
KCB = 7.02
KBC = 8.76
Then,
KCBEIC
KCB =
=
LBC
KBCEIC
KBC =
=
LBC
7.02E c
8.76E c
1
112122 3 d
12
= 0.117E
40
1
112122 3 d
12
= 0.146E
40
The fixed end moments are given by
1FEM2 CB = - 0.086211.521402 2 = - 206.88 k # ft
1FEM2 BC = 0.113311.521402 2 = 271.92 k # ft
For member BC, applying Eq. 13–8,
MN = KN[uN + CNuF - c1HCN 2] + 1FEM2 N
MCB = 0.117E[uc + 0.735102 - 011 + 0.7352] + 1 - 206.882 = 0.117Euc - 206.88(1)
MBC = 0.146E[0 + 0.589uC - 011 + 0.5892] + 271.92 = 0.085994EuC + 271.92
(2)
Since member AC is prismatic, Eq. 11–8 is applicable,
MN = 2EK12uN + uF - 3c2 + 1FEM2 N
1
112122 3
12
MAC = 2E D
T [2102 + uC - 3102] + 0 = 0.03333EuC(3)
40
1
112122 3
12
T [2uC + 0 - 3102] + 0 = 0.06667EuC(4)
MCA = 2E D
40
635
B
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–4.
(Continued)
Moment equilibrium of joint C gives
MCA + MCB = 0
0.06667EuC + 0.117EuC - 206.88 = 0
uC =
1126.39
E
Substitute this result into Eqs. (1) to (4),
MCB = -75.09 k # ft = - 75.1 k # ft
MBC = 368.78 k # ft = 369 k # ft
MAC = 37.546 k # ft = 37.5 k # ft
MCA = 75.09 k # ft = 75.1 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
MCB = -75.1 k # ft
MBC = 369 k # ft
MAC = 37.5 k # ft
MCA = 75.1 k # ft
636
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–5. Use the moment-distribution method to determine the
moment at each joint of the symmetric bridge frame. Supports
at F and E are fixed and B and C are fixed ­connected. Use
Table 13–2. The modulus of elasticity is constant and the members are each 1 ft thick. The haunches are parabolic.
2 ft
4 ft
4 ft
D
A
6 ft
B
9 ft 8 ft
C
25 ft
For span AB,
aA =
6
= 0.2
30
aB =
9
= 0.3
30
rA = rB =
4 - 2
= 1
2
From Table 13.2,
CAB = 0.683
CBA = 0.598
kAB
= 6.73
kBA = 7.68
KAB
6.73EI
=
= 0.2243EI
30
KBA =
7.68EI
= 0.256EI
30
K′BA = 0256EI[1 - 10.683210.5982]
= 0.15144 EI
1FEM2 AB = -0.09111421302 2 = - 327.96 k # ft
1FEM2 BA = 0.10421421302 2 = 375.12 k # ft
For span CD,
CDC = 0.683
CCD = 0.598
KDC = 6.73
KCD = 7.68
KDC = 0.2243EI
KCD = 0.256EI
K′CD = 0.15144EI
1FEM2 CD = -375.12 k # ft
1FEM2 DC = 327.96 k # ft
For span BC,
aB = aC =
8
= 0.2
40
rB = rC =
4 - 2
= 1
2
From Table 13.2,
CBC = CCB = 0.619
kBC = kCB = 6.41
KBC = KCB =
6.41EI
= 0.16025EI
40
1FEM2 BC = - 0.09561421402 2 = - 611.84 k # ft
1FEM2 CB = 611.84 k # ft
637
2 ft
E
F
30 ft
Solution
4 k>ft
40 ft
30 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–5.
(Continued)
For span BF,
CBF = 0.5
KBF =
4EI
= 0.16EI
25
1FEM2 BF = 1FEM2 FB = 0
For span CE,
CCE = 0.5
KCE = 0.16EI
1FEM2 CE = 1FEM2 EC = 0
Joint
A
F
Member
AB
FB
BF
BA
BC
CB
CD
0
0.3392
0.3211
0.3397
0.3397
0.5
0.598
0.619
0.619
375.12
-611.84
611.84
-375.12
76.01
80.41
-80.41
-76.01
224.00
-49.77
49.77
-224.00
- 55.95
-59.19
59.19
55.95
36.64
-36.64
DF
1
COF
0.683
FEM
- 327.96
Dist.
327.96
CO
80.30
40.15
Dist.
- 59.09
CO
C
- 29.55
Dist.
CO
- 12.42
- 11.77
-12.45
12.45
7.71
-7.71
- 2.61
- 2.48
-2.62
2.62
1.62
-1.62
- 6.21
Dist.
- 1.31
CO
Dist.
- 0.55
- 0.52
- 0.27
CO
Dist.
CO
0
2.76
-0.55
0.55
0.34
-0.34
- 0.11
- 0.11
-0.12
0.12
0.07
-0.07
- 0.03
- 0.02
-0.02
0.02
5.49
604
- 0.05
Dist.
Σ
B
-609
E
D
CE
EC
DC
0.3211
0.3392
0
1
0.598
0.5
0.683
327.96
80.30
327.96
-40.15
59.09
29.55
11.77
12.42
2.48
2.61
6.21
1.31
0.52
0.55
0.27
0.11
0.11
0.02
0.03
0.05
609
-604
2.76 5.49 604
-609
-5.49
-2.76
k # ft
Ans.
0
Ans:
Σ
0
638
609
-604
-5.49
-2.76
k # ft
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–6.
Solve Prob. 13–5 using the slope-deflection equations.
2 ft
4 ft
4 ft
D
A
6 ft
B
9 ft 8 ft
C
25 ft
See Prob 13–5 for the tabulated data.
MN = KN[uN + CNuF - c11 - CN 2] + 1FEM2 N
For span AB,
MAB = 0.2243EI1uA + 0.683uB - 02 - 327.96
MAB = 0.2243EIuA + 0.15320EIuB - 327.96(1)
MBA = 0.256EI1uB + 0.598uA - 02 + 375.12
MBA = 0.256EIuB + 0.15309EIuA + 375.12(2)
For span BC,
MBC = 0.16025EI1uB + 0.619uC - 02 - 611.84
MBC = 0.16025EIuB + 0.099194EIuC - 611.84(3)
MCB = 0.16025EI1uC + 0.619uB - 02 + 611.84
MCB = 0.16025EIuC + 0.099194EIuB + 611.84(4)
For span CD.
MCD = 0.256EI1uC + 0.598uD - 02 - 375.12
MCD = 0.256EIuC + 0.15309EIuD - 375.12(5)
MDC = 0.2243EI1uD + 0.683uC - 02 + 327.96
MDC = 0.2243EIuD + 0.15320EIuC + 327.96(6)
For span BF,
MBF = 2E a
1
b 12uB + 0 - 02 + 0
25
MBF = 0.16EIuB(7)
MFB = 2E a
1
b 12102 + uB - 02 + 0
25
MFB = 0.08EIuB(8)
For span CE,
MCE = 2E a
1
b 12uC + 0 - 02 + 0
25
MCE = 0.16EIuC(9)
MEC = 2E a
1
b 12102 + uC - 02 + 0
25
MEC = 0.08EIuC(10)
639
2 ft
E
F
30 ft
Solution
4 k>ft
40 ft
30 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–6.
(Continued)
Equilibrium equations:
MAB = 0(11)
MDC = 0(12)
MBA + MBC + MBF = 0(13)
MCB + MCE + MCD = 0(14)
Solving Eqs. 11–14 simultaneously,
uA =
1438.53
EI
MAB = 0
uB =
MBA = 604 k # ft
MBC = - 610 k # ft
MBF = 5.53 k # ft
MFB = 2.77 k # ft
MCB = 610 k # ft
MCD = - 604 k # ft
MCE = - 5.53 k # ft
MEC = - 2.77 k # ft
MDC = 0
34.58
EI
uC =
- 34.58
EI
uD =
-1438.53
EI
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans:
MAB = 0
MBA = 604 k # ft
MBC = -610 k # ft
MBF = 5.53 k # ft
MFB = 2.77 k # ft
MCB = 610 k # ft
MCD = -604 k # ft
MCE = -5.53 k # ft
MEC = -2.77 k # ft
MDC = 0
640
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–7. Use the moment-distribution method to determine the
moment at each joint of the frame. E is constant and the members have a thickness of 1 ft. The supports at A and C are fixed.
Use Table 13–1.
2 k>ft
5 ft
C
B
2 ft
3 ft
9 ft
5 ft
3 ft
2 ft
2 ft
Solution
A
3.2
aB = aC =
= 0.2
16
rB = rC =
5 - 2.5
= 1
2.5
CBC = CCB = 0.619
kBC = kCB = 6.41
1FEM2 BC = - 0.14591221162 = -4.6688 k # ft
1FEM2 CB = 4.6688 k # ft
kBCEIC
KBC = KCB =
=
L
4EI
=
L
KBA = KCD =
1DF2 BA = 1DF2 CD =
6.411E2a
4E c
1
b11212.52 3
12
= 0.5216E
16
1
112132 3 d
12
= 0.6E
15
0.6E
= 0.535
0.5216E + 0.6E
1DF2 BC = 1DF2 CB = 0.465
Joint
A
Member
AB
BA
BC
CB
CD
DF
0
0.535
0.465
0.465
0.535
0
COF
0.5
0.5
0.619
0.619
0.5
0.5
- 4.6688
4.6688
B
FEM
2.498
1.249
C
2.171
- 2.171
- 1.344
1.344
0.7191 0.6249
0.359
0.207
0.103
0.059
0.029
0.017
-1.249
- 0.6249 -0.7191
- 0.387
0.387
0.180
- 0.180
- 0.111
0.111
0.052
- 0.052
- 0.032
0.032
-0.359
-0.207
-0.103
-0.059
-0.029
-0.017
- 0.015
- 0.009
0.009
0.005
0.004
- 0.004
- 0.002
0.002
0.001
0.001
- 0.001
-0.001
3.51
- 3.51
3.51
-3.51
0.002
1.750
-2498
0.015
0.008
Σ
D
DC
-0.008
-0.005
0.002
-1.75 k # ft
641
Ans.
Ans:
MAB = 1.75 k # ft; MBA = 3.51 k # ft;
MBC = -3.51 k # ft; MCB = 3.51 k # ft;
MCD = -3.51 k # ft; MDC = - 1.75 k # ft;
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–8. Solve Prob. 13–7 using the slope-deflection equations.
2 k>ft
5 ft
C
B
2 ft
3 ft
2 ft
2 ft
Solution
aB = aC =
3.2
= 0.2
16
rB = rC =
5 - 2.5
= 1
2.5
A
CBC = CCB = 0.619
kBC = kCB = 6.41
1FEM2 BC = 0.14591221162 = -4.6688 k # ft
1FEM2 CB = -4.6688 k # ft
kBCEIc
KBC = KCB =
=
L
6.411E2a
1
b11212.52 2
12
= 0.5216E
16
MN = KN[uN + CNuF - c11 + CN 2] + 1FEM2 N
MAB =
2EI
10 + uB - 02 + 0
15
MBA =
2EI
12uB + 0 - 02 + 0
15
MCD =
2EI
12uC + 0 - 02 + 0
15
MDC =
2EI
10 + uC - 02 + 0
15
MBC = 0.5216E1uB + 0.6191uC 2 - 02 - 4.6688
MCB = 0.5216E1uC + 0.6191uB 2 - 02 + 4.6688
Equilibrium.
MBA + MBC = 0
MCB + MCD = 0
Or
2E a
1
b112132 3
12
12uB 2 + 0.5216E[uB + 0.619uC] - 4.6688 = 0
15
1.1216uB + 0.32287uC =
2E a
4.6688
E
(1)
1
b 112132 3
12
12uC 2 + 0.5216E[uC + 0.619uB] + 4.6688 = 0
15
1.1216uC + 0.32287uB = -
4.6688
(2)
E
642
9 ft
3 ft
5 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–8.
(Continued)
Solving Eqs. 1 and 2:
uB = -uC =
5.84528
E
MAB = 1.75 k # ft
MBA = 3.51 k # ft
MBC = - 3.51 k # ft
MCB = 3.51 k # ft
MCD = -3.51 k # ft
MDC = -1.75 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
MAB = 1.75 k # ft
MBA = 3.51 k # ft
MBC = - 3.51 k # ft
MCB = 3.51 k # ft
MCD = - 3.51 k # ft
MDC = - 1.75 k # ft
643
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–9. Use the moment-distribution method to determine
the moment at each joint of the frame. Assume that E is constant and the members have a thickness of 1 ft. The supports
at A and D are pinned and the joints at B and C are fixed connected. Use Table 13.1.
6k
2.5 ft
9 ft
6k
12 ft
9 ft
B
C
2.5 ft
6 ft
Solution
Using Table 13–1,
1 ft
6 ft
1 ft
1 ft
20 ft
aB = aC =
6
= 0.2
30
rB = rC =
2.5 - 1
= 1.5
1
CBC = CCB = 0.691
A
D
KBC = KCB = 9.08
b =
9
= 0.3
30
FEMBC = - 10.209721621302 - 0.05151621302
= -47.016
FEMCB = 0.05151621302 + 10.209721621302
= 47.016
3EI
20
DFBA =
= 0.3314
3EI
9.08EI
+
20
30
9.08EI
30
DFBC =
= 0.6686
3EI
9.08EI
+
20
30
Joint
Member
DF
COF
FEM
A
AB
1
B
BA
0.3314
0.5
15.581
7.199
3.326
1.536
0.710
0.328
0.152
0.070
0.032
0.015
0.007
0
0.003
29.0
BC
0.6686
0.691
- 47.016
31.435
- 21.722
14.523
- 10.035
6.710
- 4.636
3.100
- 2.142
1.432
- 0.990
0.662
- 0.457
0.306
- 0.211
0.141
- 0.098
0.065
- 0.045
0.030
- 0.021
0.014
- 0.010
0.006
- 29.0
C
CB
0.6686
0.691
47.016
- 31.435
21.722
- 14.523
10.035
- 6.710
4.636
- 3.100
2.142
- 1.432
0.990
- 0.662
0.457
- 0.306
0.211
- 0.141
0.098
- 0.065
0.045
- 0.030
0.021
- 0.014
0.010
- 0.006
29.0
D
CD
DC
0.3314 1
0.5
-15.581
-7.199
-3.326
-1.536
-0.710
-0.328
-0.152
-0.070
-0.032
-0.015
-0.007
-0.003
-29.0
0 k # ft
644
Ans.
Ans.
MBA = 29.0 k # ft; MBC = -29.0 k # ft;
MCB = 29.0 k # ft; MCD = -29.0 k # ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–10.
6k
Solve Prob. 13–9 using the slope-deflection equations.
2.5 ft
9 ft
6k
12 ft
9 ft
B
C
2.5 ft
6 ft
1 ft
6 ft
1 ft
1 ft
20 ft
A
D
Solution
See the first part of Prob. 13-9.
cAB = cBD = cDC = 0
Applying Eq. 13–8,
MBA =
3EIC
1uB 2
20
MBC =
9.08EIC
1uB + 0.6911uC 22 - 47.016
30
MCB =
9.08EIC
1uC + 0.691uB 2 + 47.016
30
MCD =
3EIC
1uC 2
20
Moment equilibrium at B and C:
MBA + MBC = 0
0.45267EICuB + 0.20914EICuC = 47.016
MCB + MCD = 0
0.20914EICuB + 0.45267EICuC = - 47.016
uB =
193.06516
EIC
uC =
- 193.06516
EIC
Thus,
MBA = 29.0 k # ft
MBC = - 29.0 k # ft
MCB = 29.0 k # ft
MCD = - 29.0 k # ft
Ans.
Ans.
Ans.
Ans.
Ans.
MBA = 29.0 k # ft; MBC = - 29.0 k # ft;
MCB = 29.0 k # ft; MCD = - 29.0 k # ft
645
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–11. Use the moment-distribution method to determine
the moment at each joint of the frame. The supports at A and
C are pinned and the joints at B and D are fixed c­ onnected.
Assume that E is constant and the members have a thickness
of 1 ft. The haunches are tapered, so use Table 13.1.
500 lb>ft
2.5 ft
B
D
6 ft
20 ft
1 ft
A
Solution
For span BD,
aB = aD =
6
= 0.2
30
rB = rD =
2.5 - 1
= 1.5
1
From Table 13.1,
CBD = CDB = 0.691
kBD = kDB = 9.08
KBD = KDB =
kEIC
9.08EI
=
= 0.30267EI
L
30
1FEM2 BD = -0.102110.521302 2 = - 45.945 k # ft
1FEM2 DB = 45.945 k # ft
For span AB and CD,
KBA = KDC =
3EI
= 0.15EI
20
1FEM2 AB = 1FEM2 BA = 1FEM2 DC = 1FEM2 CD = 0
646
1 ft
18 ft
2.5 ft
6 ft
1 ft
C
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–11.
(Continued)
Joint
A
Mem.
AB
BA
0.15EI
0.30267EI
0.30267EI
0.15EI
1
0.3314
0.6686
0.6686
0.3314
0
0.691
0.691
0
K
DF
COF
B
- 45.95
45.95
- 30.72
- 21.22
21.22
7.03
14.19
- 14.19
- 9.81
9.81
CO
Dist.
3.25
CO
Dist.
1.50
CO
Dist.
0.69
CO
Dist.
0.32
aM
0
3.03
-3.03
- 2.09
2.09
1.40
-1.40
- 0.97
0.97
1
-7.03
-3.25
-1.50
-0.69
-0.32
-0.65
0.15
0.30
-0.30
- 0.21
0.21
0.07
0.14
-0.14
- 0.10
0.10
0.03
0.06
-0.06
- 0.04
0.04
0.01
0.03
-0.03
-0.01
28.3
- 28.3
28.3
-28.3
CO
Dist.
4.53
CD
-15.23
0.45
CO
Dist.
-6.56
- 4.53
DC
- 0.45
CO
Dist.
6.56
C
0.65
CO
Dist.
DB
30.72
CO
Dist.
BD
15.23
FEM
Dist.
D
-0.15
-0.07
-0.03
0
k·ft
Ans.
Ans.
ΣM
647
0
28.3
-28.3
28.3
- 28.3 0 k·ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–12. Solve Prob. 13–11 using the slope-deflection
equations.
500 lb>ft
2.5 ft
B
D
6 ft
20 ft
1 ft
A
1 ft
18 ft
2.5 ft
6 ft
1 ft
C
Solution
See Prob. 13-11 for the tabular data.
