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Chapter 1
1.1
The axial forces in the segments are
Pbr = 2P = 2(10) = 20 kN (C)
Pal = 2P = 2(10) = 20 kN (T)
π D2 − 0.082
P = σw A
330 × 10 = 110 × 10
4
D = 0.1011 m = 101.1 mm ◭
3
6
Pst = 3P = 3(10) = 30 kN (T)
Pbr
20 × 103
=
= 50 × 106 Pa (C) ◭
Abr
400 × 10−6
20 × 103
Pal
=
= 33.3 × 106 Pa (T) ◭
σ al =
Aal
600 × 10−6
Pst
30 × 103
σ st =
=
= 100 × 106 Pa (T) ◭
Ast
300 × 10−6
σ br =
1.2
1.5
Pmax = PAB = 35 kN
Pmax
35 × 103
σ max =
= 58.3 × 106 Pa
=
A
600 × 10−6
= 58.3 MPa ◭
Axial forces in the segments are (see solution of Prob.1.4)
Pbr = 2P (C)
Pal = 2P (T)
Pst = 3P (T)
Assuming that stress in bronze governs:
2P
Pbr
110 × 106 =
Abr
400 × 10−6
3
P = 22.0 × 10 N = 22.0 kN
1.3
(σ br )w =
Assuming that stress in aluminum governs:
ΣFx = 0
ΣFy = 0
Pal
2P
68 × 106 =
Aal
600 × 10−6
P = 20.4 × 103 N = 20.4 kN
(σ al )w =
PAC cos 45◦ − PAB cos 30◦ = 0
PAC sin 45◦ + PAB sin 30◦ − W = 0
Assuming that stress in steel governs:
The solution is
PAC = 0.8966W
3P
Pst
120 × 106 =
Ast
300 × 10−6
3
P = 12.0 × 10 N = 12.0 kN ◭
PAB = 0.7321W
(σ st )w =
Assuming that stress in AC governs:
PAC
0.8966W
150 × 106 =
AAC
200 × 10−6
W = 33 500 N = 33.5 kN ◭
(σ AC )w =
1.6
Assuming that stress in AB governs:
0.7321W
PAB
100 × 106 =
AAB
400 × 10−6
W = 54 600 N = 54.6 kN
(σ AB )w =
ΣFy = 0
Pwd − 2Pst sin 60◦ = 0
Pwd = 2Pst sin 60◦ = 2σ st Ast sin 60◦
π(10)2
= 2(600)
sin 60◦ = 81621 N
4
1.4
Awd =
Pwd
σ wd
πd2
81621
=
4
2
d = 228 mm ◭
1
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1.7
Pz = √
1.5
P = 0.4286P
12 + 32 + 1.52
MAB = 0
2Pz (3) − W (1.5) = 0
2(0.4286P )(3) − 600 × 9.81(1.5) = 0
P = 3433.3 N
s
r
P
3433.3
πd2
=2
=P
d=2
= 4.9 mm ◭
σw
4
πσ w
π(180)
For concrete:
P = σ co Aco = 4.5(300 × 300) = 405 000 N
1.10
For wood:
2
P = σ wd Awd = 10
π(200)
= 314 159 N ◭
4
From symmetry, the axial force in each cable is
P = 9000 N.
1.8
A=
P
4500
=
= 25 mm2 ◭
σw
180
for both cables.
1.11
For steel:
1.2
√
T (4) + T (2) − 6000(2) = 0
5.44
T = 2957 N
2957
T
2
=
= 11.6 N/mm ◭
σ =
A
π(18)2 /4
Pst
2P
= σ st
= 120 × 106
Ast
500 × 10−6
P = 30.0 × 103 N
ΣMC = 0
For timber:
Pwd
P
= σ wd
= 12 × 106
Awd
2000 × 10−6
P = 24.0 × 103 N ◭
1.12
For concrete:
5P
Pco
= σ co
= 16 × 106
Aco
8000 × 10−6
P = 25.6 × 103 N
ΣFy = 0
1.9
z
A
Pwd =
5
W
4
3
Pwd − Pst = 0
5
3
3 5
3
Pst = Pwd =
W = W
5
5 4
4
E
1.5 m
4
Pwd − W = 0
5
ΣFx = 0
B
1m 1
.5
m
R
W
3m
P
P
C
x
For steel cable:
y
2m
σ st Ast = Pst
D
160(450) =
3
W
4
W = 96 000 N
2
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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For wood strut:
σ wd Awd = Pwd
5
7.2(14400) = W
4
ΣMC = 0
+
PAB = 66.67 kN
W = 82944 N ◭
600PAB − 40(1000) = 0
66.67
PAB
= 94.3 × 103 kPa
= π
AAB
2
2
(0.05 − 0.04 )
4
= 94.3 MPa ◭
σ AB =
1.13
1.16
When P is maximized, both cables are stressed to the
limiting values. The corresponding forces in the cables are
PAC = (σ AC )w A = 100 × 106 (400 × 10−6 )
From FBD of cylinder:
3
N2 − M g = 0
5
ΣFy = 0
N2 =
From FBD of bar AB:
5
ΣMA = 0
4
M g − 8PBC = 0
3
σw =
PBC
ABC
5
Mg
3
PBC =
= 40 × 103 N = 40 kN
PBD = (σ BD )w A = (50 × 106 )(400 × 10−6 )
= 20 × 103 N = 20 kN
5
Mg
6
(5/6)M g
100 × 10−6
6000
M=
= 612 kg ◭
9.81
50 × 106 =
M g = 6000 N
ΣF = 0
40 + 20 − 9.81 − P = 0
ΣMA = 0
P = 50.19 kN ◭
2(20) − 1.0(9.81) − x (50.19) = 0
x = 0.602 m = 602 mm ◭
1.14
1.17
ΣMB = 0
1.5
PCD = 1625 N
σ CD =
4
PCD
5
− 1.5 (300) − 3(500) = 0
From FBD of truss:
PCD
1625
=
= 9.2 N/mm2 ◭
ACD
π(15)2 /4
1.15
ΣMA = 0
12RF − 8(30) = 0
RF = 20 kN
C
From FBD of joint F :
PAB
RC
400
3
PDF = 33.33 kN
− PDF + 20 = 0
5
33.33
PDF
=
= 18.52 N/mm2 (C) ◭
=
A
1800
ΣFy = 0
A
σ DF
600
40 kN
3
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From FBD of joint D:
From FBD of portion ABC:
PBD = PDF = 33.33 kN (due to symmetry)
3
ΣFy = 0
2 (33.33) − PDE = 0
5
PDE = 40 kN
PDE
40
2
σ DE =
=
= 22.2 N/mm (T) ◭
A
1800
From FBD of joint E:
3
ΣFy = 0
40 − 30 − PCE = 0
5
PCE = 16.667 kN
16.667
PCE
=
= 9.26 N/mm2 (T) ◭
σ CE =
A
1800
ΣMD = 0
7(40) + 4(70) − 4
1
√ PCF
2
=0
PCF = 198.0 kN
PCF
198.0 × 103
σ CF =
= 141.4 × 106 Pa (C) ◭
=
A
1400 × 10−6
4
PBD = 50 kN
PBD = 0
ΣMC = 0
3(40) − 3
5
PBD
50 × 103
σ BD =
= 35.7 × 106 Pa (T) ◭
=
A
1400 × 10−6
From FBD of member AB:
ΣMD = 0
7(40) − 4PBC = 0
PBC = 70 kN
PBC
70 × 103
σ BC =
= 50.0 × 106 Pa (C) ◭
=
A
1400 × 10−6
1.18
1.20
By symmetry, the reaction at the support E is 140 kN ↑.
Using the FBD of the section shown:
3
√ PCD (6) + 140(12) − 140(6) = 0
ΣMG = 0
10
PCD = −147.6 kN = 147.6 kN (C)
147.6 × 103
PCD
=
= 1.476 × 10−3 m2
ACD =
σ comp
100 × 106
Using the FBD of joint C:
ΣFy = 0
− 150 − PCE = 0
PCE = −150 kN = 150 kN (C)
PCE
150 000
ACE =
=
= 1071.4 mm2 ◭
σ comp
140
= 1476 mm2 ◭
ΣMD = 0
140(6) − PGF (4) = 0
PGF = 210 kN (T)
PGF
210 × 103
AGF =
=
= 1.500 × 10−3 m2
σ tens
140 × 106
= 1500 mm2 ◭
2
√ PGD (18) + 140(12) − 140(6) = 0
ΣMO = 0
13
PGD = −84.13 kN = 84.13 kN (C)
PGD
84.13 × 103
AGD =
=
= 0.841 × 10−3 m2
σ comp
100 × 106
Using the FBD of the section:
3
150(4) − PBE (8) = 0
5
PBE = 125 kN = 125 kN (T)
125 000
PBE
=
= 625 mm2 ◭
ABE =
σ tens
200
3
ΣMB = 0
− 150(4) − 150(8) − PEF (8) = 0
5
PEF = −375 kN = 375 kN (C)
375 000
PEF
=
AEF =
= 2678.6 mm2 ◭
σ comp
140
ΣMD = 0
= 841 mm2 ◭
1.19
4
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1.22
62.5 × 103
PBE
=
= 625 × 10−6 m2
ABE =
(σ T )w
100 × 106
ABF
ACF
= 625 mm2 ◭
PBF
42.72 × 103
=
=
= 427 × 10−6 m2
(σ T )w
100 × 106
A
60 × 120
=
= 11201.2 mm2
◦
sin 40
sin 40◦
V = P cos 40◦ = 0.7660P
A′ =
= 427 mm2 ◭
52.5 × 103