For span AB,
MN = 3E
I
[u - c] + 1FEM2 N
L N
MBA = 3E a
MBA =
I
b[uB - 0] + 0
20
3EI
u
20 B
(1)
For span BD,
MN = KN[uN + CNuF - c11 + CN 2] + 1FEM2 N
MBD = 0.30267EI1uB + 0.691uD - 02 - 45.945
MBD = 0.30267EIuB + 0.20914EIuD - 45.945
(2)
MDB = 0.30267EI1uD + 0.691uB - 02 + 45.945
MDB = 0.30267EIuD + 0.20914EIuB - 45.945
(3)
For span DC,
MN = 3E
I
[u - c] + 1FEM2N
L N
MDC = 3E a
MDC =
I
b[uD - 0] + 0
20
3EI
u
20 D
(4)
Equilibrium equations:
MBA + MBD = 0(5)
MDB + MDC = 0(6)
Solving Eqs. 5–6 simultaneously,
uB =
188.67
EI
uD =
MBA = 28.3 k # ft
MBD = - 28.3 k # ft
MDB = 28.3 k # ft
MDC = - 28.3 k # ft
MAB = MCD = 0
188.67
EI
Ans.
Ans.
MBA = 28.3 k # ft
MBD = -28.3 k # ft
MDB = 28.3 k # ft
MDC = -28.3 k # ft
MAB = MCD = 0
Ans.
Ans.
Ans.
Ans.
648
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–1. Determine the stiffness matrix K for the truss. AE is
constant.
2
1
4 kN 1
1
4
2
2m
5
4
908
8
3
3
2m
Solution
Member 1:
lx =
1-0
= 0.7071
12
1
0.3536
- 0.3536
k1 = AE ≥
-0.3536
0.3536
ly =
2
- 0.3536
0.3536
0.3536
- 0.3536
1-2
= - 0.7071
12
3
- 0.3536
0.3536
0.3536
- 0.3536
4
0.3536
-0.3536
¥
-0.3536
0.3536
1
2
3
4
Member 2:
lx =
2-1
= 0.7071
12
3
0.3536
- 0.3536
k2 = AE ≥
-0.3536
0.3536
ly =
4
- 0.3536
0.3536
0.3536
- 0.3536
0-1
= - 0.7071
12
5
- 0.3536
0.3536
0.3536
- 0.3536
6
0.3536
-0.3536
¥
-0.3536
0.3536
3
4
5
6
Member 4:
lx =
0-1
= - 0.7071
12
3
0.3536
0.3536
k4 = AE ≥
- 0.3536
- 0.3536
ly =
4
0.3536
0.3536
- 0.3536
- 0.3536
0-1
= - 0.7071
12
7
-0.3536
- 0.3536
0.3536
0.3536
8
-0.3536
-0.3536
¥
0.3536
0.3536
3
4
7
8
Member 3:
lx =
0-2
= -1
2
5
0.5
0
k3 = AE ≥
- 0.5
0
ly = 0
6
0
0
0
0
7
- 0.5
0
0.5
0
8
0
0
¥
0
0
5
6
7
8
649
2
6
4
7
3
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–1.
(Continued)
Member 5:
lx = 0
ly =
1
0
0
k5 = AE ≥
0
0
0 - 2
= -1
2
2
0
0.5
0
- 0.5
7
0
0
0
0
8
0
-0.5
¥
0
0.5
1
2
7
8
Structure stiffness matrix:
K = k1 + k2 + k3 + k4 + k5
0.3536
- 0.3536
- 0.3536
0.3536
K = AE H
0
0
0
0
- 0.3536
0.8536
0.3536
- 0.3536
0
0
0
- 0.5
-0.3536
0.3536
1.0607
- 0.3536
-0.3536
0.3536
- 0.3536
- 0.3536
0.3536
- 0.3536
-0.3536
1.0607
0.3536
- 0.3536
- 0.3536
- 0.3536
0
0
-0.3536
0.3536
0.8536
- 0.3536
- 0.5
0
0
0
0.3536
-0.3536
- 0.3536
0.3536
0
0
0
0
- 0.3536
-0.3536
- 0.5
0
0.8536
0.3536
0
- 0.5
-0.3536
-0.3536
X
0
0
0.3536
0.8536
0.3536
- 0.3536
-0.3536
0.3536
K = AE H
0
0
0
0
-0.3536
0.8536
0.3536
-0.3536
0
0
0
- 0.5
-0.3536
0.3536
1.0607
- 0.3536
-0.3536
0.3536
-0.3536
-0.3536
0.3536
-0.3536
-0.3536
1.0607
0.3536
- 0.3536
- 0.3536
-0.3536
0
0
-0.3536
0.3536
0.8536
- 0.3536
- 0.5
0
Ans.
Ans.
650
0
0
0.3536
-0.3536
- 0.3536
0.3536
0
0
0
0
- 0.3536
-0.3536
- 0.5
0
0.8536
0.3536
0
- 0.5
-0.3536
- 0.3536
X
0
0
0.3536
0.8536
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–2. Determine the force in members 1 and 5. AE is
constant.
2
1
4 kN 1
1
4
2
2m
5
4
908
8
3
3
2m
Solution
4 1
0 6
0 2
Dk = C 0 S 7 Qk = E 0 U 3
0 8
0 4
0 5
Using the structure stiffness matrix of Prob. 14–1 and applying
Q = KD, we have
4
0.3536
0
- 0.3536
0
- 0.3536
0
0.3536
H X = AE H
0
0
Q6
0
Q7
0
Q8
0
- 0.3536
0.8536
0.3536
- 0.3536
0
0
0
- 0.5
- 0.3536
0.3536
1.0607
- 0.3536
-0.3536
0.3536
- 0.3536
- 0.3536
0.3536
- 0.3536
-0.3536
1.0607
0.3536
- 0.3536
- 0.3536
- 0.3536
0
0
-0.3536
0.3536
0.8536
- 0.3536
- 0.5
0
- 0.3536
0.8536
0.3536
-0.3536
0
-0.3536
0.3536
1.0607
- 0.3536
-0.3536
0.3536
- 0.3536
- 0.3536
1.0607
0.3536
D1
0
0
0
D2
0
- 0.3536 U E D3 U + E 0 U
0.3536
D4
0
0.8536
D5
0
0
0
0.3536
-0.3536
- 0.3536
0.3536
0
0
0
0
- 0.3536
-0.3536
- 0.5
0
0.8536
0.3536
Partition matrix:
4
0.3536
0
- 0.3536
E 0 U = AE E -0.3536
0
0.3536
0
0
4 = AE10.3536D1 - 0.3536D2 - 0.3536D3 + 0.3536D4 2(1)
0 = AE1 -0.3536D1 + 0.8536D2 + 0.3536D3 - 0.3536D4 2(2)
0 = AE1 -0.3536D1 + 0.3536D2 + 1.0607D3 - 0.3536D4 - 0.3536D5 2(3)
0 = AE10.3536D1 - 0.3536D2 - 0.3536D3 + 1.0607D4 + 0.3536D5 2(4)
0 = AE1 -0.3536D3 + 0.3536D4 + 0.8536D5 2(5)
Solving the above equations yields:
D1 =
38.627
8.00
9.656
- 9.656
8.00
, D2 =
, D3 =
, D4 =
, D5 =
AE
AE
AE
AE
AE
651
2
6
4
7
3
0
D1
- 0.5
D2
D3
-0.3536
-0.3536
D
X H 4X
D5
0
0
0
0.3536
0
0.8536
0
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–2.
(Continued)
For member 1,
lx = 0.7071, ly = - 0.7071, L = 1.414 m
AE
q1 =
[ - 0.7071 0.7071 0.7071
1.414
= - 5.66 kN = 5.66 kN 1C2
38.627 1
1
8.000 2
- 0.7071]
D
T
AE
9.656 3
-9.656 4
Ans.
For member 5,
lx = 0 ly = - 1 L = 2 m
38.627 1
AE
1
8.00 2
q5 =
D
[0 1 0 - 1]
T
0
7
2
AE
0
8
= 4.00 kN 1T2
Ans.
652
Ans.
q1 = 5.66 kN 1C2
q5 = 4.00 kN 1T2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–3. Determine the stiffness matrix K for the truss. Take
A = 0.0015 m2 and E = 200 GPa for each member.
2m
5
2m
4
10
6
4
5
4
3
9
8
2
1
1
7
6
3
2m
2
5
3
1
2
30 kN
Solution
The origin of the global coordinate system will be set at joint 1 .
For member 1 , L = 2 m. lx =
5
1
0.0015[2001109 2] 0
k1 =
D
2
-1
0
0 - 2
0 - 0
= - 1 ly =
= 0
2
2
6 7 8 5
-1
0
1
0
0
0
0
0
For member 2 , L = 2 m. lx =
0 5
0 6
T
0 7
0 8
0
0
0
0
-1
0
1
0
For member 3 , L = 212m. lx =
6 1
0 1
0 2
T
0 5
0 6
=
2 1501106 2
0
D
-1501106 2
0
2 - 4
12
= 2
212
ly =
0
0
0
0
k3 =
212
0.5
- 0.5
D
- 0.5
0.5
-0.5
0.5
0.5
- 0.5
0 5
0 6
T
0 7
0 8
-0.5
0.5
0.5
- 0.5
0.5 1
- 0.5 2
T
- 0.5 3
0.5 4
=
5 6
-1501106 2
0
1501106 2
0
0 1
0 2
T
0 5
0 6
2 - 0
12
= 2
212
1 2 3 4 1
0.0015[2001109 2]
7 8
-1501106 2
0
1501106 2
0
0
0
0
0
2 - 4
0 - 0
= - 1 ly =
= 0
2
2
1 2 5
1
0.0015[2001109 2]
0
k2 =
D
2
-1
0
=
6 1501106 2
0
D
-1501106 2
0
2 3 4
53.0331106 2
-53.0331106 2
D
-53.0331106 2
53.0331106 2
653
- 53.0331106 2
53.0331106 2
53.0331106 2
- 53.0331106 2
- 53.0331106 2
53.0331106 2
53.0331106 2
- 53.0331106 2
53.0331106 2 1
- 53.0331106 2 2
T
- 53.0331106 2 3
53.0331106 2 4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–3.
(Continued)
5
2 - 2
2 - 0
= 0 ly =
= 1
2
2
6 3
4 5
6 3 4
0
0
D
0
0
0
1
0
-1
For member 4 , L = 2 m. lx =
k4 =
0.0015[2001109 2]
2
0
0
0
0
0 5
-1 6
T
0 3
1 4
0
1501106 2
0
- 1501106 2
0 - 2
12
= 2
212
For member 5 , L = 212 m. lx =
3
=
0
0
D
0
0
ly =
0
0
0
0
0 - 2
12
= 2
212
4 7 8 3
0.5
0.0015[2001109 2] 0.5
D
k5 =
- 0.5
212
- 0.5
0.5
0.5
- 0.5
- 0.5
- 0.5
- 0.5
0.5
0.5
For member 6 , L = 2 m. lx =
- 0.5 3
-0.5 4
T
0.5 7
0.5 8
=
5
0
-1501106 2 6
T
3
0
1501106 2 4
4 7 8
53.0331106 2
53.0331106 2
D
-53.0331106 2
- 53.0331106 2
0 - 2
2 - 2
= -1 ly =
= 0
2
2
53.0331106 2
53.0331106 2
-53.0331106 2
- 53.0331106 2
3 4 9 10 3 4 9 10
1
0.0015[2001109 2] 0
D
k6 =
2
-1
0
0
0
0
0
-1
0
1
0
0 3
0 4
T
0 9
0 10
=
1501106 2
0
D
-1501106 2
0
654
0
0
0
0
-1501106 2
0
1501106 2
0
0 3
0 4
T
0 9
0 10
-53.0331106 2
-53.0331106 2
53.0331106 2
53.0331106 2
-53.0331106 2 3
-53.0331106 2 4
T
53.0331106 2 7
53.0331106 2 8
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–3.
(Continued)
Structure stiffness matrix is a 10 * 10 matrix since the highest code number is 10. Thus,
1 2 3 4 5 6 7 8 9 10
K =
203.033
- 53.033
- 53.033
53.033
- 150
I
0
0
0
0
0
-53.033
53.033
53.033
- 53.033
0
0
0
0
0
0
- 53.033
53.033
256.066
0
0
0
-53.033
-53.033
-150
0
53.033
- 53.033
0
256.66
0
- 150
-53.033
-53.033
0
0
-150
0
0
0
300
0
-150
0
0
0
0
0
0
- 150
0
150
0
0
0
0
0
0
-53.033
-53.033
-150
0
203.033
53.033
0
0
0
0
- 53.033
- 53.033
0
0
53.033
53.033
0
0
0
0
-150
0
0
0
0
0
150
0
0 1
0 2
0 3
0 4
0 5
Y 1106 2 N>m
0 6
0 7
0 8
0 9
0 10
Ans.
Ans.
1 2 3 4 5 6 7 8 9 10
K =
203.033
- 53.033
- 53.033
53.033
- 150
I
0
0
0
0
0
- 53.033
53.033
53.033
- 53.033
0
0
0
0
0
0
-53.033
53.033
256.066
0
0
0
- 53.033
- 53.033
- 150
0
53.033
- 53.033
0
256.66
0
- 150
-53.033
-53.033
0
0
655
- 150
0
0
0
300
0
- 150
0
0
0
0
0
0
- 150
0
150
0
0
0
0
0
0
-53.033
-53.033
-150
0
203.033
53.033
0
0
0
0
- 53.033
-53.033
0
0
53.033
53.033
0
0
0
0
-150
0
0
0
0
0
150
0
0 1
0 2
0 3
0 4
0 5
Y 1106 2 N>m
0 6
0 7
0 8
0 9
0 10
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2m
*14–4. Determine the vertical displacement at joint 2 and
the force in member 5. Take A = 0.0015 m2 and E = 200 GPa.
5
2m
4
10
6
4
5
4
3
9
8
2
1
7
6
3
2m
2
5
1
3
2
1
30 kN
Solution
Here,
0
1
- 301103 2 2
0
3
V
Qk = F
0
4
0
5
0
6
0 7
0 8
Dk = D T
0 9
0 10
Then, applying Q = KD,
0
203.033
-301103 2
- 53.033
0
- 53.033
53.033
0
0
- 150
I
Y = I
0
0
Q7
0
Q8
0
Q9
0
Q10
0
- 53.033
53.033
53.033
- 53.033
0
0
0
0
0
0
-53.033
53.033
256.066
0
0
0
- 53.033
- 53.033
- 150
0
53.033
- 53.033
0
256.066
0
- 150
- 53.033
-53.033
0
0
- 150
0
0
0
300
0
-150
0
0
0
656
0
0
0
- 150
0
150
0
0
0
0
0
0
- 53.033
- 53.033
- 150
0
203.033
53.033
0
0
0
0
- 53.033
-53.033
0
0
53.033
53.033
0
0
0
0
-150
0
0
0
0
0
150
0
0
D1
0
D2
0
D3
0
D4
0
D
6
Y 110 2 I 5 Y
D6
0
0
0
0
0
0
0
0
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–4.
(Continued)
From the matrix partition, Qk = K11Du + K12Dk is given by
0
203.033
- 301103 2
- 53.033
0
- 53.033
F
V = F
0
53.033
0
- 150
0
0
- 53.033
53.033
53.033
- 53.033
0
0
-53.033
53.033
256.066
0
0
0
53.033
-53.033
0
256.066
0
-150
- 150
0
0
0
300
0
0
D1
0
0
D2
0
D
0
0
V 1106 2 F 3 V + F V
- 150
D4
0
D5
0
0
150
D6
0
Expanding this matrix equality,
0 = [203.033D1 - 53.033D2 - 53.033D3 + 53.033D4 - 150D5]1106 2(1)
- 301103 2 = [ - 53.033D1 + 53.033D2 + 53.033D3 - 53.033D4]1106 2(2)
0 = [ -53.033D1 + 53.033D2 + 256.066D3]1106 2(3)
0 = [53.033D1 - 53.033D2 + 256.066D4 - 150D6]1106 2(4)
0 = [ -150D1 + 300D5]1106 2(5)
0 = [ -150D4 + 150D6]1106 2(6)
Solving Eqs. (1) to (6),
D1 = -0.0004 m D2 = - 0.0023314 m D3 = 0.0004 m D4 = -0.00096569 m
Ans.
D5 = -0.0002 m D6 = - 0.00096569 m = 0.000966 mT Ans.
Force in member 5 : Here lx = -
12
12
, ly = and L = 212 m.
2
2
Applying Eqs. 14–23,
1q5 2 F =
0.0015[2001109 2
212
12
c
2
12
2
12
2
0.0004
3
12
- 0.00096569 4
d D
T
2
0
7
0
8
Ans.
= -42.4 kN = 42.4 kN 1C2
657
Ans.
D1 = -0.0004 m
D2 = -0.0023314 m
D3 = 0.0004 m
D4 = -0.00096569 m
D5 = -0.0002 m
D6 = 0.000966 mT
1q5 2 F = -42.4 kN
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–5. Determine the stiffness matrix K for the truss. Take
A = 0.75 in2, E = 29(103) ksi. Assume all joints are pin
connected.
4
3
2
1
4 ft
6
2
5
2
1
3
3 ft
3
8
4
1
2k
7
4 ft
Solution
Member 2:
Member 1:
lx =
0 - 4
1
7 - 3
1
= ; ly =
=
412
12
412
12
1
0.0884
AE -0.0884
D
k1 =
12 -0.0884
0.0884
lx =
2
3
4
- 0.0884
0.0884
0.0884
- 0.0884
- 0.0884
0.0884
0.0884
- 0.0884
0.0884 1
-0.0884 2
T
-0.0884 3
0.0884 4
0 - 4
3 - 3
= -1; ly =
= 0
4
4
1
0.25
AE
0
k2 =
D
12 - 0.25
0
2
0
0
0
0
5
- 0.25
0
0.25
0
6
0 1
0 2
T
0 5
0 6
Member 3:
lx =
0 - 4
0 - 3
= -0.8; ly =
= - 0.6
5
5
1
0.128
AE 0.096
D
k3 =
12 -0.128
- 0.096
2
7
8
0.096
0.072
- 0.096
- 0.072
-0.128
- 0.096
0.128
0.096
- 0.096 1
-0.072 2
T
0.096 7
0.072 8
Structure stiffness matrix: K = k1 + k2 + k3
Substituting A = 0.75 in2, E = 291103 2 ksi,
845.3289
13.79611
- 160.2039
160.2039
K = H
- 453.125
0
- 232
- 174
13.79611
290.7039
160.2039
- 160.2039
0
0
- 174
- 130.5
- 160.2039
160.2039
160.2039
- 160.2039
0
0
0
0
160.2039
- 160.2039
- 160.2039
160.2039
0
0
0
0
- 453.125
0
0
0
453.125
0
0
0
0
0
0
0
0
0
0
0
- 232
- 174
0
0
0
0
232
174
-174
-130.5
0
0
X k>in.
0
0
174
130.5
Ans.
Ans.
845.3289
13.79611
- 160.2039
160.2039
K = H
- 453.125
0
- 232
- 174
13.79611
290.7039
160.2039
-160.2039
0
0
-174
- 130.5
-160.2039
160.2039
160.2039
-160.2039
0
0
0
0
658
160.2039
- 160.2039
- 160.2039
160.2039
0
0
0
0
- 453.125
0
0
0
453.125
0
0
0
0
0
0
0
0
0
0
0
- 232
- 174
0
0
0
0
232
174
-174
-130.5
0
0
X k>in.
0
0
174
130.5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4
14–6. Determine the vertical deflection of joint 1 and the
force in member 2 of the truss in Prob. 14–5.
3
2
1
4 ft
6
2
5
2
1
3
3 ft
3
8
4
1
2k
7
4 ft
Solution
0 3
0 4
0 5
Dk = F V
0 6
0 7
0 8
Qk = c
0 1
d
- 2 2
Using the structure stiffness matrix of Prob. 14–5 and
applying Q = KD,
0
845.3289
-2
13.79611
Q3
- 160.2039
Q
160.2039
H 4X = H
Q5
- 453.125
Q6
0
Q7
- 232
Q8
- 174
13.79611
290.7039
160.2039
-160.2039
0
0
-174
-130.5
- 160.2039
160.2039
160.2039
-160.2039
0
0
0
0
160.2039
-160.2039
-160.2039
160.2039
0
0
0
0
- 453.125
0
0
0
453.125
0
0
0
0
0
0
0
0
0
0
0
- 232
- 174
0
0
0
0
232
174
-174
D1
-130.5
D2
0
0
0
0
XH X
0
0
0
0
0
174
130.5
0
Partition matrix:
0 = 845.3289D1 + 13.79611D2
- 2 = 13.79611D1 + 290.7039D2
Solving the above linear equations:
D1 = 0.11237110 - 3 2 in.; D2 = - 6.8852110 - 3 2 in. = 6.8852110 - 3 2 in.T Ans.
To find force in member 2:
lx =
0 - 4
3 - 3
= - 1; ly =
= 0
4
4
q2 =
0.75129110322
48
[1 0
0.11237 1
6.8852 2
- 1 0]110 - 3 2 D
T
0
5
0
6
Ans.
q2 = 50.9 lb 1T2
Ans.
D2 = 6.8852110 - 3 2 in.T
659
q2 = 50.9 lb 1T2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–7. Determine the stiffness matrix K for the truss. AE is
constant.
3m
6
5
3
2
2
3
3 kN
4m
1
2
1
4
3
458
Solution
The origin of the global coordinate system is set at joint 1 .
4
For member 2, L = 5 m. Referring to Fig. a, ux″ = 180° - 45° - sin - 1 a b = 81.87° uy″ = 171.87°.
5
Thus, lx″ = cos ux″ = cos 81.87° = 0.14142 and ly″ = cos uy″ = cos 171.87° = - 0.98995.
Also, lx =
0 - 3
0 - 4
= - 0.6 and ly =
= - 0.8.
5
5
1
0.072
0.096
k2 = AE D
0.01697
- 0.11879
2
0.096
0.128
0.02263
- 0.15839
3
0.01697
0.02263
0.004
- 0.028
4
- 0.11879 1
- 0.15839 2
T
-0.028 3
4
0.196
For member 1 , L = 4 m. Referring to Fig. b, ux″ = 45° and uy″ = 135°.
Thus, lx″ = cos 45° =
12
12
and ly″ = cos 135° = .
2
2
Also, lx = 0 and ly = - 1.
5 6 3 4
0
0
0
0
5
0
0.25
0.17678 -0.17678 6
k1 = AE D
T
0
0.17678
0.125
-0.125 3
0 -0.17678 - 0.125
0.125
4
For member 3 , L = 3 m, lx = 1, and ly = 0.
5 6 1 2
0.33333
0 - 0.33333 0 5
0
0
0
0 6
k3 = AE D
T
-0.33333 0
0.33333
0 1
0
0
0
0 2
660
1
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The structure stiffness matrix is a 6 * 6 matrix since the highest
code number is 6. Thus,
0.40533
0.096
0.01697
0.096
0.128
0.02263
0.01697
0.02263
0.129
K = AE F
- 0.11879 - 0.15839 - 0.153
-0.33333
0
0
0
0
0.17678
-0.11879
- 0.15839
-0.153
0.321
0
-0.17678
- 0.33333
0
0
0
0.33333
0
0
0
0.17678
V
- 0.17678
0
0.25
Ans.
Ans.
0.40533
0.096
0.01697
0.096
0.128
0.02263
0.01697
0.02263
0.129
K = AE F
- 0.11879 - 0.15839 -0.153
-0.33333
0
0
0
0
0.17678
661
-0.11879
-0.15839
-0.153
0.321
0
- 0.17678
- 0.33333
0
0
0
0.33333
0
0
0
0.17678
V
- 0.17678
0
0.25
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*14–8. Determine the vertical displacement of joint 2 and
the support reactions. AE is constant.
3m
6
2
2
3
5
1
3
3 kN
4m
1
2
1
4
Solution
3
0
1
0 4
Here, Qk = £ - 31103 2 § 2 and Dk = £ 0 § 5
0
3
0 6
458
Applying Q = KD,
0
0.40533
- 31103 2
0.096
0
0.01697
F
V = AE F
Q4
- 0.11879
Q5
- 0.33333
Q6
0
0.096
0.128
0.02263
- 0.15839
0
0
0.01697
0.02263
0.129
-0.153
0
0.17678
-0.11879
- 0.15839
- 0.153
0.321
0
-0.17678
- 0.33333
0
0
0
0.33333
0
0
D1
D2
0
0.17678
D
V F 3V
0
- 0.17678
0
0
0.25
0
From the matrix partition, Qk = K11Du + K12Dk.
0
0.40533
£ -31103 2 § = AE £ 0.096
0.01697
0
0.096
0.128
0.02263
0.01697 D1
0
0.02263 § £ D2 § + £ 0 §
0.129
D3
0
Expanding this matrix equality,
0 = AE10.40533D1 + 0.096D2 + 0.01697D3 2(1)
-31103 2 = AE10.096D1 + 0.128D2 + 0.02263D3 2(2)
0 = AE10.01697D1 + 0.02263D2 + 0.0129D3 2(3)
Solving Eqs. (1) to (3),
D1 =
6.7501103 2
AE
D3 =
4.24261103 2
AE
D2 =
- 29.2501103 2
AE
=
29.31103 2
AE
Ans.
T
Again, the matrix partition Qu = K21Du + K22Dk gives
Q4
- 0.11879
£ Q5 § = AE £ - 0.33333
Q6
0
- 0.15839
0
0
Q4 = 8.1821103 2 N = 3.18 kNQ
Q6 = 750 N c
- 0.153
6.7501103 2
0
1
0
§
£ - 29.2501103 2 § + £ 0 §
AE
0.17678
4.24261103 2
0
Q5 = - 2.2501103 2 N = - 2.25 kN = 2.25 kN d
Ans.
Ans.
Ans.
D1 =
6.7501103 2
AE
Q4 = 3.18 kNQ
Q6 = 750 N c
662
D3 =
4.24261103 2
AE
Q5 = 2.25 kN d
D2 =
29.31103 2
AE
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–9. Determine the stiffness matrix K for the truss. AE is
constant.
2
8
47
2
3
1
5
4
Solution
2
0
0.3333
0
-0.3333
3
0
0
0
0
1
4
1
0
- 0.3333 2
T
3
0
0.3333 4
2
0
0
0
0
7
- 0.25
0
0.25
0
8
0 1
0 2
T
0 7
0 8
Member 3:
lx =
0 - 4
= -1 ly = 0
4
3
0.25
0
k3 = AE D
-0.25
0
4
0
0
0
0
6
- 0.25
0
0.25
0
5
0 3
0 4
T
0 6
0 5
Member 4:
lx = 0 ly =
3 - 0
= 1
3
6
0
0
k4 = AE D
0
0
5
0
0.3333
0
-0.3333
7
0
0
0
0
8
6
0
- 0.3333 5
T
7
0
0.3333 8
Member 5:
lx =
0 - 4
3 - 0
= -0.8 ly =
= 0.6
5
5
3
0.128
- 0.096
k5 = AE D
-0.128
0.096
4
- 0.096
0.072
0.096
- 0.072
3
2
10 kN
0 - 4
= -1 ly = 0
4
1
0.25
0
k2 = AE D
-0.25
0
6
4m
Member 2:
lx =
4
3
0 - 3
lx = 0 ly =
= -1
3
1
0
0
k1 = AE D
0
0
6
5
Member 1:
3m
1
7
-0.128
0.096
0.128
-0.096
8
0.096 3
- 0.072 4
T
-0.096 7
0.072 8
663
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–9.
(Continued)
Member 6:
lx =
0 - 4
0 - 3
= -0.8 ly =
= - 0.6
5
5
1
0.128
0.096
k6 = AE D
-0.128
- 0.096
2
0.096
0.072
- 0.096
- 0.072
6
- 0.128
- 0.096
0.128
0.096
5
-0.096 1
-0.072 2
T
0.096 6
0.072 5
Structure stiffness matrix:
K = k1 + k2 + k3 + k4 + k5 + k6
0.378
0.096
0
0
K = AE H
-0.096
- 0.128
-0.25
0
113.4
28.8
0
0
K = H
- 28.8
- 38.4
- 75
0
0.096
0.4053
0
- 0.3333
- 0.072
-0.096
0
0
28.8
121.6
0
- 100
-21.6
- 28.8
0
0
0
0
113.4
- 28.8
0
- 75
- 38.4
28.8
0
0
0.378
- 0.096
0
- 0.25
- 0.128
0.096
0
- 0.3333
-0.096
0.4053
0
0
0.096
-0.072
0
- 100
- 28.8
121.6
0
0
28.8
- 21.6
-28.8
- 21.6
0
0
121.6
28.8
0
-100
- 0.096
-0.072
0
0
0.4053
0.096
0
-0.3333
- 0.128
-0.096
-0.25
0
0.096
0.378
0
0
-0.25
0
-0.128
0.096
0
0
0.378
-0.096
0
0
0.096
-0.072
X
- 0.3333
0
-0.096
0.4053
- 38.4
-28.8
- 75
0
28.8
113.4
0
0
-75
0
-38.4
28.8
0
0
113.4
-28.8
0
0
28.8
- 21.6
X 1106 2 N>m
- 100
0
-28.8
121.6
113.4
28.8
0
0
K = H
- 28.8
- 38.4
- 75
0
28.8
121.6
0
- 100
- 21.6
- 28.8
0
0
0
0
113.4
-28.8
0
-75
-38.4
28.8
Ans.
Ans.
664
0
-100
-28.8
121.6
0
0
28.8
- 21.6
- 28.8
-21.6
0
0
121.6
28.8
0
- 100
- 38.4
- 28.8
- 75
0
28.8
113.4
0
0
- 75
0
- 38.4
28.8
0
0
113.4
- 28.8
0
0
28.8
- 21.6
X 1106 2 N>m
-100
0
- 28.8
121.6
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–10. Determine the force in member 6. Take A = 0.0015 m2
and E = 200 GPa for each member.
2
8
47
2
3
1
5
4
3m
1
6
5
4
3
1
6
3
2
4m
10 kN
Solution
Using the structure stiffness matrix of Prob. 14.9 and applying
Q = KD, we have
0 1
0 6
0 2
Dk = C 0 S 7 Qx = E 0 U 3
0 8
- 10 4
0 5
0
0.378
0.096
0
0
- 0.096 - 0.128 - 0.25
0
D1
0
0.096
0.4053
0
- 0.3333 -0.072 -0.096
0
0
D2
D3
0
0
0
0.378
-0.096
0
- 0.25 - 0.128 0.096
-10
0
-0.3333 -0.096 0.4053
0
0
0.096
- 0.072
D
X H 4X
H
X = AE H
D5
0
-0.096 -0.072
0
0
0.4053
0.096
0
-0.3333
Q6
- 0.128 -0.096 -0.25
0
0.096
0.378
0
0
0
Q7
- 0.25
0
- 0.128 0.096
0
0
0.378
- 0.096
0
Q8
0
0
0.096
- 0.072 -0.3333
0
-0.096 0.4053
0
Partition matrix:
0
0.378
0
0.096
E 0 U = AE E 0
- 10
0
0
- 0.096
0.096
0.4053
0
- 0.3333
-0.072
0
0
0.378
- 0.096
0
0
- 0.3333
- 0.096
0.4053
0
- 0.096
D1
0
- 0.072
D2
0
0 U E D3 U + E 0 U
0
D4
0
0.4043
D5
0
0 = AE10.378D1 + 0.096D2 - 0.096D5 2(1)
0 = AE10.096D1 + 0.4053D2 - 0.0333D4 - 0.072D5 2(2)
0 = AE10.378D3 - 0.096D4 2(3)
- 10 = AE1 - 0.3333D2 - 0.096D3 + 0.4053D4 2(4)
0 = AE1 -0.096D1 - 0.072D2 + 0.4053D3 2(5)
Solving the above equations yields:
D1 =
23.3333
- 105
- 30
-118.125
-13.125
, D2 =
, D3 =
, D4 =
, D5 =
AE
AE
AE
AE
AE
Ans.
For member 6,
lx = - 0.8, ly = - 0.6, L = 5 m
AE
q6 =
[0.8 0.6
5
- 0.8
23.3333 1
1
- 105 2
- 0.6]
D
T
0
6
AE
- 13.125 5
= - 7.292 kN = 7.29 kN 1C2
Ans.
665
Ans.
q6 = 7.29 kN 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–11. Determine the force in member 1 if this member was
10 mm too long before it was fitted into the truss. For the
solution remove the 10-kN load. Take A = 0.0015 m2 and
E = 200 GPa for each member.
2
8
47
2
3
1
5
4
3m
1
6
5
4
3
1
6
3
2
4m
10 kN
Solution
1Q1 2 0
0 1
0
1
AE10.012 - 1 2
1Q2 2 0
- 0.003333 2
D
T =
D
T = AE D
T
1Q3 2 0
3
0 3
0
3
1Q4 2 0
1 4
0.003333 4
Use the structure stiffness matrix of Prob. 14-9.
0
0.378
0
0.096
0
0
0
0
H X = AE H
0
- 0.096
Q6
- 0.128
Q7
- 0.25
Q8
0
0.096
0.4053
0
- 0.3333
- 0.072
- 0.096
0
0
0
0
0.378
-0.096
0
- 0.25
- 0.128
0.096
0
-0.3333
-0.096
0.4053
0
0
0.096
-0.072
- 0.096
- 0.072
0
0
0.4053
0.096
0
-0.3333
- 0.128
- 0.096
- 0.25
0
0.096
0.378
0
0
-0.25
0
-0.128
0.096
0
0
0.378
-0.096
0
D1
0
0
D2
-0.003333
D3
0.096
0
-0.072
D4
0.003333
X H X + AE H
X
D5
- 0.3333
0
0
0
0
-0.096
0
0
0.4053
0
0
0 = 0.378D1 + 0.096D2 + 0D3 + 0D4 - 0.096D5 + 0(1)
0 = 0.096D1 + 0.4053D2 + 0D3 - 0.3333D4 - 0.072D5 - 0.003333(2)
0 = 0D1 + 0D2 + 0.378D3 - 0.096D4 + 0D5 + 0(3)
0 = 0D1 = -0.3333D2 - 096D3 + 0.4053D4 + 0D5 + 0.003333(4)
0 = - 0.096D1 - 0.072D2 + 0D3 + 0D4 + 0.4053D5 + 0(5)
Solving the above equations yields:
D1 = -0.0011111, D2 = 0.005
D3 = -0.0011111, D4 = - 0.004375
D5 = 0.000625
The force in member 1:
lx = 0, ly = - 1, L = 3 m
q1 =
0.0015120021109 2
3
[0 1 0
- 0.001111 1
0.0050000 2
- 1] D
T - 10.00152120021109 210.0033332
- 0.001111 3
- 0.004375 4
Ans.
= - 62.5 kN = 62.5 kN 1C2
666
Ans.
q1 = 62.5 kN 1C2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*14–12. Determine the stiffness matrix K for the truss. AE is
constant.
2
3
1
8k
1
2
4
10 ft
6
4
3
5
2
4
3
5
10
8
5
9
Member 1: lx =
10 - 0
10 - 20
= 0.7071 ly =
= -0.7071
1200
1200
0.03536
- 0.03536
k1 = AE D
- 0.03536
0.03536
- 0.03536
0.03536
0.03536
-0.03536
-0.03536
0.03536
0.03536
-0.03536
Member 2: lx = 0 ly =
10 - 20
= -1
10
0
0
k2 = AE D
0
0
0
- 0.10
T
0
0.10
0
0.10
0
-0.10
0
0
0
0
Member 3: lx = 0 ly =
0 - 10
= -1
10
0
0
k3 = AE D
0
0
0
- 0.10
T
0
0.10
0
0.10
0
-0.10
Member 4: lx =
0.10
0
k4 = AE D
- 0.10
0
0
0
0
0
0.03536
- 0.03536
T
- 0.03536
0.03536
10 - 0
= 1 ly = 0
10
0
0
0
0
- 0.10
0
0.10
0
0
0
T
0
0
Member 5:
lx =
0 - 10
0 - 10
= - 0.7071 ly =
= - 0.7071
1200
1200
0.03536
0.03536
k5 = AE D
- 0.03536
- 0.03536
0.03536
0.03536
-0.03536
- 0.03536
-0.03536
-0.03536
0.03536
0.03536
-0.03536
-0.03536
T
0.03536
0.03536
667
1
7
10 ft
Solution
10 ft
6
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–12.