|PCF |
=
= 656 × 10−6 m2
=
(σ C )w
80 × 106
τw =
= 656 mm2 ◭
V
A′
1250 =
0.7660P
11201.2
P = 18278.7 kN ◭
1.23
1.21
A
0.05 × 0.1
=
= 5.321 × 10−3 m2
cos 20◦
cos 20◦
N = P cos 20◦ = 0.9397P
V = P sin 20◦ = 0. 3420P
A′ =
From FBD of truss
ΣMF = 0
12RA − 8(36) = 0
Assuming that compression governs:
RA = 24 kN
N
0.9397P
18 × 106 =
′
A
5.321 × 10−3
3
P = 101.9 × 10 N = 101.9 kN
From FBD of portion ABC
ΣMB = 0:
4(24) − 4PCE = 0
σw =
ΣME = 0:
Assuming that shear governs:
8(24) − 4(36) + 4
ΣFx = 0:
1
√ PBD
5
+4
2
√ PBD
5
V
0. 3420P
4 × 106 =
A′
5.321 × 10−3
P = 62.2 × 103 N = 62.2 kN ◭
=0
τw =
1
2
√ PBD + √ PBE + PCE = 0
5
2
Solution is
1.24
PCE = 24.0 kN PBD = −8.944 kN PBE = −22. 63 kN
ACE =
Equivalent joint
24.0 × 103
PCE
= 1.2 mm2 ◭
=
(σ T )w
20 × 103
From Eqs. (1.5):
3
ABD =
ABE =
8.944 × 10
|PBD |
= 0.745 mm2 ◭
=
(σ C )w
12 × 103
4000
P
2
cos2 θ =
cos2 75◦ = 1.07 N/mm ◭
A
180 × 22.5
P
4000
2
τ =
sin 2θ =
sin 150◦ = 0.25 N/mm ◭
2A
2(180 × 22.5)
σ =
|PBE |
22.63 × 103
= 1.886 mm2 ◭
=
(σ C )w
12 × 103
5
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1.25
1.28
Assuming that tension in wood governs:
(σ w )wd =
P
A
20 =
P
0.12
P = 0.2 MN
P
πd2
400 × 103
(300 × 106 )
=
2
4
2
d = 0.0291 m = 29.1 mm ◭
τ wA =
Assuming that tension in glue governs:
P
P
cos2 50◦
cos2 θ
8=
A
0.12
P = 0.1936 MN
(σ w )gl =
1.29
Assuming that shear in glue governs:
P
sin 2θ
2A
P = 0.244 MN
(τ w )gl =
12 =
P
sin 100◦
2(0.1)2
Shear in rivet determines upper limit on P :
P = τ As = 60 × 106
The largest allowable P is determined by tension in glue:
P = 193.6 kN ◭
π(0.02)2
= 18 850 N
4
Plate thickness is determined by bearing stress:
P = σ b Ab
18 850 = (120 × 106 )(0.02t)
t = 7.85 × 10−3 m = 7.85 mm ◭
1.26
1.30
(a)
P = τ A = τ (πd × t)
= 350 × 106 π(0.02)(0.025) = 550 × 103 N ◭
τ=
P
50 × 103
=
= 53.1 × 106 Pa = 53.1 MPa ◭
3A
π(0.02)2
3
4
(b)
1.27
σb =
P = τA
σw
πd2
= τ (πd × t)
4
(c)
σw d
= tτ
4
P
50 × 103
=
= 18.18 × 106 Pa
t(w − d)
0.025(0.13 − 0.02)
= 18.18 MPa ◭
σ =
(a) d = 75 mm
t=
50 × 103
P
=
= 33.3 × 106 Pa = 33.3 MPa ◭
3td
3(0.025)(0.02)
σw d
300 (75)
=
= 23.4 mm ◭
4τ
4 (240)
(b) t = 7.5 mm
d=
4 (7.5) (240)
4tτ
=
= 24 mm ◭
σw
300
6
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1.31
1.34
Assuming that shear in rivets governs:
π(0.02)2
(40 × 106 )
4
= 37.7 × 103 N = 37.7 kN ◭
P = 3Aτ w = 3
Assuming that bearing stress governs:
1
1
− √ PAB − √ PBC + F = 0
2
5
1
2
ΣFy = 0
− √ PAB + √ PBC = 0
2
5
√
√
2 2
5
PAB =
F = 0.9428F
PBC =
F = 0.7454F
3
3
ΣFx = 0
P = 3td (σ b )w = 3(0.025)(0.02)(90 × 106 )
= 135 × 103 = 135 kN
Assuming that tension in plates governs:
P = t(w − d)σ w = 0.025(0.13 − 0.02)(120 × 106 )
= 330 × 103 N = 330 kN
τ w A = PAB
(30000)
π(0.035)2
= 0.9428F
4
F = 30.6 kN ◭
1.32
1.35
F =
T
2200
=
= 73.33 × 103 N