(Continued)
Member 6: lx = 0 ly =
0 - 10
= -1
10
0
0
k6 = AE D
0
0
0
- 0.10
T
0
0.10
0
0.10
0
- 0.10
0
0
0
0
Structure stiffness matrix:
K = k1 + k2 + k3 + k4 + k5 + k6
0.03536
- 0.03536
0
0
- 0.03536
K = AE I
0.03536
0
0
0
0
- 0.03536
0.13536
0
- 0.10
0.03536
- 0.03536
0
0
0
0
0
0
0.10
0
-0.10
0
0
0
0
0
0
-0.10
0
0.20
0
0
0
0
0
- 0.10
-0.03536
0.03536
-0.10
0
0.17071
0
0
0
- 0.03536
- 0.03536
- 0.03536
0.13536
0
- 0.10
0.03536
-0.03536
0
0
0
0
0
0
0.10
0
-0.10
0
0
0
0
0
0.03536
-0.03536
0
0
0
0.17071
0
- 0.10
- 0.03536
-0.03536
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
- 10
0
- 0.10
0
0
0
0
0
0
- 0.03536
-0.03536
0
0
0.03536
0.03536
0
0
0
-0.10
-0.03536
Y
-0.03536
0
0
0.03536
0.03536
Ans.
Ans.
0.03536
- 0.03536
0
0
- 0.03536
K = AE I
0.03536
0
0
0
0
0
-0.10
0
0.20
0
0
0
0
0
- 0.10
668
- 0.03536
0.03536
- 0.10
0
0.17071
0
0
0
- 0.03536
- 0.03536
0.03536
-0.03536
0
0
0
0.17071
0
- 0.10
-0.03536
- 0.03536
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
- 10
0
- 0.10
0
0
0
0
0
0
- 0.03536
- 0.03536
0
0
0.03536
0.03536
0
0
0
-0.10
- 0.03536
Y
-0.03536
0
0
0.03536
0.03536
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
14–13. Determine the horizontal displacement of joint 3
and the force in member 1. AE is constant.
3
1
8k
1
2
4
10 ft
6
4
3
5
2
4
3
5
10
8
5
9
10 ft
6
1
7
10 ft
Solution
0
0
Dk = D T
0
0
8
0
0
Qk = F V
0
0
0
Using the structure stiffness matrix of Prob. 14-5 and applying:
Q = KD, we have
- 0.03536
0.13536
0
- 0.10
0.03536
-0.03536
0
0
0
0
8
0.03536
0
- 0.03536
0
0
0
0
0
- 0.03536
I
Y = AE I
0
0.03536
Q7
0
Q8
0
Q9
0
Q10
0
0
0
0.10
0
- 0.10
0
0
0
0
0
0
-0.10
0
0.20
0
0
0
0
0
- 0.10
-0.03536
0.03536
-0.10
0
0.17071
0
0
0
-0.03536
- 0.03536
0.03536
- 0.03536
0
0
0
0.17071
0
- 0.10
- 0.03536
-0.03536
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
- 10
0
- 0.10
0
0
Partition matrix:
8
0.03536
0
- 0.03536
0
0
F V = AE F
0
0
0
-0.03536
0
0.03536
- 0.03536
0.13536
0
-0.10
0.03536
- 0.03536
0
0
0.10
0
- 0.10
0
0
-0.10
0
0.20
0
0
- 0.03536
0.03536
- 0.10
0
0.17071
0
8 = AE10.03536D1 - 0.03536D2 - 0.03536D5 + 0.0353D6 2
0.03536
D1
0
-0.03536
D2
0
0
D3
0
VF V + F V
0
D4
0
0
D5
0
0.17071
D6
0
0 = AE1 - 0.03536D1 + 0.13536D2 - 0.10D4 + 0.3536D5 - 0.03536D6 2
0 = AE10.10D3 - 0.10D5 2
0 = AE1 - 0.10D2 + 0.20D4 2
0 = AE1 - 0.03536D1 + 0.03536D2 - 0.10D3 + 0.17071D5 2
0 = AE10.03536D1 - 0.03536D2 + 0.17071D6 2
669
0
0
0
0
- 0.03536
-0.03536
0
0
0.03536
0.03536
0
D1
0
D2
0
D3
- 0.10
D4
-0.03536
D
Y I 5Y
D6
- 0.03536
0
0
0
0
0.03536
0
0.13536
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–13.
(Continued)
Solving the above equations yields:
D1 =
932.548
933
=
AE
AE
D2 =
160.00
386.274
80.00
386.274
-160.00
, D3 =
, D4 =
, D5 =
, D6 =
AE
AE
AE
AE
AE
Ans.
For member 1,
lx = 0.7071, ly = - 0.7071, L = 14.142 ft
AE
q1 =
[ -0.7071 0.7071 0.7071
14.142
932.548
1
160.00
- 0.7071]
D
T
AE 386.274
-160.00
Ans.
= 11.3 k 1C2
Ans.
933
AE
q1 = 11.3 k 1C2
D1 =
670
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–1. Determine the reactions at the supports. Assume ② is a
roller. EI is constant.
1
4
5
5 kN>m
6
2
1
1
3
2m
2
2
3
2m
Solution
Member 2 :
Member 1 :
1
4
2
4
1.5
2
- 1.5
1
-1.5
- 1.5
1.5
- 1.5
1.5
1.5 1
1.5
1
2
T k2 = EI D
- 1.5
-1.5 4
1.5
2
3
-5
1.5
- 1.667
1.5
1.667
1.5
F
V = EI F
Q4 - 5
- 1.5
Q5
0
Q6
0
1.5
2
1
- 1.5
0
0
1.5
1
4
0
- 1.5
1
1.5
1.5
k1 = EI D
- 1.5
1.5
3
-1.5
- 1.5
0
3
-1.5
1.5
0
0
- 1.5
- 1.5
1.5
- 1.5
3
5
1.5
2
-1.5
1
-1.5
- 1.5
1.5
- 1.5
6
1.5 4
1
3
T
-1.5 5
2
6
0
D1
0
D2
D
1
V F 3V
1.5
0
- 1.5
0
2
0
Solving,
-5
= 1.5D1 + 1.5D2 + 1.5D3
EI
D1 =
- 20
EI
- 1.667
= 1.5D1 + 2D2 + 1D3
EI
D2 =
11.67
EI
1.667
= 1.5D1 + 1D2 + 4D3
EI
5.0
D3 =
EI
Q4 - 5 = -1.5EI a
- 20.0
11.67
b - 1.5 EI a
b + 0
EI
EI
Q4 = 17.5 kN c Ans.
Q5 = 0 + 0 - 1.5EI a
Q6 = 0 + 0 + 1EI a
5.0
b = -7.50 kN = 7.50 kNT EI
5.0
b = 5.00 kN # mB
EI
Ans.
Ans.
Ans.
Q4 = 17.5 kN c
Q5 = 7.50 kNT
Q6 = 5.00 kN # mB
671
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–2.
6
Determine the reactions at the supports. EI is constant.
5
12 kN
4
6 kN>m
3
1
1
2m
Solution
Member 1:
1.5
1.5
k1 = EI D
- 1.5
1.5
1.5
2
- 1.5
1
-1.5
-1.5
1.5
-1.5
1.5
1
T
- 1.5
2
1.5
2
- 1.5
1
-1.5
-1.5
1.5
-1.5
1.5
1
T
- 1.5
2
Member 2:
1.5
1.5
k2 = EI D
- 1.5
1.5
3
2
-1
1
-2
0
F
V = EI F
Q4 - 6
- 1.5
Q5 - 12
1.5
Q6 - 6
0
1
4
1
- 1.5
0
1.5
0
1
2
0
-1.5
1.5
-1.5
- 1.5
0
1.5
-1.5
0
1.5
0
- 1.5
-1.5
3
-1.5
0
D1
D2
1.5
1.5
D
V F 3V
0
0
- 1.5
0
1.5
0
3
= 2D1 + 1D2
EI
-1
= 1D1 + 4D2 + 1D3
EI
-2
= 1D2 + 2D3
EI
Solving these equations yields
D1 =
1.75
EI
D2 =
- 0.50
EI
D3 =
- 0.75
EI
Q4 - 6.0 = -1.5EI a
1.75
-0.50
b - 1.5EI a
b + 0
EI
EI
Ans.
Q4 = 4.125 kN
672
1
2
2
2
1m
3
1m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–2
(Continued)
Q5 - 12.0 = 1.5EI a
1.75
-0.75
b + 0 - 1.5EI a
b
EI
EI
Ans.
Q5 = 15.75 kN
Q6 - 6.0 = 0 + 1.5EI a
-0.50
-0.75
b + 1.5EI a
b
EI
EI
Ans.
Q6 = 4.125 kN
Check for equilibrium:
a + ΣM2 = 0;
4.125122 + 12112 - 4.125122 - 12112 = 0
+ c ΣFy = 0;
4.125 + 15.75 + 4.125 - 12 - 12 = 0
(Check)
(Check)
Ans.
Q4 = 4.125 kN;
Q5 = 15.75 kN;
Q6 = 4.125 kN
673
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–3.
Determine the reactions at the supports. EI is constant.
4
6
5
7
2
1
M0
1
1
3
2m
2
2
3
1m
Solution
Member 2:
Member 1:
6
1.5
1.5
k1 = EI D
- 1.5
1.5
7
4
3
1.5
2
- 1.5
1
-1.5
-1.5
1.5
-1.5
1.5 6
1
7
T
- 1.5 4
2
3
2
4
0
6
-6
0
0
0
0
2
- 1.5
0
1.5
1
4
12
6
k2 = EI D
- 12
6
2
5
6
4
-6
2
-12
-6
12
-6
1
6 4
2 2
T
-6 5
4 1
Q = KD
-M
4
0
2
0
0
G 0 W = EI G 6
Q5
-6
Q6
0
Q7
0
6
6
-1.5
13.5
-12
- 1.5
-1.5
-6
-6
0
- 12
12
0
0
0
0
1.5
-1.5
0
1.5
1.5
0
D1
0
D2
1
D3
-1.5 W G D4 W
0
0
1.5
0
2
0
-M
= 4D1 + 2D2 + 6D4
EI
0 = 2D1 + 4D2 + 6D4
0 = 2D3 - 1.5D4
0 = 6D1 + 6D2 - 1.5D3 + 13.5D4
Solving the above equations yields
D1 =
- 3M
EI
Q5 = - 6EI a
D2 =
- 2.5M
EI
D3 =
2M
EI
D4 =
2.667M
EI
-3M
-2.5M
2.667M
b - 6EI a
b + 0 - 12EI a
b = M c
EI
EI
EI
Ans.
2M
2.667M
b - 1.5EI a
b = -M = M T EI
EI
Ans.
2M
2.667M
b - 1.5EI a
b = -2M = 2MA
EI
EI
Ans.
Q6 = 0 + 0 + 1.5EI a
Q7 = 0 + 0 + 1EI a
Ans.
Q5 = M c
Q6 = M T
Q7 = 2MA
674
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*15–4. Determine the reactions at the supports ①, ②, and ③.
Assume ① is pinned and ② and ③ are rollers. EI is constant.
1
2
1
1
12 ft
Solution
The FEMs are shown on the figure.
- 24
Qk = C 24 S
- 64
k1 = EI c
D1
Dk = C D2 S
D3
0.3333
0.16667
0.16667
d
0.3333
k2 = EI c
0.3333
0.16667
0.16667
d
0.3333
K = k1 + k2
0.3333
K = EI C 0.16667
0
0.16667
0.6667
0.16667
0
0.16667 S
0.3333
Q = KD
- 24
0.3333
C 24 S = EI C 0.16667
- 64
0
0.16667
0.6667
0.16667
0
D1
0.16667 S C D2 S
D3
0.3333
- 24 = EI30.3333D1 + 0.16667D2 4
24 = EI30.16667D1 + 0.6667D2 + 0.16667D3 4
- 64 = EI30.16667D2 + 0.3333D3 4
Solving:
D1 = -140>EI
D2 = 136>EI
D3 = -260>EI
q = kD
c
q1
0.3333
d = EI c
q2
0.16667
0.16667 - 140>EI
dc
d
0.3333
136>EI
q1 = 0.33331- 1402 + 0.1666711362 = - 24 k # ft
q2 = 0.166671-1402 + 0.333311362 = 22 k # ft
c
q2
0.3333
d = EI c
0.16667
q3
8k
2 k>ft
0.16667
136>EI
dc
d
0.3333 -260>EI
q2 = 0.333311362 + 0.166671-2602 = 2 k # ft
q3 = 0.1666711362 + 0.33331-2602 = - 64 k # ft
675
2
3
2
12 ft
3
8 ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*15–4
(Continued)
Member 1:
M1 = - 24 + 24 = 0
M2 = 22 - 24 = - 2 k # ft
b + ΣM1 = 0;
F′1122 - 24162 - 2 = 0
F′ = 12.1667 k
+ T ΣFy = 0;
F1 - 24 + 12.1667 = 0; F1 = 11.833 k
Member 2:
M2 = 2 + 0 = 2 k # ft
M3 = - 64 + 0 = - 64 k # ft
b + ΣM2 = 0;
64 - 2 - F ″′1122 = 0
F ″′ = 5.1667 k
+ T ΣFy = 0;
- 5.1667 + F ″ = 0
F ″ = 5.1667 k
Thus,
F1 = 11.8 k
Ans.
F2 = 12.1667 - 5.1667 = 7.00 k
Ans.
F3 = 5.1667 + 8 = 13.2 k
Ans.
Ans.
F1 = 11.8 k
F2 = 7.00 k
F3 = 13.2 k
676
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–5. Determine the moments at ② and ③. Assume ② and ③
are rollers and ① and ④ are pins. EI is constant.
5
6
1
1
10 ft
Solution
Member 1:
0.012
0.06
k1 = EI D
-0.012
0.06
0.06
0.40
- 0.06
0.20
-0.012
-0.06
0.012
-0.06
0.06
0.20
T
- 0.06
0.40
0.06
0.40
- 0.06
0.20
-0.012
-0.06
0.012
-0.06
0.06
0.20
T
- 0.06
0.40
Member 2:
0.012
0.06
k2 = EI D
-0.012
0.06
677
2
4
3
2
1
8
7
3 k>ft
2
10 ft
3
3
10 ft
4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–5
(Continued)
Member 3:
0.012
0.06
k3 = EI D
-0.012
0.06
0.06
0.40
- 0.06
0.20
- 0.012
-0.06
0.012
-0.06
0.40
0.20
0
0
K = EI H
0.06
-0.06
0
0
0.20
0.80
0.20
0
0.06
0
- 0.06
0
0
0.20
0.80
0.20
0
0.06
0
- 0.06
0
0
Dk = D T
0
0
- 25.0
0
Qk = D
T
0
25.0
0.06
0.20
T
- 0.06
0.40
0
0
0.20
0.40
0
0
0.06
-0.06
0.06
0.06
0
0
0.012
-0.012
0
0
- 0.06
0
0.06
0
- 0.012
0.024
-0.012
0
0
- 0.06
0
0.06
0
- 0.012
0.024
-0.012
0
0
-0.06
-0.06
X
0
0
-0.012
0.012
Apply Q = KD.
-25.0
0.40
0
0.20
0
0
25.0
0
H
X = EI H
Q5 - 15.0
0.06
Q6 - 30.0
- 0.06
Q7 - 30.0
0
Q8 - 15.0
0
0.20
0.80
0.20
0
0.06
0
-0.06
0
0
0.20
0.80
0.20
0
0.06
0
-0.06
0
0
0.20
0.40
0
0
0.06
- 0.06
0.06
0.06
0
0
0.012
-0.012
0
0
0
0.20
0.80
0.20
0
D1
0
0
D
0
T D 2T + D T
0.20
D3
0
0.40
D4
0
- 0.06
0
0.06
0
- 0.012
0.024
-0.012
0
Partition matrix:
-25.0
0.40
0
0.20
D
T = EI D
0
0
25.0
0
0.20
0.80
0.20
0
-25.0 = EI10.40D1 + 0.20D2 2(1)
0 = EI10.20D1 + 2.80D2 + 0.20D3 2(2)
678
0
-0.06
0
0.06
0
- 0.012
0.024
-0.012
0
D1
0
D2
- 0.06
D3
-0.06
D
X H 4X
0
0
0
0
- 0.012
0
0.012
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–5
(Continued)
0 = EI10.20D2 + 0.80D3 + 0.20D4 2(3)
25.0 = EI10.20D3 + 0.40D4 2(4)
Solving the above equations yields
D1 =
- 75.00
EI
D2 =
25.00
EI
D3 =
- 25.00
EI
D4 =
75.00
EI
For member 1,
q3
0.012
q1
0.06
D T = EI D
q6
-0.012
q2
0.06
q5 = 12.0 k,
0.06
0.40
- 0.06
0.20
q1 = 0,
M2 = q2 = -30.0 k # ft
- 0.012
- 0.06
0.012
- 0.06
0.06
0
15.0
0.20
1 -75.00
25.0
D
T
T + D
T
-0.06 EI
0
15.0
0.40
25.00
-25.0
q6 = 18.0 k
Ans.
For member 2,
q6
0.012
q2
0.06
D T = EI D
q7
-0.012
q3
0.06
q6 = 15.0 k,
0.06
0.40
- 0.06
0.20
- 0.012
- 0.06
0.012
- 0.06
0.06
0
15.0
0.20
1
25.00
25.0
T
D
T + D
T
-0.06 EI
0
15.0
0.40
-25.00
-25.0
q7 = 15.0 k,
M2 = q2 = 30.0 k # ft
Ans.
q3 = M3 = -30.0 k # ft
Ans.
Check for equilibrium:
a + ΣM1 = 0;
18.01102 - 30.0152 - 30.0 = 0
+ c ΣFy = 0;
a + ΣM2 = 0;
12.0 + 18.0 - 30.0 = 0
+ c ΣFy = 0;
15.0 + 15.0 - 30.0 = 0
1check2
1check2
30.0 + 15.01102 - 30.0152 - 30.0 = 0
1check2
1check2
679
Ans.
M2 = M3 = 30.0 k # ft
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–6. Determine the internal moment in the beam at ① and
②. EI is constant. Assume ② and ③ are rollers.
10 kN>m
3
1
2
1
1
6m
2
2
3
3m
Solution
The FEMs are shown on the figure.
Qk = c
- 7.5
d
7.5
k1 = EI c
Dk = [0]
0.6667
0.3333
0.3333
d
0.6667
k2 = EI c
0.6667
1.3333
0
0.3333
0 S
0.6667
1.3333
0.6667
0.6667
d
1.3333
K = k1 + k2
2
K = EI C 0.6667
0.3333
Q = KD
-7.5
2
C 7.5 S = EI C 0.6667
Q3
0.3333
0.6667
1.3333
0
0.3333
D1
0 S C D2 S
0
0.6667
-7.5 = EI32D1 + 0.6667D2 4
7.5 = EI30.6667D1 + 1.3333D2 4
Q3 = EI10.333321D1 2
Solving:
D1 = - 6.75>EI
D2 = 9.00>EI
Q3 = - 2.25
Thus, at node 1,
M3 = 2.25 kN # m
Ans.
q = k1D
c
q3
0.6667
d = EI c
q1
0.3333
q3 = - 2.25 kN # m
0.3333
0
dc
d
0.6667 - 6.75>EI
q1 = - 4.50 kN # m
Thus, at node 2,
M1 = 4.50 kN # m
Ans.
Ans.
M3 = 2.25 kN # m;
M1 = 4.50 kN # m
680
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–7. Determine the reactions at the supports. EI is a
constant.
6
4
3
12 kN>m
5
1
1
2
1
2
5m
2
3
2.5 m
Solution
Member 2:
Member 1:
6
0.096
0.24
k1 = EI D
- 0.096
0.24
5
0.24
0.8
- 0.24
0.4
4
- 0.096
- 0.24
0.096
- 0.24
2
0.24 6
0.4 5
T
- 0.24 4
0.8 2
0
1.6
25.0
0.8
Q3
- 0.96
F
V = EI F
Q4 - 30.0
0.96
Q5 - 25.0
0
Q6 - 30.0
0
0.8
2.4
-0.96
0.72
0.4
0.24
- 0.96
- 0.96
0.768
-0.768
0
0
4
0.768
0.96
k2 = EI D
-0.768
0.96
0.96
0.72
-0.768
0.864
- 0.24
- 0.096
0
0.4
0
- 0.24
0.8
0.24
2
0.96
1.6
- 0.96
0.8
0
D1
0.24
D2
0
0
VF V
- 0.096
0
0
0.24
0.096
0
3
-0.768
-0.96
0.768
-0.96
1
0.96 4
0.8 2
T
-0.96 3
1.6 1
0 = 1.6D1 + 0.8D2 D1 =
25.0
12.5
= 0.8D1 + 2.4D2 D2 =
EI
EI
Q5 - 25.0 = 0 + 0.4EI a
- 6.25
12.5
b - 0.96EI a
b = - 6.00 kN
Q3 = -0.96EI a
EI
EI
Ans.
= 6 kNT Q5 = 30 kN # mB
12.5
b
EI
Q6 - 30.0 = 0 + 0.24EI a
-6.25
12.5
b + 0.72a
b
Q4 - 30.0 = 0.96EI a
EI
EI
Q4 = 33 kN c Ans.
-6.25
EI
Q6 = 33.0 kN c Ans.
12.5
b
EI
Ans.
a + ΣM1 = 0; 30.0 + 33.0152 - 60.012.52 - 617.52 = 0(Check)
+ c ΣFy = 0; 33.0 + 33.0 - 60.0 - 6 = 0(Check)
Ans.
Q3 = 6 kNT
Q4 = 33 kN c
Q5 = 30 kN # mB
Q6 = 33.0 kN c
681
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*15–8. Determine the moments at the supports. Assume ② is
a roller. EI is constant.
3
25 kN>m
4
1
5
2
1
1
2
6m
Solution
Member 1:
0.16667
0.66667
- 0.16667
0.33333
- 0.05556
- 0.16667
0.05556
- 0.16667
0.16667
0.33333
T
-0.16667
0.66667
0.02344
0.09375
k2 = EI D
-0.02344
0.09375
0.09375
0.50
- 0.09375
0.25
- 0.02344
- 0.09375
0.02344
- 0.09375
0.09375
0.25
T
-0.09375
0.50
1.16667
- 0.07292
0.16667
K = EI F
0.33333
- 0.09375
0.25
-0.07292
0.07899
- 0.05556
- 0.16667
- 0.02344
0.09375
0.16667
- 0.05556
0.05556
0.16667
0
0
0.33333
-0.16667
0.16667
0.6667
0
0
-0.09375
-0.02344
0
0
0.02344
- 0.09375
0.25
0.09375
0
V
0
-0.09375
0.5
- 0.07292
0.07899
- 0.05556
- 0.16667
- 0.02344
0.09375
0.16667
- 0.05556
0.05556
0.16667
0
0
0.33333
-0.16667
0.16667
0.66667
0
0
- 0.09375
-0.02344
0
0
0.02344
-0.09375
0.05556
0.16667
k1 = EI D
- 0.05556
0.16667
Member 2:
0
0
Dk = E 0 U
0
0
Q3 = [75]
Use Eq. Q = KD.
75
1.1667
Q2 - 75.0
-0.07292
Q3 - 75.0
0.16667
F
V = EI F
Q4 - 75.0
0.33333
Q5
- 0.09375
Q6
0.25
682
0.25
D1
0.09375
0
0
0
VF V
0
0
- 0.09375
0
0.5
0
6
2
8m
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*15–8
(Continued)
Partition matrix:
75.0 = EI11.166672Dk
D1 =
64.286
EI
Q2 - 75.0 = - 0.07292EI a
64.286
b
EI
Q2 = 70.31 kN
Q3 - 75.0 = 0.16667EI a
64.286
b
EI
Q3 = 85.71 kN
Q4 - 75.0 = 0.33333EI a
64.286
b
EI
Q4 = M1 = 96.43 kN # m = 96.4 kN # m
Q3 = -0.09375EI a
Ans.
64.286
b = - 6.03 kN = 6.03 kN
EI
Q4 = M3 = 0.25EI a
64.286
b = 16.07 kN # m = 16.1 kN # m
EI
Ans.
Check for equilibrium:
a + ΣM2 = 0; 150132 + 16.07 + 96.43 - 85.71162 - 6.03182 = 0(Check)
+ c ΣFy = 0; 85.71 + 70.31 - 6.03 - 150 = 0(Check)
Ans.
Q4 = M1 = 96.43 kN # m = 96.4 kN # m
Q4 = M3 = 0.25EI a
683
64.286
b = 16.1 kN # m
EI
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–9. Determine the reactions at the supports. There is a
slider at ①. EI is constant.
1
3
w
4
1
1
2
2
L
Solution
- wL
2
2
Q2 - wL
12
D
T
Q3 - wL
2
2
Q4 + wL
12
6
L2
4
L
-6
L2
2
L
12
L3
6
L2
= EI E - 12
L3
6
L2
- 12
L3
-6
L2
12
L3
-6
L2
6
L2
D1
2
L
0
-6 U D
T
2
0
L
4
0
L
- wL
12
= 3 D1 + 0 + 0 + 0
2EI
L
D1 =
Q2 -
- wL4
24EI
wL2
6
- wL4
b
= 2 EI a
12
24EI
L
Q2 = -
Q3 -
wL2
6
Ans.
Ans.
Ans.
wL
- 12
wL4
b
=
EI a
3
2
24EI
L
Q3 = wL
Q4 +
2
4
wL
6
- wL
b
= 2 EI a
12
24EI
L
Q4 = -
wL2
3
Check for equilibrium:
a + ΣM1 = 0; wL1L2 -
wL2
wL2
L
- wL a b = 0
6
3
2
+ c ΣFy = 0; wL - wL = 0
(Check)
(Check)
Ans.
Q2 = 684
wL2
wL2
; Q3 = wL; Q4 = 6
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–10. Determine the moments at ① and ③. Assume ② is a
roller and ① and ③ are fixed. Also, here EI is constant.
4
2
3 k>ft
6
2 k>ft
5
1
1
36 ft
1
2
2
3
24 ft
Solution
Member 1:
12
1362 3
6
2
k1 = EI E 1362
- 12
1362 3
6
1362 2
- 12
1362 3
-6
1362 2
12
1362 3
-6
1362 2
6
1362 2
4
36
-6
1362 2
2
36
6
1362 2
2
36
-6 U
1362 2
4
36
Member 2:
12
1242 3
6
2
k2 = EI E 1242
- 12
1242 3
6
1242 2
6
1242 2
4
24
-6
1242 2
2
24
- 12
1242 3
-6
1242 2
12
1242 3
-6
1242 2
5
18
72.0
-6
Q2 - 36.0
1242 2
2
Q3 + 144
24
F
V = EI G 5
Q4 - 72.0
864
2
Q5 - 216
36
Q6 - 36.0
6
6
1242 2
2
24
U
-6
1242 2
4
24
-6
1242 2
12
1242 3
-6
1242 2
- 12
1242 3
2
24
-6
1242 2
4
24
6
1242 2
0
0
1362 2
0
0
5
864
- 12
1242 3
6
1242 2
35
31 104
-6
1362 2
- 12
1362 3
2
36
0
0
-6
1362 2
4
36
6
1362 2
6
1362 2
D1
0
0
0
WF V
- 12
0
1362 3
6
0
1362 2
0
12
0
1362 3
5
EI(D1)
18
259.2
D1 =
EI
2EI 259.2
a
b
Q3 + 144 =
24
EI
72.0 =
Q3 = - 122.4 k # ft = 122 k # ft
Ans.
2EI 259.2
Q5 - 216 =
a
b
36
EI
Q5 = 230.4 k # ft = 230 k # ft
Ans.
Ans.
Q3 = 122 k # ft;
Q5 = 230 k # ft
685
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–11. Determine the moments at ① and ③ if the support
② settles 0.1 ft. Assume ② is a roller and ① and ③ are fixed.
EI = 9500 k # ft2.
4
2
3 k>ft
6
2 k>ft
5
1
1
36 ft
1
2
2
3
24 ft
Solution
Member 1:
12
1362 3
6
2
k1 = EI E 1362
- 12
1362 3
6
1362 2
- 12
1362 3
-6
1362 2
12
1362 3
-6
1362 2
6
1362 2
4
36
-6
1362 2
2
36
6
1362 2
2
36
-6 U
1362 2
4
36
Member 2:
12
1242 3
6
2
k2 = EI E 1242
- 12
1242 3
6
1242 2
6
1242 2
4
24
-6
1242 2
2
24
- 12
1242 3
-6
1242 2
12
1242 3
-6
1242 2
5
18
72.0
-6
Q2 - 36.0
1242 2
2
Q + 144
24
F 3
V = EI G 5
Q4 - 72.0
864
2
Q5 - 216
36
Q6 - 36.0
6
1362 2
72.0 = 9500c
6
1242 2
2
24
U
-6
1242 2
4
24
-6
1242 2
12
1242 3
-6
1242 2
- 12
1242 3
2
24
-6
1242 2
4
24
6
1242 2
0
0
0
0
5
864
- 12
1242 3
6
1242 2
35
31 104
-6
1362 2
- 12
1362 3
2
36
0
0
-6
1362 2
4
36
6
1362 2
6
1362 2
D1
0
0
0
WF V
- 12
0
1362 3
6
0
1362 2
0
12
0
1362 3
5
5
D +
1 - 0.12 d
18 1
864
D1 = 0.029368 rad
Q3 + 144 = 9500c
2
6
10.0293682 +
1 - 0.12 d
24
1242 2
Q3 = - 130.65 k # ft = 131 k # ft
Q5 - 216 - 9500c
Ans.
2
-6
10.0293682 +
1 - 0.12 d
36
1362 2
Q5 = 235.90 k # ft = 236 k # ft
Ans.
Ans.
Q3 = 131 k # ft;
Q5 = 236 k # ft
686
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–1. Determine the structure stiffness matrix K for the
frame. Take E = 29(103) ksi, I = 650 in4, A = 20 in2 for each
member.
2
6k
4k
1
5
3
6
1
1
4
2
2
12 ft
8
3
10 ft
Solution
Member 1.
lx =
10 - 0
= 1
10
ly = 0
3
20(29)(10 )
AE
=
= 4833.33 k>in.
L
10(12)
3
12(29)(10 )(650)
12EI
=
= 130.90 k>in.
3
L
(10)3(12)3
3
6(29)(10 )(650)
6EI
=
= 7854.17 k
L2
(10)2(12)2
3
4(29)(10 )(650)
4EI
=
= 628 333.33 k # in.
L
(10)(12)
3
2(29)(10 )(650)
2EI
=
= 314 166.67 k # in.
L
(10)(12)
1
4833.33
0
0
k1 = F
-4833.33
0
0
2
0
130.90
7854.17
0
-130.90
7854.17
3
0
7854.17
628 333.33
0
-7854.17
314 166.67
4
-4833.33
0
0
4833.33
0
0
5
0
- 130.90
- 7854.17
0
130.90
-7854.17
6
1
0
7854.17 2
314 166.67 3
V
4
0
-7854.17 5
628 333.33 6
Member 2.
lx = 0
- 12 - 0
= -1
12
ly =
3
(20)(29)(10 )
AE
=
= 4027.78 k>in.
L
(12)(12)
3
6(29)(10 )(650)
6EI
=
= 5454.28 k
L2
(12)2(12)2
3
12(29)(10 )(650)
12EI
=
= 75.75 k>in.
3
L
(12)3(12)3
3
4(29)(10 )(650)
4EI
=
= 523 611.11 k # in.
L
(12)(12)
3
2(29)(10 )(650)
2EI
=
= 261 805.55 k # in.
L
(12)(12)
687
9
7
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–1.
(Continued)
4
75.75
0
5454.28
k2 = F
- 75.75
0
5454.28
5
0
4027.78
0
0
- 4027.78
0
6
5454.28
0
523 611.11
-5454.28
0
261 805.55
7
- 75.75
0
-5454.28
75.75
0
-5454.28
8
0
-4027.78
0
0
4027.78
0
9
5454.28 4
5
0
261 805.55 6
V
-5454.28 7
8
0
523 611.11 9
0
7854.17
628 333.33
0
- 7854.17
314 166.67
0
0
0
-4833.33
0
0
4909.08
0
5454.28
-75.75
0
5454.28
0
- 130.90
- 7854.17
0
4158.68
- 7854.17
0
- 4027.78
0
0
7854.17
314 166.67
5454.28
- 7854.17
1 151 944.44
-5454.28
0
261 805.55
Structure Stiffness Matrix.
4833.33
0
0
- 4833.33
K = I
0
0
0
0
0
0
130.90
7854.17
0
-130.90
7854.17
0
0
0
0
0
0
- 75.75
0
- 5454.28
75.75
0
- 5454.28
0
0
0
0
- 4027.78
0
0
4027.78
0
0
0
0
5454.28
0
Y k>in.
261 805.55
-5454.28
0
523 611.11
Ans.
Ans.
4833.33
0
0
- 4833.33
K = I
0
0
0
0
0
0
130.90
7854.17
0
-130.90
7854.17
0
0
0
0
7854.17
628 333.33
0
- 7854.17
314 166.67
0
0
0
-4833.33
0
0
4909.08
0
5454.28
-75.75
0
5454.28
0
-130.90
-7854.17
0
4158.68
-7854.17
0
-4027.78
0
688
0
7854.17
314 166.67
5454.28
-7854.17
1 151 944.44
- 5454.28
0
261 805.55
0
0
0
- 75.75
0
- 5454.28
75.75
0
- 5454.28
0
0
0
0
-4027.78
0
0
4027.78
0
0
0
0
5454.28
0
Y k>in.
261 805.55
-5454.28
0
523 611.11
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
16–2. Determine the components of displacement at ①. Take
E = 29(103) ksi, I = 650 in4, A = 20 in2 for each member.
6k
1
4k
Solution
5
3
6
1
1
4
2
2
12 ft
-4 1
-6 2
0 7
0 3
Dk = C 0 S 8 Qk = F
V
0 4
0 9
0 5
0 6
See Prob. 16–1.
8
3
9
7
10 ft
-4
4833.33
-6
0
0
0
0
-4833.33
I 0 Y = I
0
0
0
Q7
0
Q8
0
Q9
0
0
130.90
7854.17
0
- 130.90
7854.17
0
0
0
0
7854.17
628 333.33
0
-7854.17
314 166.67
0
0
0
- 4833.33
0
0
4909.08
0
5454.28
-75.75
0
5454.28
0
-130.90
- 7854.17
0
4158.68
- 7854.17
0
- 4027.78
0
0
7854.17
314 166.67
5454.28
- 7854.17
1 151 944.44
-5454.28
0
261 805.55
Partition Matrix.
-4
4833.33
-6
0
0
0
F
V + F
0
- 4833.33
0
0
0
0
0
130.90
7854.17
0
-130.90
7854.17
0
7854.17
628 333.33
0
- 7854.17
314 166.67
- 4833.33
0
0
4909.08
0
5454.28
0
- 130.90
-7854.17
0
4158.68
-7854.17
0
D1
0
D2
7854.17
0
314 166.67
D3
0
VF V + F V
D4
5454.28
0
- 7854.17
D5
0
1 151 944.44
D6
0
0
0
0
- 75.75
0
- 5454.28
75.75
0
-5454.28
0
0
0
0
-4027.78
0
0
4027.78
0
0
D1
0
D2
D3
0
5454.28
D4
Y I D5 Y
0
261 805.55 D6
0
-5454.28
0
0
523 611.11
0
- 4 = 4833.33D1 - 4833.33D4
- 6 = 130.90D2 + 7854.17D3 - 130.90D5 + 7854.17D6
0 = 7854.17D2 + 628 333.33D3 - 7854.17D5 + 314 166.67D6
0 = - 4833.33D1 + 4909.08D4 + 5454.28D6
0 = - 130.90D2 - 7854.17D3 + 4158.68D5 - 7854.17D6
0 = 7854.17D2 + 314 166.67D3 + 5454.28D4 - 7854.17D5 + 1 151 944.44D6
Solving the above equations yields
D1 = -0.608 in. = 0.608 in. d Ans.
D2 = -1.11 in. = 1.11 in. T D3 = 0.00999 rad B
Ans.
D4 = -0.607 in.
Ans.
D5 = -0.00149 in.
Ans.
D6 = 0.00770 rad
Ans.
Ans.
689
Ans.
D1 = -0.608 in. = 0.608 in. d
D2 = -1.11 in. = 1.11 in. T
D3 = 0.00999 rad B
D4 = -0.607 in.