r
0.03
F = τ wA
73.33 × 103 = (60 × 106 ) (0.07b)
b = 0.01746 m = 17.46 mm ◭
ΣFy = 0
ΣMA = 0
1.33
By − 19.62 = 0
6By − 8Bx − 3(19.62) = 0
Solution is
ΣMB = 0
ΣFx = 0
ΣFy = 0
Bx = 7. 358 kN
By = 19. 62 kN
p
B = 7.3582 + 19.622 = 20.95 kN
360Ax − 1080(1000) = 0
Ax − RB cos 45◦ = 0
20.95 × 103
πd2
2
4
d = 14.91 × 10−3 m = 14.91 mm
Ay + RB sin 45◦ − 1000 = 0
τw =
Ax = 3000 N
Ay = −2000 N
RB = 4242.5 N
p
(3000)2 + (−2000)2 = 3605.6 N
RA =
πd2
π
A =
= (0.015)2 = 1.767 × 10−4 m2
4
4
B
2A
60 × 106 =
RA
3605.6
= 20.4 MPa ◭
=
A
1.767 × 10−4
RB
4242.6
τB =
= 24 MPa ◭
=
A
1.767 × 10−4
τA =
7
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1.36
1.38
MD = 0 240(30 sin 60◦ ) − 200P = 0
ΣFx = 0 P cos 30◦ − Bx = 0
ΣMC = 0 400(P cos 30◦ ) − 480(P sin 30◦ ) − 480By = 0
P = 31.18 kN
ΣFx = 0 Dx − P − 30 cos 60◦ = 0
Dx = 46.18 kN
◦
ΣFy = 0 Dy − 30 sin 60 = 0
Dy = 25.98 kN
p
D = 46.182 + 25.982 = 52.99 kN
Solution is
Bx = 0.8660P
By = 0.2217P
p
B = P 0.86602 + 0.22172 = 0.8939P
(a)
31.18 × 103
πd2
4
d = 19.92 × 10−3 m = 19.92 mm ◭
σw =
P
A
100 × 106 =
τw =
(b)
B
2A
120 =
0.8939P
π(0.030)2
2
4
P = 189.8 kN ◭
1.39
3
τ=
D
52.99 × 10
= 84.3 × 106 Pa = 84.3 MPa ◭
=
2A
π(0.02)2
2
4
ΣMB = 0
4.924(RA sin 60◦ ) − 0.8682(RA cos 60◦ ) − 2.462(10000) = 0
1.37
ΣFx = 0
ΣFy = 0
RA cos 60◦ − Bx = 0
RA sin 60◦ − 10000 + By = 0
Solution is
ΣMB = 0
ΣMC = 0
ΣFx = 0
T = 17.276P
150P − 50(T sin 10◦ ) = 0
RA = 6427.8 N
Bx = 3213.9 N
By = 4433.4 N
p
B = (3213.9)2 + (4433.4)2 = 5475.8 N
200P − 50By = 0
Bx + T cos 10◦ = 0
Bx = −17.014P
By = 4P
τ=
Assuming that cable governs: T = σ w Acab
17.276P = (140 × 106 )
π(0.003)2
4
B
=
2A
5475.8
= 2.85 MPa ◭
π(.035)2
2
4
P = 57.3 N
Assuming that pin B governs: RB = τ w Apin
P
p
π(0.006)2
(−17.014)2 + 42 = (28 × 106 )
4
P = 45.3 N ◭
8
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1.40
1.43
Strength of boards is
Total area of the bolts is
π(0.01)2
= 314.2 × 10−6 m2
A=4
4
P = σ bd Abd = 4.2(172.5 × 52.5) × 10−6 = 38 kN
For plywood:
P
= σ pw Apw
2
38
= 7.2 × 103 (0.1725t) t = 0.015 m
2
= 15 mm ◭
30 × 103 cos 40◦
P
σ =
= 73.1 × 106 Pa ◭
=
A
314.2 × 10−6
(30 × 103 ) sin 40◦
V
= 61.4 × 106 Pa ◭
=
τ =
A
314.2 × 10−6
For glue (double shear):
P
= τ Agl
2
38
= 0.3 × 103 (0.1725b) b = 0.367 m
2
= 367 mm ◭
1.44
1.41
P = τ As
σA = σ b Ab
π(0.02)2
π(d2 − 0.022 )
(150 × 106 )
= (13 × 106 )
4
4
d = 0.0708 m = 70.8 mm ◭
6 × 103 = 1.8 × 106 (0.1b)
b = = 0.0333 m = 33.3 mm ◭
P = σ b Ab
6 × 103 = 5.5 × 106 (0.1a)
a = 0.01091 mm = 10.91mm ◭
1.45
1.42
d = 19 mm (rivet diameter); t = 6 mm for BC and t = 13
mm for BE (wall thickness). Because the gusset plate is
thicker than the members, bearing between the rivets and
the plate does not have to be considered.
First find the maximum safe shear force V in each bolt.