D5 = -0.00149 in.
D6 = 0.00770 rad
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–3. Determine the structure stiffness matrix K for the
frame. Take E = 29(103) ksi, I = 450 in4, A = 8 in2 for each
member. All joints are fixed connected.
5
2
3
4k
1
2
12 ft
6
2
1
4
3
3
8
1
11
9
4
7
8 ft
Solution
Member 1:
lx = 0
12 - 0
= 1
12
ly =
3
8(29)(10 )
AE
=
= 1611.11
L
(12)(12)
3
12(29)(10 )(450)
12EI
=
= 52.44
3
L
(123)(123)
3
6(29)(10 )(450)
6EI
=
= 3776.04
L2
(122)(122)
3
4(29)(10 )(450)
4EI
=
= 362 500
L
(12)(12)
3
2(29)(10 )(450)
2EI
=
= 181 250
L
(12)(12)
52.44
0
- 3776.04
k1 = F
-52.44
0
- 3776.04
0
1611.11
0
0
- 1611.11
0
- 3776.04
0
362 500
3776.04
0
181 250
-52.44
0
3776.04
52.44
0
3776.04
ly = 0
8(29)(10 )
AE
=
= 2416.67
L
8(12)
0
- 1611.11
0
0
1611.11
0
-3776.04
0
181 250
V
3776.04
0
362 500
Member 2:
lx =
3
8 - 0
= 1
8
3
3
12(29)(10 )(450)
12EI
=
= 177.00
3
L
(83)(123)
3
4(29)(10 )(450)
4EI
=
= 543 750
L
(8)(12)
2416.67
0
0
k2 = F
-2416.67
0
0
0
177.00
8496.09
0
- 177.00
8496.09
6(29)(10 )(450)
6EI
=
= 8496.09
2
L
(82)(122)
3
2(29)(10 )(450)
2EI
=
= 271 875
L
(8)(12)
0
8496.09
543 750
0
- 8496.09
271 875
-2416.67
0
0
2416.67
0
0
0
- 177.00
-8496.09
0
177.00
-8496.09
690
0
8496.09
271 875
V
0
- 8496.09
543 750
50 k ? ft
12
10
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–3.
(Continued)
Member 3:
lx = 0
ly =
0 - 12
= -1
12
AE
= 1611.11
L
12EI
= 52.44
L3
6EI
= 3776.04
L2
4EI
= 362 500
L
2EI
= 181 250
L
52.44
0
3776.04
k3 = F
- 52.44
0
3776.04
0
1611.11
0
0
-1611.11
0
3776.04
0
362 500
-3776.04
0
181 250
- 52.44
0
-3776.04
52.44
0
-3776.04
0
-1611.11
0
0
1611.11
0
3776.04
0
181 250
V
-3776.04
0
362 500
Structure stiffness matrix:
2469.11
0
3776.04 -2416.67
0
0
-52.44
0
3776.04
0
0
0
0
1788.11 8496.09
0
-177.00 8496.09
0
- 1611.11
0
0
0
0
3776.04 8496.09 906 250
0
-8496.09 271 875 -3776.04
0
181 250
0
0
0
- 2416.67
0
0
2469.11
0
3776.04
0
0
0
-52.44
0
3776.04
0
- 177.00 - 8496.09
0
1788.11 - 8496.09
0
0
0
0
-1611.11
0
0
8496.09 271 875 3776.04 - 8496.09 906 250
0
0
0
-3776.04
0
181 250
K =
Ans.
- 52.44
0
- 3776.04
0
0
0
52.44
0
-3776.04
0
0
0
0
- 1611.11
0
0
0
0
0
1611.11
0
0
0
0
3776.04
0
181 250
0
0
0
-3776.04
0
362 500
0
0
0
0
0
0
-52.44
0
-3776.04
0
0
0
52.44
0
- 3776.04
0
0
0
0
-1611.11
0
0
0
0
0
1611.11
0
0
0
0
3776.04
0
181 250
0
0
0
-3776.04
0
362 500
Ans.
2469.11
0
3776.04 - 2416.67
0
0
- 52.44
0
3776.04
0
0
0
0
1788.11 8496.09
0
- 177.00 8496.09
0
-1611.11
0
0
0
0
3776.04 8496.09 906 250
0
- 8496.09 271 875 -3776.04
0
181 250
0
0
0
- 2416.67
0
0
2469.11
0
3776.04
0
0
0
-52.44
0
3776.04
0
- 177.00 - 8496.09
0
1788.11 -8496.09
0
0
0
0
-1611.11
0
0
8496.09 271 875 3776.04 -8496.09 906 250
0
0
0
-3776.04
0
181 250
K =
- 52.44
0
- 3776.04
0
0
0
52.44
0
-3776.04
0
0
0
0
- 1611.11
0
0
0
0
0
1611.11
0
0
0
0
3776.04
0
181 250
0
0
0
-3776.04
0
362 500
0
0
0
0
0
0
- 52.44
0
- 3776.04
0
0
0
52.44
0
-3776.04
0
0
0
0
-1611.11
0
0
0
0
0
1611.11
0
0
0
0
3776.04
0
181 250
0
0
0
-3776.04
0
362 500
691
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5
2
*16–4. Determine the horizontal displacement of
joint ②. Also compute the support reactions. Take
E = 29(103) ksi, I = 450 in4, A = 8 in2 for each member. All
joints are fixed connected.
3
4k
1
2
12 ft
6
2
1
4
3
3
8
1
50 k ? ft
11
9
4
7
12
10
8 ft
Solution
0
0
0
Dk = F V
0
0
0
4
0
0
Qk = F
V
0
0
- 600
4
2469.11
0
3776.04 - 2416.67
0
0
- 52.44
0
3776.04
0
0
0
D1
0
0
1788.11 8496.09
0
-177.00 8496.09
0
-1611.11
0
0
0
0
D2
D3
0
3776.04 8496.09 9062.50
0
-8496.09 271 875 -3776.04
0
181 250
0
0
0
0
-2416.67
0
0
2469.11
0
3776.04
0
0
0
-52.44
0
3776.04 D4
D5
0
0
- 177.00 - 8496.09
0
1788.11 -8496.09
0
0
0
0
-1611.11
0
- 600
0
8496.09 271 875 3776.04 - 8496.09 906 250
0
0
0
-3776.04
0
181 250 D6
=
Q7
-52.44
0
- 3776.04
0
0
0
52.44
0
-3776.04
0
0
0
0
Q8
0
-1611.11
0
0
0
0
0
1611.11
0
0
0
0
0
Q9
3776.04
0
181 250
0
0
0
-3776.04
0
362 500
0
0
0
0
Q10
0
0
0
-52.44
0
-3776.04
0
0
0
52.44
0
- 3776.04 0
Q11
0
0
0
0
0
-1611.11
0
0
0
0
0
1611.11
0
Q12
0
0
0
3776.04
0
181 250
0
0
0
-3776.04
0
362 500 0
Partition matrix:
4
2469.11
0
0
0
3776.04
F
V = F
0
- 2416.67
0
0
- 600
0
0
1788.11
8496.09
0
- 177.00
8496.09
3776.04
8496.09
906 250
0
- 8496.09
271 875
- 2416.67
0
0
2469.11
0
3776.04
0
- 177.00
-8496.09
0
1788.11
- 8496.09
4 = 2469.11D1 + 3776.04D3 - 2416.67D4
0 = 1788.11D2 + 8496.09D3 - 177.00D5 + 8496.09D6
0 = 3776.04D1 + 8496.09D2 + 906 250D3 - 8496.09D5 + 271 875D6
0 = - 2416.67D1 + 2469.11D4 + 3776.04D6
0 = - 177.00D2 - 8496.09D3 + 1788.11D5 - 8496.09D6
- 600 = 8496.09D2 + 271 875D3 + 3776.04D4 - 8496.09D5 + 906 250D6
692
0
D1
0
8496.09
D2
0
271 875
D
0
V F 3V + F V
3776.04
D4
0
-8496.09
D5
0
906 250
D6
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–4.
(Continued)
Solving the above equations yields
Ans.
D1 = 0.08029 in. = 0.0803 in.
D2 = 0.005056 in.
D3 = -0.0001121 rad
D4 = 0.08020 in.
D5 = -0.005056 in.
D6 = -0.001054 rad
Q7
-52.44
Q8
0
Q9
3776.04
F
V = F
0
Q10
0
Q11
Q12
0
0
-1611.11
0
0
0
0
- 3776.04
0
181 250
0
0
0
0
0
0
-52.44
0
3776.04
0
0
0
0
-1611.11
0
Q7 = -3.79 k
Ans.
Q8 = -8.15 k
Ans.
Q9 = 23.6 k # ft
Ans.
Q10 = - 0.214 k
Ans.
Q11 = 8.15 k
Ans.
Q12 = 9.27 k # ft
0
0.08029
0
0
0.005056
0
0
-0.0001121
0
V F
V + F V
- 3776.04
0.08020
0
0
-0.005056
0
181 250
-0.001054
0
Ans.
Ans.
D1 = 0.0803 in.
Q7 = - 3.79 k
Q8 = - 8.15 k
Q9 = 23.6 k # ft
Q10 = - 0.214 k
Q11 = 8.15 k
Q12 = 9.27 k # ft
693
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–5. Determine the structure stiffness matrix K for each
member of the frame. Take E = 29(103) ksi, I = 700 in4,
A = 30 in2 for each member.
8k
2
3
6 ft
1
9
5
Member 1:
lx = 0
3
30(29)(10 )
AE
=
= 6041.67
L
(12)(12)
12 - 0
= 1
12
ly =
3
12(29)(10 )(700)
12EI
=
= 81.58
3
L
(12)3(12)3
3
4(29)(10 )(700)
4EI
=
= 563 888.89
L
(12)(12)
81.58
0
- 5873.84
k1 = F
- 81.58
0
- 5873.84
0
6041.67
0
0
-6041.67
0
3
6(29)(10 )(700)
6EI
=
= 5873.84
2
L
(122)(122)
3
2(29)(10 )(700)
2EI
=
= 281 944.44
L
(12)(12)
- 5873.84
0
563 888.89
5873.84
0
281 944.44
-81.58
0
5873.84
81.58
0
5873.84
0
-6041.67
0
0
6041.67
0
- 5873.84
0
281 944.44
V
5873.84
0
563 888.89
0
- 81.58
- 5843.84
0
81.58
- 5873.74
0
5873.84
281 944.44
V
0
-5873.84
563 888.89
Member 2:
lx =
12 - 0
= 1
12
ly = 0
AE
= 6041.67
L
12EI
= 81.58
L3
6EI
= 5873.84
L2
4EI
= 563 888.89
L
2EI
= 281 944.44
L
6041.67
0
0
k2 = F
- 6041.67
0
0
0
81.58
5873.84
0
-81.58
5873.84
0
5873.84
563 888.89
0
- 5873.84
281 944.44
- 6041.67
0
0
6041.67
0
0
694
6
2
12 ft
Solution
4
1
2
1
7
6 ft
8
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–5.
(Continued)
Structure stiffness matrix:
6123.25
0
5873.84
0
K = I 5873.84
- 6041.67
0
- 81.58
0
0
6123.25
5873.84
5873.84
0
0
- 81.58
0
-6041.67
5873.84
5873.84
1 127 777.78
281 944.44
281 944.44
0
-5873.84
-5873.84
0
0
5873.84
281 944.44
563 888.89
0
0
-5873.84
0
0
5873.84
0
281 944.44
0
563 888.89
0
0
- 5873.84
0
-6041.67
0
0
0
0
6041.67
0
0
0
0
-81.58
-5873.84
-5873.84
0
0
81.58
0
0
-81.58
0
-5873.84
0
-5873.84
0
0
81.58
0
0
- 6041.67
0
0
Y
0
0
0
0
6041.67
Ans.
Ans.
6123.25
0
5873.84
0
K = I 5873.84
-6041.67
0
- 81.58
0
0
6123.25
5873.84
5873.84
0
0
- 81.58
0
- 6041.67
5873.84
5873.84
1 127 777.78
281 944.44
281 944.44
0
- 5873.84
- 5873.84
0
0
5873.84
281 944.44
563 888.89
0
0
- 5873.84
0
0
695
5873.84
0
281 944.44
0
563 888.89
0
0
-5873.84
0
-6041.67
0
0
0
0
6041.67
0
0
0
0
- 81.58
- 5873.84
- 5873.84
0
0
81.58
0
0
- 81.58
0
- 5873.84
0
- 5873.84
0
0
81.58
0
0
- 6041.67
0
0
0
Y
0
0
0
6041.67
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8k
16–6. Determine the internal loadings at the ends of each
member. Take E = 29(103) ksi, I = 700 in4, A = 30 in2 for
each member.
2
3
6 ft
7
6 ft
4
6
1
2
2
12 ft
3
1
9
5
1
8
Solution
See Prob. 16–5.
0
Dk = C 0 S
0
0
-4
- 144
Qk = F
V
144
0
0
0
6123.25
-4
0
- 144
5873.84
144
0
I 0 Y = I 5873.84
0
- 6041.67
Q7 - 4
0
- 81.58
Q8
Q9
0
0
5873.84
0
5873.84
6123.25
5873.84
5873.84
0
5873.84 1 127 777.78 281 944.44 281 944.44
5873.84
281 944.44 563 888.89
0
0
281 944.44
0
563 888.89
0
0
0
0
-81.58
-5873.84
-5873.84
0
0
- 5873.84
0
- 5873.84
- 6041.67
0
0
0
- 6041.67
0
0
0
0
6041.67
0
0
0
0
6123.25
5873.84
5873.84
0
0
- 6041.67
D1
0
0
D2
0
0
D3
0
VF V + F V
0
D4
0
D5
0
0
6041.67
D6
0
0
-81.58
-5873.84
-5873.84
0
0
81.58
0
0
-81.58
0
-5873.84
0
-5873.84
0
0
81.58
0
Partition matrix:
0
6123.25
-4
0
- 144
5873.84
V = F
F
144
0
0
5873.84
0
- 6041.67
5873.84
5873.84
1 127 777.78
281 944.44
281 944.44
0
0
5873.84
281 944.44
563 888.89
0
0
5873.84
0
281 944.44
0
563 888.89
0
0 = 6123.25D1 + 5873.84D3 + 5873.84D5 - 6041.67D6
-4 = 6123.25D2 + 5873.84D3 + 5873.84D4
- 144 = 5873.84D1 + 5873.84D2 + 1 127 777.78D3 + 281 944.44D4 + 281 944.44D5
144 = 5873.84D2 + 281 944.44D3 + 563 888.89D4
0 = 5873.84D1 + 281 944.44D3 + 563 888.89D5
0 = - 6041.67D1 + 6041.67D6
Solving,
D1 = 0.07289 in.
D2 = - 0.0006621 in.
D3 = -0.0005062 rad
D4 = 0.0005153 rad
D5 = - 0.0005062 rad
D6 = 0.07289 in.
696
0
D1
-6041.67 D2
D3
0
0
D4
0
Y I D5 Y
0
D6
0
0
0
0
6041.67
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–6.
(Continued)
Support reactions:
Q7 - 4 = 0 - 81.58( - 0.0006621) - 5873.84( - 0.0005062) - 5873.84(0.0005153) + 0 + 0
Ans.
Q7 = 4.00 k
Q8 = -81.58( - 0.07289) + 0 - 5873.84( - 0.0005062) - 5873.84( - 0.0005062)
Ans.
Q8 = 0
Q9 = 0 - 6041.67( - 0.0006621) + 0 + 0 + 0 + 0
Ans.
Q9 = 4.00 k
Check equilibrium:
+ c ΣFy = 0;
a + ΣM3 = 0;
-8 + 4 + 4 = 0
(Check)
4(12) - 8(6) = 0
(Check)
Ans.
Q7 = 4.00 k;
Q8 = 0;
Q9 = 4.00 k
697
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–7. Determine the structure stiffness matrix K for
the frame. Assume ③ is pinned and ① is fixed. Take
E = 200 MPa, I = 300(106) mm4, A = 21(103) mm2 for each
member.
5m
8
2
300 kN ? m
9
7
1
3
1
1
2
4m
3
Solution
For member 1,
lx =
5 - 0
= 1
5
ly = 0
(0.021)(200)(10 )
AE
=
= 840 000 N>m
L
5
(12)(200)(106)(300)(10-6)
12EI
=
= 5760 N>m
3
L
53
6(200)(106)(300)(10-6)
6EI
=
= 14 400 N
2
L
52
2(200)(106)(300)(10-6)
2EI
=
= 24 000 N # m
L
5
6
4(200)(106)(300)(10-6)
4EI
=
= 48 000 N # m
L
5
7
840 000
0
0
k1 = F
- 840 000
0
0
8
0
5760
14 400
0
- 5760
14 400
9
0
14 400
48 000
0
-14 400
24 000
1
- 840 000
0
0
840 000
0
0
2
0
-5760
-14 400
0
5760
- 14 400
3
7
0
14 400 8
24 000 9
V
1
0
- 14 400 2
48 000 3
For member 2,
lx = 0
ly =
0 - ( - 4)
4
= 1
(0.021)(200)(10 )
AE
=
= 1 050 000 N>m
L
4
(12)(200)(106)(300)(10-6)
12EI
=
= 11 250 N>m
3
L
43
6(200)(106)(300)(10-6)
6EI
=
= 22 500 N
2
L
42
2(200)(106)(300)(10-6)
2EI
=
= 30 000 N # m
L
4
6
4(200)(106)(300)(10-6)
4EI
=
= 60 000 N # m
L
4
5
11 250
0
- 22 500
k2 = F
- 11 250
0
- 22 500
6
0
1 050 000
0
0
- 1 050 000
0
4
- 22 500
0
60 000
22 500
0
30 000
1
-11 250
0
22 500
11 250
0
22 500
2
0
-1 050 000
0
0
1 050 000
0
698
3
-22 500 5
6
0
30 000 4
V
22 500 1
2
0
60 000 3
2
6
4
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–7.
(Continued)
Structure Stiffness Matrix.
851 250
0
22 500
22 500
K = I - 11 250
0
- 840 000
0
0
0
1 055 760
- 14 400
0
0
- 1 050 000
0
- 5760
-14 400
22 500
-14 400
108 000
30 000
- 22 500
0
0
14 400
24 000
22 500
0
30 000
60 000
- 22 500
0
0
0
0
-11 250
0
-22 500
-22 500
11 250
0
0
0
0
0
-1 050 000
0
0
0
1 050 000
0
0
0
-840 000
0
0
0
0
0
840 000
0
0
0
-5760
14 400
0
0
0
0
5760
14 400
-11 250
0
-22 500
-22 500
11 250
0
0
0
0
0
- 1 050 000
0
0
0
1 050 000
0
0
0
-840 000
0
0
0
0
0
840 000
0
0
0
-14 400
24 000
0
0
Y N>m
0
0
14 400
48 000
Ans.
Ans.
851 250
0
22 500
22 500
K = I - 11 250
0
- 840 000
0
0
0
1 055 760
- 14 400
0
0
- 1 050 000
0
- 5760
- 14 400
22 500
-14 400
108 000
30 000
-22 500
0
0
14 400
24 000
22 500
0
30 000
60 000
-22 500
0
0
0
0
699
0
-5760
14 400
0
0
0
0
5760
14 400
0
-14 400
24 000
0
0
Y N>m
0
0
14 400
48 000
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–8. Determine the support reactions at ① and ③. Take
E = 200 MPa, I = 300(106) mm4, A = 21(103) mm2 for each
member.
5m
8
2
300 kN ? m
9
7
3
1
1
2
1
4m
3
0 5
0 6
Dk = E 0 U 7
0 8
0 9
0 1
0 2
Qk = D
T
300 3
0 4
See Prob. 16–7.
0
851 250
0
0
300
22 500
0
22 500
I Q5 Y = I - 11 250
Q6
0
Q7
- 840 000
Q8
0
Q9
0
0
1 055 760
- 14 400
0
0
- 1 050 000
0
- 5760
- 14 400
22 500
- 14 400
108 000
30 000
-22 500
0
0
14 400
24 000
22 500
0
30 000
60 000
-22 500
0
0
0
0
-11 250
0
-22 500
-22 500
11 250
0
0
0
0
Partition matrix:
0
851 250
0
0
D
T = D
300
22 500
0
22 500
0
1 055 760
- 14 400
0
22 500
- 14 400
108 000
30 000
22 500
D1
0
0
D2
0
TD T + D T
D3
30 000
0
60 000
D4
0
0 = 851 250D1 = 22 500D3 + 22 500D4
0 = 1 055 760D2 - 14 400D3
300 = 22 500D1 - 14 400D2 + 108 000D3 + 30 000D4
0 = 22 500D1 + 30 000D3 + 60 000D4
Solving,
D1 = - 0.00004322 m
D2 = 0.00004417 m
D3 = 0.00323787 rad
D4 = - 0.00160273 rad
Check equilibrium:
+
S
a Fx = 0;
36.30 - 36.30 = 0 (Check)
+ c a Fy = 0;
46.37 - 46.37 = 0 (Check)
a + a M○
1 = 0;
300 + 77.07 - 36.30(4) - 46.37(5) = 0 (Check)
700
0
-1 050 000
0
0
0
1 050 000
0
0
0
-840 000
0
0
0
0
0
840 000
0
0
0
-5760
14 400
0
0
0
0
5760
14 400
0
D1
-14 400 D2
24 000
D3
0
D4
0
YI 0 Y
0
0
0
0
14 400
0
48 000
0
2
6
4
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–8.
(Continued)
Q5
- 11 250
Q6
0
E Q7 U = E -840 000
Q8
0
Q9
0
0
- 1 050 000
0
-5760
-14 400
- 22 500
0
0
14 400
24 000
-22 500
0
-0.00004322
0
0
0.00004417
0
U + E0U
UE
0.00323787
0
0
-0.00160273
0
0
Q5 = -36.3 kN = 36.3 kN d Ans.
Q6 = -46.4 kN = 46.4 kNT Ans.
Q7 = 36.3 kN S Ans.
Q8 = 46.4 kN c Ans.
Q9 = 77.1 kN m B
Ans.
Ans.
Q5 = 36.3 kN d
Q6 = 46.4 kNT
Q7 = 36.3 kN S
Q8 = 46.4 kN c
Q9 = 77.1 kN m B
701
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–9. Determine the structure stiffness matrix K for the
frame. Assume ① and ③ are pins. Take E = 29(103) ksi,
I = 600 in4, A = 10 in2 for each member.
9
2
200 lb>ft
3
5
8
1
1
1
2
12 ft
2
8 ft
7
3
Solution
Member 1:
lx =
3
12 - 0
= 1
12
(10)(29)(10 )
AE
=
= 2013.89
L
(12)(12)
ly = 0
3
3
12(29)(10 )(600)
12EI
=
= 69.93
3
L
(123)(12)
6(29)(10 )(600)
6EI
=
= 5034.72
2
L
(122)(122)
3
3
4(29)(10 )(600)
4EI
=
= 483 333,33
L
(12)(12)
2(29)(10 )(600)
2EI
=
= 241 666.67
L
(12)(12)
2013.89
0
0
k1 = F
- 2013.89
0
0
- 2013.89
0
0
2013.89
0
0
0
69.93
5034.72
0
-69.93
5034.72
0
5034.72
483 333.33
0
- 5034.72
241 666.67
0
-69.93
-5034.72
0
69.93
- 5034.72
0
5034.72
241 666.67
V
0
- 5034.72
483 333.33
Member 2:
lx = 0
3
10(29)(10 )
AE
=
= 3020.83
L
8(12)
-8 - 0
= -1
8
ly =
3
12(29)(10 )(600)
12EI
=
= 236.00
3
L
(83)(123)
3
4(29)(10 )(600)
4EI
=
= 725 000
L
8(12)
236.00
0
11 328.13
k2 = F
-236.00
0
11 328.13
0
3020.83
0
0
- 3020.83
0
3
6(29)(10 )(600)
6EI
=
= 11 328.13
2
L
(82)(12)2
3
2(29)(10 )(600)
2EI
=
= 362 500
L
8(12)
11 328.13
0
725 000
- 11 328.13
0
362 500
-236.00
0
-11 328.13
236.00
0
- 11 328.13
0
- 3020.83
0
0
3020.83
0
702
11 328.13
0
362 500
V
- 11 328.13
0
725 000
4
6
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–9.
(Continued)
Structure stiffness matrix:
2249.89
0
11 328.13
11 328.13
K = I
0
- 236.00
0
- 2013.89
0
0
3090.76
- 5034.72
0
- 5034.72
0
- 3020.83
0
- 69.93
11 328.13
- 5034.72
1 208 333.33
362 500
241 666.67
-11 328.13
0
0
5034.72
11 328.13
0
362 500
725 000
0
-11 328.13
0
0
0
0
- 5034.72
241 666.67
0
483 333.33
0
0
0
5034.72
-236.00
0
-11 328.13
-11 328.13
0
236.00
0
0
0
0
- 3020.83
0
0
0
0
3020.83
0
0
- 2013.89
0
0
0
0
0
0
2013.89
0
0
- 69.93
5034.72
0
5034.72 Y 0
0
0
69.93
Ans.
-236.00
0
-11 328.13
- 11 328.13
0
236.00
0
0
0
0
-3020.83
0
0
0
0
3020.83
0
0
- 2013.89
0
0
0
0
0
0
2013.89
0
0
- 69.93
5034.72
0
5034.72 Y
0
0
0
69.93
Ans.
2249.89
0
11 328.13
11 328.13
K = I
0
- 236.00
0
- 2013.89
0
0
3090.76
-5034.72
0
- 5034.72
0
- 3020.83
0
- 69.93
11 328.13
-5034.72
1 208 333.33
362 500
241 666.67
-11 328.13
0
0
5034.72
11 328.13
0
362 500
725 000
0
-11 328.13
0
0
0
703
0
- 5034.72
241 666.67
0
483 333.33
0
0
0
5034.72
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9
16–10. Determine the internal loadings at the ends of each
member. Assume
①
and
③
are
pins. Take
E = 29(103) ksi, I = 600 in4, A = 10 in2 for each member.
2
200 lb>ft
3
5
8
1
1
1
2
12 ft
2
8 ft
7
3
4
6
Solution
See Prob. 16-9.
0
0
Dk = D T
0
0
0
- 1.2
Qk = E 28.8 U
0
- 28.8
0
2249.89
0
11 328.13
11 328.13
0
-236.00
0
- 2013.89
0
D1
- 1.20
0
3090.76
-5034.72
0
-5034.72
0
- 3020.83
0
-69.93
D2
362 500
241 666.67 -11 328.13
0
0
5034.72 D3
28.8
11 328.13 - 5034.72 1 208 333.33
0
11 328.13
0
362 500
725 000
0
-11 328.13
0
0
0
D4
I - 28.8 Y K = I
0
- 5034.72 241 666.67
0
483 333.33
0
0
0
5034.72 Y I D5 Y
Q6
0
236.00
0
0
0
0
- 236.00
0
- 11 328.13 -11 328.13
Q7
0
-3020.83
0
0
0
0
3020.83
0
0
0
Q8
- 2013.89
0
0
0
0
0
0
2013.89
0
0
Q9 - 1.20
0
- 69.93
5034.72
0
5034.72
0
0
0
69.93
0
Partition matrix:
0
2249.89
- 1.20
0
E 28.8 U = E 11 328.13
0
11 328.13
- 28.8
0
0
3090.76
- 5034.72
0
- 5034.72
11 328.13
-5034.72
1 208 333.33
362 500
241 666.67
11 328.13
0
362 500
725 000
0
0
D1
0
-5034.72
D2
0
241 666.67 U E D3 U = E 0 U
0
D4
0
483 333.33
D5
0
0 = 2249.89D1 + 11 328.13D3 + 11 328.13D4
- 1.2 = 3090.76D2 - 5034.72D3 - 5034.72D5
28.8 = 11 328.13D1 - 5034.72D2 + 1 208 333.33D3 + 362 500D4 + 241 666.67D5
0 = 11 328.13D1 + 362 500D3 + 725 000 D4
- 28.8 = - 5034.72D2 + 241 666.67 D3 + 483 333.33D5
D1 = - 0.0001290 in.
D2 = - 0.000455 in.
D4 = - 0.0000216 rad
D5 = 0.0000879 rad
D3 = 0.0000472 rad
704
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–10.
(Continued)
For member 1,
qNx′
2013.89
qNy′
0
qNz′
0
F
V = F
qFx′
- 2013.89
qFy′
0
qFz′
0
0
69.93
5034.72
0
- 69.93
5034.72
0
5034.72
483 333.33
0
-5034.72
241 666.67
-2013.89
0
0
2013.89
0
0
0
-69.93
- 5034.72
0
69.93
- 5034.72
0
1
5034.72
0
241 666.67
0
VF
0
0
-5034.72
0
483 333.33
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
-0.0000879
VF
V
0
-0.000129
0
- 0.000455
1
0.0000472
0
1.20
28.8
V
+ F
0
1.20
- 28.8
qNx′ = 0.260 k
Ans.
qNy′ = 1.03 k
Ans.
qNz′ = 0
Ans.
qFx′ = -0.206 k
Ans.
qFy′ = 1.37 k
Ans.
qFz′ = -2.08 k # ft
Ans.
For member 2,
qNx′
3020.83
0
0
- 3020.83
0
0
0 -1 0 0 0 0
- 0.000129
qNy′
0
236.00
11 328.13
0
-236.00
11 328.13
1 0 0 0 0 0
-0.000455
q
0
11 328.13
725 000
0
- 11 328.13
326 500
0 0 1 0 0 0
0.0000472
F Nz′ V = F
VF
VF
V
qFx′
-3020.83
0
0
3020.83
0
0
0 0 0 0 -1 0
0
qFy′
0
- 236.00 -11 328.13
0
236.00
- 11 328.13
0 0 0 1 0 0
0
qFz′
0
11 328.13
342 500
0
-11 328.13
725 000
0 0 0 0 0 1
-0.0000216
0
0
0
+ F V
0
0
0
qNx′ = 1.37 k
Ans.
qNy′ = 0.260 k
Ans.
qNz′ = 2.08 k # ft
Ans.
qFx′ = -1.37 k
Ans.
qFy′ = -0.260 k
Ans.
qFz′ = 0
Ans.
705
Ans.
For member 1:
qNx′ = 0.260 k;
qNz′ = 0;
qFy′ = 1.37 k;
qNy′ = 1.03 k;
qFx′ = - 0.206 k;
qFz′ = - 2.08 k # ft
For member 2:
qNx′ = 1.37 k;
qNz′ = 2.08 k # ft;
qFy′ = -0.260 k;
qNy′ = 0.260 k;
qFx′ = - 1.37 k;
qFz′ = 0
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–11. Determine the structure stiffness matrix K
for each member of the frame. Take E = 29(103) ksi,
I = 700 in4, A = 30 in2 for each member. Joint ① is pin
connected.
2 k>ft
2
7
4
5k
6
1
2
12 ft
2
3
3
12 ft
1
9
5
1
Solution
Member 1: l x =
0 - 0
= 0;
12
ly =
81.581
0
- 5873.843
k1 = F
- 81.581
0
- 5873.843
0
6041.667
0
0
- 6041.667
0
- 5873.843
0
563 888.88
5873.843
0
281 944.44
Member 2: l x =
12 - 0
= 1;
12
6041.667
0
0
k2 = F
-6041.667
0
0
0
81.581
5873.843
0
- 81.581
5873.843
12 - 0
= 1
12
-81.581
0
5873.843
81.581
0
5873.843
ly =
0
5873.843
563 888.88
0
- 5873.843
281 944.44
8
0
- 6041.667
0
0
6041.667
0
- 5873.843
0
281 944.44
V
5873.843
0
563 888.88
Ans.
0
5873.843
281 944.44
V
0
-5873.843
563 888.88
Ans.
12 - 12
= 0
12
- 6041.667
0
0
6041.667
0
0
0
-81.581
- 5873.843
0
81.581
-5873.843
Ans.
81.581
0
-5873.843
k1 = F
- 81.581
0
-5873.843
0
6041.667
0
0
-6041.667
0
6041.667
0
0
k2 = F
- 6041.667
0
0
0
81.581
5873.843
0
-81.581
5873.843
706
- 5873.843
0
563 888.88
5873.843
0
281 944.44
0
5873.843
563 888.88
0
- 5873.843
281 944.44
- 81.581
0
5873.843
81.581
0
5873.843
0
- 6041.667
0
0
6041.667
0
- 5873.843
0
281 944.44
V
5873.843
0
563 888.88
-6041.667
0
0
6041.667
0
0
0
-81.581
-5873.843
0
81.581
- 5873.843
0
5873.843
281 944.44
V
0
-5873.843
563 888.88
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2 k>ft
2
*16–12. Determine the support reactions at ① and ③. Take
E = 29(103) ksi, I = 700 in4, A = 30 in2 for each member.
Joint ① is pin connected.
7
4
5k
6
1
2
12 ft
2
3
3
12 ft
1
9
5
1
Solution
8
K = k1 + k2
6123.248
0
5873.842
0
K = I 5873.842
- 6041.667
0
-81.581
0
0
Dk = C 0 S
0
0
6123.248
5873.842
5873.842
0
0
- 81.581
0
- 6041.667
5873.842
5873.842
112 777.78
281 944.44
281 944.44
0
- 5873.842
- 5873.842
0
0
5873.842
281 944.44
563 888.88
0
0
-5873.842
0
0
5873.842
0
281 944.44
0
503 888.88
0
0
-5873.842
0
-6041.667
0
0
0
0
6041.667
0
0
0
0
-81.581
-5873.842
-5873.842
0
0
81.581
0
0
-81.581
0
-5873.842
0
-5873.842
0
0
81.581
0
0
-6041.667
0
0
Y
0
0
0
0
6041.667
5
- 12
- 288
Qk = F
V
288
0
0
Substituting into Q = KD, partition, and solving for the displacements and loads yields
D1 = 0.63626 in.
D2 = -1.159(10-3) in.
D3 = -2.716(10-3) rad
D4 = 1.8809(10-3) rad
D5 = -5.2697(10 - 3) rad
D6 = 0.6363 in.
Q7 = 17.0 k
Ans.
Q8 = -5.00 k
Ans.
Q9 = 7.00 k
Ans.
Ans.
Q7 = 17.0 k
Q8 = - 5.00 k
Q9 = 7.00 k
707
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–13. Determine the structure stiffness matrix K for the
frame. Take E = 29(103) ksi, I = 600 in4, A = 10 in2 for each
member. Assume joints ① and ③ are pinned; joint ② is fixed
connected.
5k
9
5
6 ft
2
6 ft
3
8
1
1
2
1
2
8 ft
7
3
4
Solution
Member 1: l x =
12 - 0
= 1;
12
ly =
2013.89
0
0
k1 = F
-2013.89
0
0
0
69.927
5034.722
0
-69.927
5034.722
0
5034.722
483 330
0
- 5034.722
241 667
Member 2: l x =
12 - 12
= 0;
12
236.003
0
11 328.125
k2 = F
-236.003
0
11 328.125
ly =
6
0 - 0
= 0
8
- 2013.89
0
0
2013.89
0
0
0
- 69.927
- 5034.722
0
69.927
-5034.722
0
5034.722
241 667
V
0
- 5034.722
483 330
-8 - 0
= -1
8
0
3020.833
0
0
- 3020.833
0
11 328.125
0
725 000
-11 328.125
0
362 500
- 236.003
0
- 11 328.125
236.003
0
-11 328.125
0
- 3020.833
0
0
3020.833
0
11 328.125
0
362 500
V
-11 328.125
0
725 000
0
3090.76
- 5034.722
0
- 5034.722
0
- 3020.833
0
- 69.927
11 328.125
- 5034.722
1208.33(103)
362 500
241 666.67
- 11 328.125
0
0
5034.722
11 328.125
0
362 500
725 000
0
-11 328.125
0
0
0
0
-5034.722
241 666.67
0
483 333.33
0
0
0
5034.722
-236
0
- 11 328.125
-11 328.125
0
236
0
0
0
K = k1 + k2
2249.892
0
11 328.125
11 328.125
0
K = I
- 236
0
-2013.89
0
0
-3020.833
0
0
0
0
3020.833
0
0
-2013.89
0
0
0
0
0
0
2013.89
0
0
- 69.927
5034.722
0
5034.722 Y Ans.
0
0
0
69.927
Ans.