Assuming that bearing stress governs:
(a)
V = σ b Ab = 90 (0.025 × 0.015) = 33.75 kN
nτ
Assuming that shear stress governs:
V = τ As = (60)
π
(0.025)2 = 29.45 kN
4
From FBD of joint C: PBC = 96 kN
πd2
= PBC
4
n = 4.84
n(70 × 106 )
π(0.019)2
= 96 × 103
4
T = 4V r = 4(29.45)(0.105) = 12.4 kN · m ◭
9
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nσ b (td) = PBC
n = 6.02
1.47
n(140 × 106 )(0.006 × 0.019) = 96 × 103
Use 7 rivets ◭
(b) From FBD of portion ABC:
3
ΣMA = 0
8
PBE − 4(96) = 0
5
nτ
πd2
= PBE
4
n = 4.03
n(70 × 106 )
PBE = 80 kN
π(0.019)2
= 80 × 103
4
From FBD of truss:
ΣMA = 0
nσ b (td) = PBE
n = 2.31
6
n(140 × 10 )(0.013 × 0.019) = 80 × 10
From FBD of joint F:
4
PDF + 180 = 0
ΣFy = 0
5
From FBD of portion EDF:
Use 5 rivets ◭
ΣMD = 0
1.46
4
PBC = 96 kN (from solution of Prob. 1.45)
nτ As = PBC
n(70 × 106 )
n = 5.19
(b) PBE = 80 kN (from solution of Prob. 1.45)
n = 2.00
3
√ PBD
13
+3
2
√ PBD
13
+ 6(180) − 3(200) = 0
π(0.022)2
= 80 × 103
4
n = 3.01
nσ b Ab = PBE
PDF
225 × 103
= −187.5 × 106 Pa
=−
A
1200 × 10−6
= 187.5 MPa (C) ◭
PCE
135 × 103
σ CE =
= 112.5 × 106 Pa
=−
A
1200 × 10−6
= 112.5 MPa (T) ◭
96.15 × 103
PDF
= −80.1 × 106 Pa
=−
σ BD =
A
1200 × 10−6
= 80.1 MPa (C) ◭
n(140 × 106 )(0.006 × 0.022) = 96 × 103
n(70 × 106 )
PCE = 135 kN
σ DF =
Use 6 rivets ◭
nτ As = PBE
4PCE − 3(180) = 0
PBD = −96.15 kN
π(0.022)2
= 96 × 103
4
n = 3.61
nσ b Ab = PBC
PDF = −225 kN
ΣMD = 0
Repeat Prob. 1.45 with d = 22 mm.
(a)
10RF − 4(100) − 7(200) = 0
RF = 180 kN
3
1.48
n(140 × 106 )(0.013 × 0.022) = 80 × 103
P = σw A
Use 4 rivets ◭
πd2
45 × 10 = (300 × 10 ) 2
4
3
6
d = 9.77 × 10−3 m = 9.77 mm ◭
10
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1.52
1.49
Maximum axial force equals weight of cable:
PAB = 24000 − 14000 = 10000 N
σ AB =
P
A
PBC = 14000 N (T)
Pmax = ρgAL
(0.0452 − 0.03752)
4
= 20.6 MPa (C) ◭
P
14000
σ BC =
=
= 35.2 MPa (T) ◭
A BC
(π/4) (0.02252 )
AB
Pmax
= ρgL
A
390 × 106 = 2700(9.81)L
L = 14 720 m = 14.72 km ◭
10000
= π σ max =
Result is independent of diameter of cable.
1.53
1.50
σ st Ast = σ al Aal
72
π(24)2
π(37.52 − 302 )
= σ al
4
4
= 81.9 MPa ◭
1.54
Assuming that stress in steel column governs:
P = σ st Ast =
156(1202 − 1052 ) × 10−6
4
(a) Largest bearing stress is between pin and 12-mm
thick member.
= 131.6 kN
Assuming that bearing stress on concrete governs:
2
P = σ b Apl = 7.2(210) × 10
−6
σb =
= 317.5 kN ◭
(b)
1.51
25 × 103
P
=
= 166.7 × 106 Pa ◭
td
0.012(0.0125)
Pin is in double shear.
τ=
P
25 × 103
=
= 101.9 × 106 Pa ◭
2
2 (πd /4)
2π(0.0125)2/4
(c) Largest normal stress is in 12-mm thick member at
section m-n.