2249.892
0
11 328.125
11 328.125
K = I
0
- 236
0
-2013.89
0
0
3090.76
- 5034.722
0
- 5034.722
0
- 3020.833
0
- 69.927
11 328.125
- 5034.722
1208.33(103)
362 500
241 666.67
- 11 328.125
0
0
5034.722
11 328.125
0
362 500
725 000
0
-11 328.125
0
0
0
708
0
-5034.722
241 666.67
0
483 333.33
0
0
0
5034.722
-236
0
- 11 328.125
-11 328.125
0
236
0
0
0
0
-3020.833
0
0
0
0
3020.833
0
0
-2013.89
0
0
0
0
0
0
2013.89
0
0
- 69.927
5034.722
0
5034.722 Y
0
0
0
69.927
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–14. Determine the structure stiffness matrix K for the
frame. Take E = 29(103) ksi, I = 300 in4, A = 10 in2 for each
member.
2 k>ft
2
4
2
1
2
3
9
5
1
Solution
Member Stiffness Matrices. The origin of the global coordinate system
○
will be set at joint 1 . For member 1 , L = 10 ft, l x =
ly =
10 - 0
= 1
10
3
3
0 - 0
= 0 and
10
3
10[29(10 )]
AE
=
= 2416.67 k>in.
L
10(12)
12[29(10 )](300)
12EI
=
= 60.4167 k>in.
3
L
[10(12)]3
3
6[29(10 )](300)
6EI
=
= 3625 k
2
L
[10(12)]2
4[29(10 )](300)
4EI
=
= 290 000 k # in.
L
10(12)
3
2[29(10 )](300)
2EI
=
= 145 000 k # in.
L
10(12)
8
60.4167
0
-3625
k1 = F
-60.4167
0
- 3625
9
0
2416.67
0
0
- 2416.67
0
5
-3625
0
290 000
3625
0
145 000
For member 2 , L = 20 ft, l x =
1
- 60.4167
0
3625
60.4167
0
3625
2
0
-2416.67
0
0
2416.67
0
3
- 3625 8
9
0
145 000 5
V
3625 1
2
0
290 000 3
20 - 0
10 - 10
= 1 and l y =
= 0.
20
20
3
12[29(10 )](300)
12EI
=
= 7.5521 k>in.
L3
[20(12)]3
6[29(10 )](300)
6EI
=
= 906.25 k
2
L
[20(12)]2
4[29(10 )](300)
4EI
=
= 145 000 k # in.
L
20(12)
10[29(10 )]
AE
=
= 1208.33 k>in.
L
20(12)
3
3
3
3
2[29(10 )](300)
2EI
=
= 72 500 k # in.
L
20(12)
1
1208.33
0
0
k2 = F
- 1208.33
0
0
2
0
7.5521
906.25
0
- 7.5521
906.25
3
0
906.25
145 000
0
-906.25
72 500
6
-1208.33
0
0
1208.33
0
0
7
0
- 7.5521
- 906.25
0
7.5521
- 906.25
4
0
1
906.25 2
72 500 3
V
0
6
- 906.25 7
145 000 4
709
20 ft
1
10 ft
7
8
6
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–14.
(Continued)
Structure Stiffness Matrix. It is a 9 * 9 matrix since the highest
code number is 9. Thus,
1268.75
0
3625
0
K = I 3625
- 1208.33
0
-60.4167
0
0
2424.22
906.25
906.25
0
0
-7.5521
0
- 2416.67
3625
906.25
435 000
72 500
145 000
0
- 906.25
-3625
0
0
906.25
72 500
145 000
0
0
-906.25
0
0
3625
0
145 000
0
290 000
0
0
-3625
0
- 1208.33
0
0
0
0
1208.33
0
0
0
0
- 7.5521
- 906.25
- 906.25
0
0
7.5521
0
0
- 60.4167
0
-3625
0
-3625
0
0
60.4167
0
0
-2416.67
0
0
Y k>in.
0
0
0
0
2416.67
Ans.
Ans.
1268.75
0
3625
0
K = I 3625
- 1208.33
0
-60.4167
0
0
2424.22
906.25
906.25
0
0
- 7.5521
0
- 2416.67
3625
906.25
435 000
72 500
145 000
0
-906.25
- 3625
0
0
906.25
72 500
145 000
0
0
-906.25
0
0
710
3625
0
145 000
0
290 000
0
0
- 3625
0
-1208.33
0
0
0
0
1208.33
0
0
0
0
- 7.5521
- 906.25
- 906.25
0
0
7.5521
0
0
-60.4167
0
-3625
0
-3625
0
0
60.4167
0
0
-2416.67
0
0
Y k>in.
0
0
0
0
2416.67
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–15. Determine the support reactions at ① and ③. Take
E = 29(103) ksi, I = 300 in4, A = 10 in2 for each member.
2 k>ft
2
4
2
1
2
3
6
3
20 ft
1
10 ft
7
9
5
1
8
-1208.33
0
0
0
0
1208.33
0
0
0
0
-7.5521
- 906.25
-906.25
0
0
7.5521
0
0
Solution
Known Nodal Loads and Deflections. The nodal loads acting
on the unconstrained degree of freedom (code number 1, 2, 3, 4,
5, and 6) are shown in Figs. a and b.
0
1
- 25 2
0 7
- 1200 3
Qk = F
and
Dk = C 0 S 8
V
0
4
0 9
0
5
0
6
Load–Displacement Relation. Applying Q = KD,
0
1268.75
- 25
0
-1200
3628
0
0
I 0 Y = I 3625
0
- 1208.33
Q7
0
Q8
- 60.4167
Q9
0
0
2424.22
906.25
906.25
0
0
-7.5521
0
- 2416.67
3625
906.25
435 000
72 500
145 000
0
- 906.25
-3625
0
0
906.25
72 500
145 000
0
0
-906.25
0
0
3625
0
145 000
0
290 000
0
0
- 3625
0
-60.4167
0
- 3625
0
- 3625
0
0
60.4167
0
0
D1
- 2416.67 D2
0
D3
D4
0
Y I D5 Y
0
D6
0
0
0
0
0
2416.67
0
From the matrix partition, Qk = K11Du + K12Dk,
0 = 1268.75D1 + 3625D3 + 3625D5 - 1208.33D6(1)
- 25 = 2424.22D2 + 906.25D3 + 906.25D4(2)
- 1200 = 3625D1 + 906.25D2 + 435 000D3 + 725 00D4 + 145 000D5(3)
0 = 906.25D2 + 72 500D3 + 145 000D4(4)
0 = 3625D1 + 145 000D3 + 290 000D5(5)
0 = - 1208.33D1 + 1208.33D6(6)
Solving Eqs. (1) to (6),
D1 = 1.32
D2 = - 0.008276
D4 = 0.005552
D5 = - 0.011
D3 = - 0.011
D6 = 1.32
Using these results and applying Qu = K21Du + K22Dk,
Q7 = - 7.5521( - 0.008276) - 906.25( -0.011) - 906.25(0.005552) = 5
Q8 = - 60.4167(1.32) - 3625( -0.011) - 3625( -0.011) = 0
Ans.
Q9 = - 2416.67( - 0.008276) = 20 k c Ans.
Superposition of the above Q7 to (Q7)0 shown in Fig. a:
Q7 = 5 + 15 = 20 k c Ans.
711
Ans.
Q8 = 0
Q9 = 20 k c
Q7 = 20 k c
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5
2
*16–16. Determine the reactions at the supports ① and
④. Joints ① and ④ are pin connected and ② and ③ are fixed
connected. Take E = 29(103) ksi, I = 700 in4, A = 15 in2 for
each member.
6
3
2
1
2
3
10 ft
400 lb>ft
1
3
12
10
8
1
7
11
2.0
4672.22
0
0
40
8458.33
0
-4531.25
0
0
0
0
=
0
0
-40
8458.33
Q9
0
Q10
0
Q11 + 2
-140.97
Q12
0
4
9
8 ft
Solution
0
0
Dk = D T
0
0
4
2
0
40
0
X
Dk = H
0
0
0
- 40
0
3900.34
13 216.15
0
-275.34
13 216.15
0
0
0
0
0
-3625
2
4672.22
0
0
40
8458.33
0
- 4531.25
H
X = H
0
0
0
0
0
0
- 40
8458.33
8458.33
13 216.15
1 522 500
0
-13 216.15
422 916.67
0
338 333.33
0
0
-8458.33
0
0
3900.34
13 216.15
0
- 275.34
13 216.15
0
0
-4531.25
0
0
4672.22
0
8458.33
8458.33
0
-140.97
0
0
0
0
-275.34
-13 216.15
0
3900.34
-13 216.15
0
0
0
-3625
0
0
8458.33
13 216.15
1 522 500
0
-13 216.15
422 916.67
0
338 333.33
0
13 216.15
422 916.67
8458.33
-13 216.15
1 522 500
338 333.33
0
-8458.33
0
0
0
- 4531.25
0
0
4672.22
0
8458.33
8458.33
0
0
0
0
8458.33
0
338 333.33
676 666.67
0
-8458.33
0
0
0
0
- 275.34
-13 216.15
0
3900.34
-13 216.15
0
0
8458.33
0
338 333.33
0
0
0
0
676 666.67
0
0
-8458.33
0
0
13 216.15
422 916.67
8458.33
- 13 216.15
1 522 500
338 333.33
0
0
0
0
-140.97
0
-8458.33
-8458.33
0
140.97
0
0
0
0
0
0
8458.33
0
338 333.33
676 666.67
0
2.0 = 4672.22D1 + 8458.33D3 - 4531.25D4 + 8458.33D8
0 = 3900.34D2 + 13 216.15D3 - 275.34D5 + 13 216.15D6
40 = 8458.33D1 + 13 216.15D2 + 1 522 500D3 - 13 216.15D5 + 422 916.67D6 + 338 333.33D8
0 = - 4531.25D1 + 4672.22D4 + 8458.33D6 + 8458.33D7
0 = - 275.34D2 - 13 216.15D3 + 3900.34D5 - 13 216.15D6
0 = 13 216.15D2 + 422 916.67D3 + 8458.33D4 - 13 216.15D5 + 1 522 500D6 + 338 333.33D7
0 = 8458.33D4 + 338 333.33D6 + 676 666.67D7
- 40 = 8458.33D1 + 338 333.33D3 + 676 666.67D8
712
0
0
0
0
-3625
0
0
0
0
3625
0
0
-140.97
0
-8458.33
0
0
0
0
-8458.33
0
0
140.97
0
0
-3625
0
0
0
0
0
0
0
0
0
3625
D1
D2
D3
D4
D5
D6
D7
D8
0
0
0
0
8458.33
D1
0
0
D2
0
338 333.33
D3
0
0
D4
0
XH X = H X
D5
0
0
D6
0
0
D7
0
0
676 666.67
D8
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–16.
(Continued)
Solving,
D1 = 0.04867 in.
D2 = 0.0006897 in.
D3 = -0.00007727 rad
D4 = 0.04842 in.
D5 = -0.0006897 in.
D6 = - 0.0001406 rad
D7 = -0.0005350 rad
D8 = -0.0006288 rad
Support reactions:
Q9
0
Q10
0
D
T = D
- 140.97
Q11 + 2
Q12
0
0
0
0
-3625
0
0
- 8458.33
0
-140.97
0
0
0
0
-3625
0
0
-8458.33
0
0
0
- 8458.33
0
0
0
0
0
T
- 8458.33
0
0.04867
0
0.0006897
0
- 0.00007727
0
0.04842
0
H
X + H X
- 0.0006897
0
-0.0001406
0
- 0.0005350
0
- 0.0006288
0
Q9 = -1.11 k
Ans.
Q10 = 2.50 k
Ans.
Q11 = - 2.89 k
Ans.
Q12 = - 2.50 k
Ans.
Check for equilibrium:
+
S
ΣFx = 0;
0.4(10) - 2.89 - 1.11 = 0
+ c ΣFy = 0;
2.50 - 2.50 = 0
+ ΣM1 = 0;
2.50(8) - (0.4)(10)(5) = 0
(Check)
(Check)
(Check)
Ans.
Q9 = -1.11 k;
Q11 = -2.89 k;
713
Q10 = 2.50 k;
Q12 = - 2.50 k
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–17. Determine the structure stiffness matrix k
for each member of the two-member frame. Take
E = 200 GPa, I = 350(106) mm4, A = 20(103) mm2 for each
member. Joints ① and ③ are pinned and joint ② is fixed
connected.
20 kN
2
9
3
4
1
2
12 kN>m
8
2
3
4m
1
7
1
5 6
3m
Solution
Member 1:
lx =
3 - 0
= 0.6
5
ly =
4 - 0
= 0.8
5
20(10-3)(200)(109)
AE
=
= 800(106)
L
5
12(200)(109)(350)(10-6)
12EI
=
= 6.720(106)
L3
(5)3
6(200)(103)(350)(10-6)
6EI
=
= 16.800(106)
2
L
(5)2
4(200)(109)(350)(10-6)
4EI
=
= 56.00(106)
L
5
2(200)(109)(350)(10-6)
2EI
=
= 28.00(106)
L
5
292.30
380.77
- 13.44
k1 = (106) F
- 292.30
-380.77
- 13.44
380.77
514.42
10.08
- 380.77
-514.42
10.08
-13.44
10.08
56.00
13.44
- 10.08
28.00
-292.30
- 380.77
13.44
292.30
380.77
13.44
- 380.77
-514.42
-10.08
380.77
514.42
-10.08
- 13.44
10.08
28
V
13.44
-10.08
56.00
Member 2:
lx =
7 - 3
= 1
4
ly =
4 - 4
= 0
4
20(10-3)(200)(109)
AE
=
= 1000(106)
L
4
12(200)(109)(350)(10-6)
12EI
=
= 13.125(106)
L3
(4)3
6(200)(109)(350)(10-6)
6EI
=
= 26.25(106)
2
L
(4)2
4(200)(109)(350)(10-6)
4EI
=
= 70.00(106)
L
4
2(200)(109)(350)(10-6)
2EI
=
= 35.00(106)
L
4
714
2m
2m
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–17.
(Continued)
1000
0
0
k2 = (106) F
-1000
0
0
0
13.125
26.25
0
- 13.125
26.25
0
26.25
70.00
0
- 26.25
35.00
-1000
0
0
1000
0
0
0
- 13.125
-26.25
0
13.125
- 26.25
0
26.25
35.00
V
0
-26.25
70.00
13.44
16.17
126
35
28
- 13.44
10.08
0
- 26.25
0
26.25
35
70
0
0
0
0
-26.25
13.44
-10.08
28
0
56
-13.44
10.08
0
0
- 292.3
- 380.77
- 13.44
0
- 13.44
292.3
380.77
0
0
Structure stiffness matrix:
1292.3
380.77
13.44
0
6
K = (10 ) I 13.44
-292.3
-380.77
- 1000
0
380.77
527.544
16.17
26.25
-10.08
- 380.77
- 514.42
0
-13.125
- 380.77
-514.42
10.08
0
10.08
380.77
514.42
0
0
-1000
0
0
0
0
0
0
1000
0
0
-13.125
-26.25
-26.25
0
Y
0
0
0
13.125
Ans.
Ans.
1292.3
380.77
13.44
0
6
K = (10 ) I 13.44
- 292.3
- 380.77
- 1000
0
380.77
527.544
16.17
26.25
-10.08
- 380.77
- 514.42
0
-13.125
13.44
16.17
126
35
28
-13.44
10.08
0
-26.25
715
0
26.25
35
70
0
0
0
0
-26.25
13.44
-10.08
28
0
56
-13.44
10.08
0
0
- 292.3
- 380.77
- 13.44
0
- 13.44
292.3
380.77
0
0
- 380.77
-514.42
10.08
0
10.08
380.77
514.42
0
0
-1000
0
0
0
0
0
0
1000
0
0
-13.125
- 26.25
- 26.25
0
Y
0
0
0
13.125
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–18. Determine the support reactions at ① and ③. Take
E = 200 GPa, I = 350(106) mm4, A = 20(103) mm2 for each
member. Joints ① and ③ are pinned and joint ② is fixed.
20 kN
2
9
3
4
1
2
12 kN>m
8
2
3
4m
1
7
1
5 6
3m
2m
2m
Solution
24 000
1292.3
- 28 000
380.77
15 000
13.44
10 000
0
I -25 000 Y = (106) I 13.44
Q6 - 24 000
-292.3
-380.77
Q7 - 18 000
- 1000
Q8
0
Q9 - 10 000
380.77
527.54
16.17
26.25
- 10.08
- 380.77
- 514.42
0
-13.125
13.44
16.17
126
35
28
- 13.44
10.08
0
- 26.25
0
26.25
35
70
0
0
0
0
-26.25
13.44
-10.08
28
0
56
- 13.44
10.08
0
0
- 292.3
- 380.77
- 13.44
0
- 13.44
292.3
380.77
0
0
- 380.77
-514.42
10.08
0
10.08
380.77
514.42
0
0
-1000
0
0
0
0
0
0
1000
0
0
D1
-13.125 D2
-26.25
D3
-26.25
D4
0
Y I D5 Y
0
0
0
0
0
0
13.125
0
24(10-3) = 1292.3D1 + 380.77D2 + 13.44D3 + 13.44D5
- 28(10-3) = 380.77D1 + 527.54D2 + 16.17D3 + 26.25D4 -10.08D5
15(10-3) = 13.44D1 + 16.17D2 + 126D3 + 35D4 + 28D5
10(10-3) = 26.25D2 + 35D3 + 70D4
- 25(10-3) = 13.44D1 - 10.08D2 + 28D3 + 56D5
D1 = 56.50(10-6) m
D2 = - 116.06(10-6) m
D3 = 244.01(10-6) rad
D4 = 64.38(10-6) rad
D5 = - 602.9(10-6) rad
Q6 + 24 000 = [ - 292.3(56.50) -380.77( -116.06) - 13.44(244.01)
+ 0- 13.44( - 602.9)]
Ans.
Q6 = 8.50 kN
Q7 - 18 000 = [ - 380.77(56.50) - 514.42( - 116.06) + 10.08(244.01)
+ 0 + 10.08( - 602.9)]
Q7 = 52.6 kN
Ans.
Q8 = [ - 1000(56.50) + 0 + 0 + 0 + 0]
Q8 = 56.5 kN
Ans.
Q9 - 10 000 = [0- 13.125( -116.06) -26.25(244.01) -26.25(64.38) + 0]
Q9 = 3.43 kN
Ans.
Q6 = 8.50 kN; Q7 = 52.6 kN;
Q8 = 56.5 kN; Q9 = 3.43 kN
Ans.
716