A=
2475
= 2530.3 mm2 (inclined cross-sectional area)
cos 12◦
(a)
σ = 72 MPa = 72 N/mm
N
σ=
A
σ=
2
P cos 12◦
72 =
2530.3
25 × 103
P
=
= 166.7 × 106 Pa ◭
t (b − d)
0.012(0.025 − 0.0125)
1.55
P = 186.3 kN ◭
Assuming that shear stress in rivets governs:
(b)
P = 4τ
V
P sin 12◦
186.3 × 103 sin 12◦
τ=
=
=
= 15.3 MPa ◭
A
A
2530.3
πd2
π
= 4(84) (0.0302 ) = 237.5 kN ◭
4
4
Assuming that bearing stress governs:
P = 4σ b (td) = 4(84) (0.035 × 0.030) = 352.8 kN
11
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1.56
1.59
Assuming that normal stress in wood governs:
ΣFx = 0
ΣFy = 0
P = σ(tb) = 4.8 (60 × 120) = 34.56 kN
+ ← PAC sin 30◦ + PBC sin 70◦ = 0
+ ↓ PAC cos 30◦ − PBC cos 70◦ + P = 0
PAC = −0.9542P = 0.9542P (C)
Assuming that bearing stress on wood governs:
PBC = 0.5077P (T)
Boom in compression:
P = σ b (td) = 9 (60 × 22.5) = 12.15 kN ◭
σ w A = PAC
108(1202 − 1052 ) = 0.9542P
P = 382 kN
1.57
Cable in tension:
σ w A = PBC
150
π(11.25)2
= 0.5077P
4
P = 29.4 kN
Pin in double shear:
N is carried by surface between cast iron pieces; V is
carried by key.
2τ w A = PAC
2(81.6)π
(15)2
= 0.9542P
4
P = 30.2 kN
V = 360 sin 55◦ = 294.9 kN
The largest safe load is P = 29.4 kN determined by cable
tension. ◭
V = τ (wb)
294.9 = 300(75b)
b = 13.1 mm ◭
h
h
294.9 = 240 75
V = σb w
2
2
h = 32.8 mm ◭
C1.1
1.58
From FBD of joint A:
By symmetry, the reaction at the support A is 10 kN ↑.
Using the FBD of the section shown:
ΣFy = 0
P
h
−
=0
PAC √
2
b 2 + h2
ΣFx = 0
h
+ PAB = 0
PAC √
2
b + h2
√
b 2 + h2
PAC = P
(T)
2h
ΣME = 0
− 10(2L/ tan 50o ) − PBC (L) = 0
PBC = −20/ tan 50o = −16.78 kN
16 780
PBC
=
= 4.4 MPa (C) ◭
σ BC =
ABC
3780
ΣFy = 0
10 − PBE sin 50o = 0 PBE = 13 050 kN
PBE
13 050
σ BE =
=
= 3.5 MPa (T) ◭
ABE
3780
PAB = −P
b
b
=P
(C)
2h
2h
12
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Computations:
√
Axial forces
b
b 2 + h2
(from
PAB (h) := P ·
PAC (h) := P ·
2·h
2 · h equilibrium)
C1.1 MathCad worksheet
Given:
P := 120 · 103 · N
2
σ t : = 18 · 103 · N/m
b :=6· m
2
σ c : = 12 · 103 · N/m
Computations:
√
b 2 + h2
b
PAC (h) := P ·
PAB (h) := P ·
2·h
2·h
V (h) := 2 · AAC (h) ·
√
Volume of
b2 + h2 + 2 · AAB (h) · b
material
Plot range
and
increment
2
1.8
Plot range
and
increment
h := 0.5 · b, 0.52 · b..4 · b
√
Volume of
b2 + h2 + 2 · AAB (h) · b
material
h := 0.5 · b, 0.52 · b..4 · b
Volume (cu.m)
V (h) := 2 · AAC (h) ·
Axial forces
(from
equilibrium)
Required
crosssectional
areas
PAC (h)
PAB (h)
AAC (h) :=
AAB (h) :=
σt
σc
Required
crosssectional
areas
PAC (h)
PAB (h)
AAC (h) :=
AAB (h) :=
σc
σt
1.6
1.6
1.4
Volume (cu.m)
1.2
1.4
1
0
5
10
h (m)
1.2
15
20
Find optimal value of h:
h := 3 · b
(initial value used in solution)
d
V (h) = 0 hopt := Find(h) hopt = 7.746 m
Given
dh
1
0.8
0
5
10
h (m)
15
20
Find optimal value of h:
C1.3
h := 3 · b
(initial value used in solution)
d
V (h) = 0 hopt := Find(h) hopt = 9.487 m
Given
dh
From FBD of joint A:
C1.2
ΣFx = 0
ΣFy = 0
The forces computed in the solution of Prob. C1.1 are
reversed. Thus
√
b 2 + h2
b
PAC = P
(C)
PAB = P
(T)
2h
2h
− PAB cos θ − PAC cos α = 0
− PAB sin θ + PAC sin α − P = 0
P
P
=
(C)
cos θ tan α + sin θ
cos θ tan α + sin θ
P
PAC =
(T)
cos α tan θ + sin α
PAB = −
C1.2 MathCad worksheet
C1.3 MathCad worksheet
Given:
Given:
P := 530 · 103 · N
σ t := 125 · 106 · Pa2
P := 120 · 103 · N
2
σ t := 18 · 103 · N/m
b :=6· m
2
σ c := 12 · 103 · N/m
b := 1.8 · m
α := 30 · deg
6
σ c := 85 · 10 · Pa
13
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Computations:
6·107
P
cos(θ) · tan(α) + sin(θ)
P
PAC (θ) :=
cos(α) · tan(θ) + sin(α)
PAB (θ) :=
Required
crosssectional
areas
b
b
+ AAC (θ) ·
cos(θ)
cos(α)
Volume of
material
5.5·107
P (N)
PAC (θ)
AAC (θ) :=
σt
PAB (θ)
AAB (θ) :=
σc
V (θ) := AAB (θ) ·
Axial forces
(from
equilibrium
5·107
4.5·107
Plot range
and
increment
θ := 0 · deg, 1 · deg..75 · deg
0
10
20
30
40
Theta (deg)
50
60
C1.5 MathCad worksheet
0.045
Given:
Volume (cu.m)
0.04
L := 360 · m
0.035
γ := 150 · N/m3
d(x) : 20 · m − 0.1 · x + (0.35 · 10−3 m−1 ) · x2
0.03
Computations:
π
[d(x)2 − (d(x) − 2 · t)2 ]
4
Z x
γ
·
A(ξ)dξ
σ(x) :=
A(x) 0
0.025
0.02
t := 1.5 · m
A(x) :=
0
20
40
theta (deg)
60
80
x := 0, 0.01 · L.. L
Find optimal value of h:
θ := 40 · deg
(initial value used in solution)
d
Given
V (θ) = 0 θopt := Find(θ) θ opt = 42.59 deg
dθ
Cross-sectional area
Axial stress
Plotting range and
increment
250
200
Stress (Pa)
C1.4 MathCad worksheet
Given:
A := 400 mm
2
2
σ w := 3500 · N/mm
150
100
2
τ w := 1800 · N/mm
50
Computations:
PN (θ) :=
Allowable P if
normal
stress governs
σw · A
cos(θ)2
0
0
100
200
x (m)
300
Find x where stress is maximum:
τw · A
PS (θ) :=
sin(θ) · cos(θ)
P (θ) :=
PN (θ)
PS (θ)
Allowable P if shear
stress governs
x := 0.6 · L
if PN (θ) < PS (θ) Allowable P is the
otherwise
smaller of the two
θ := 0 · deg, 0.5 · deg..60 · deg
Given
Plotting range and
and increment
(initial value used in solution)
d
σ(x) = 0 xmax := Find(x)
dx
xmax := 239.3 m
σ max := σ(xmax )
σ max = 225.4 Pa
14
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Chapter 2
2.1
2.2
A = π(0.0140)2 /4 = 153.94 × 10−6 m2 ; L = 50.0 mm
Converting the data from P -δ to σ-ǫ :
σ (MPa) ǫ 10−3
σ (MPa) ǫ (10−3 )
A = π(15.15)2 /4 = 180.3 mm2 ; L = 60 mm
Converting the data from P -δ to σ-ǫ :
0
41.0
81.8
122
163
203
246
260
270
0
0.2
0.4
0.6
0.8
1.0
1.2
3.26
8.66
300
340
380
425
448
440
422
400
25.0
50.0
90.0
150
250
310
400
fracture
σ (MPa)
ǫ (10−3 )
0
64.1
128.7
192.7
257.6
321.7
360.5
0
1.1
2.2
3.3
4.4
5.5
7.5
388.2
399.3
402.1
404.9
410.4
404.9
377.2
10.0
12.5
30.0
40.0
50.0
60.0
fracture
×102
5
4
300
σyp
3.5
200
Stress (MPa)
Stress (MPa)
ǫ (10−3 )
(a) σ pl = 321.7 MPa ◭
(b) E = (321.7/5.5) × 109 = 58.5 × 109 Pa ◭
(c) σ yp (at 0.2% offset)= 380 MPa ◭
(d) σ u = 410.4 MPa ◭
(e) σ f = 377.2 MPa ◭
(a) σ pl = 246 MPa ◭
(b) E = (246/1.2) × 109 = 205 × 109 Pa = 205 GPa ◭
(c) σ yp = 260 MPa ◭
(d) σ u = 448 MPa ◭
(e) σ f = 400 MPa ◭
σPL
E
100
3
2.5
2
1.5
1
0.5
I
0
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Strain
0.002
0
5·10–4
0
0.001
Strain
400
Stress (MPa)
σ (MPa)
0.0015
2.3
σu
Segment AB:
P
20 × 103
=
= 10.186 × 106 Pa
AAB
(π/4)(0.05)2
= 10.186 MPa
σf
σ AB =
σyp
200
εAB = 0.0045 (from stress-strain diagram)
δ AB = LAB εAB = 400(0.0045) = 1.8 mm
0
0
0.1
0.2
Strain
0.3
Segment BC:
P
20 × 103
=
= 40.74 × 106 Pa
ABC
(π/4)(0.025)2
= 40.74 MPa
σ BC =
εBC = 0.025 (from stress-strain diagram)
δ BC = LBC εBC = 300(0.025) = 7.5 mm
Bar:
δ = δ AB + δ BC = 1.8 + 7.5 = 9.3 mm ◭
15
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2.4
2.7
50
(a) From the FBD below, P = w = weight x volume =
ρgAx.
δ =
Z L
0
=
Z L
0
P dx
=
EA
Z L
0
120
Consider the left half of the bar. The width b of the bar
varies as
b = 50 + 0.07x mm = 0.05 + 0.07x m
Q.E.D.
The elongation of the left half of the bar is
(b) Substituting ρ = M/AL gives
δ=
M gL2
M gL
=
AL 2E
2EA
b
1000
(ρgAx) dx
EA
ρgx dx
ρgL2
=
E
2E
x
Z 1.0
Z 1.0
500 × 103 dx
P
δ 1/2 =
dx =
EA
(200 × 109 ) (0.05) (0.05 + 0.07x)
0
0
Z 1.0
dx
= 50 × 10−6
0.05 + 0.07x
0
Q.E.D.
1.0
ln(0.05 + 0.07x)
0.07
0
ln(0.12)
−
ln(0.05)
= 625.3 × 10−6 m
= 50 × 10−6
0.07
= 0.6253 mm
= 50 × 10−6
2.5
P L ρgL2
+
EA
2E
7850(9.81)(150)2
(20 × 103 )(150)
+
=
9
−6
(200 × 10 )(300 × 10 )
2(200 × 109 )
δ =
Elongation of the bar is
δ = 2(0.6253) = 1.251 mm ◭
= 54.3 × 10−3 m = 54.3 mm ◭
2.8
2.6
A = π(4)2 /4 = 12.566 mm2
P = σA = 280 × 106 12.566 × 10−6 = 3518 N
(0.004) 200 × 109 12.566 × 10−6
δEA
P =
=
= 3140 N
L
3.2
d = 0.02+0.025x m
δ =
Z 0.4
A=
4P
P dx
=
EA
πE
Therefore, the maximum safe value is P = 3140 N ◭
πd2
π
= (0.02 + 0.025x)2 m2
4
4
Z 0.4
2.9
dx
2
(0.02 + 0.025x)
4 30 × 103
4P
(666.7) =
(666.7)
=
πE
π (72 × 109 )
0
0
P (1080)
P Lst
=
= 2.4 × 10−5 P
Est A
(200 × 103 ) (225)
P (180)
P Lal
=
= 1.1 × 10−5 P
δ al =
Eal A
(70 × 103 )(225)
δ st =
= 354 × 10−6 m = 0.354 mm ◭
δ = δ st + δ al
0.54 = (2.4 + 1.1) × 10−5 P
P = 15.4 kN ◭
16
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2.10
(ii) σ st ≤ 140 MPa (Pst = P )
P = σ st Ast = 140 × 106 480 × 10−6 = 67.2 × 103 N
2
2
Ast = π(18)
/4 = 254.5
mm
Acu = π 302 − 22.52 /4 = 309.3 mm2
(200 × 103 )(22.5)
Eδ
=
= 10000 MPa ◭
σ st =
L st
450
(120 × 103 )(0.225)
Eδ
=
σ cu =
= 60 MPa ◭
L cu
450
Comparing the above values, Pmax = 39.0 kN ◭ (governed
by the stress in bronze)
From the FBD: P = Pst + Pcu ,
which gives
2.14
σ br ≤ 120 MPa (|Pbr | = 2P )
2P = σ br Abr =
650 × 10−6
Equilibrium: Pst = 2P
Steel: Pst = σ st Ast
Pal = 3P
2P = (240) (12)
P = 1440 N
Aluminum: Pal = σ al Aal 3P = (120) (18)
P = 720 N
PL
PL
+
Elongation δ =
EA st
EA al
2P (2000)
3P (1500)
6 =
+
(200 × 103 ) (12) (70 × 103 ) (18)
P = 1145.5 N
2.11
A = 10(80) = 800 mm2 = 800 × 10−6 m2
Lf
π(1500.5) − π(1500.0)
=
= 333.3 × 10−6
Li
π(1500.0)
Using σ = Eǫ = P/A, we have
P = EǫA = 200 × 109 (333.3 × 10−6 )(800 × 10−6 )
Comparing the above values, Pmax = 720 N ◭
2.15
= 53.3 × 103 N = 53.3 kN ◭
Steel: Pst = P (T), L = 1 m, A = 675 mm2 ,
σ w = 120 MPa
Bronze: Pbr = 2P (C), L = 2 m, A = 900 mm2 ,
σ w = 108 MPa
Alum.: Pal = 2P (T), L = 1.5 m, A = 450 mm2 ,
σ w = 72 MPa
2.12
PL
1
=
ΣP L
EA
EA
103
[20(3.0) − 15(3.0) + 25(1.5)]
=
9
(12 × 10 )(1750 × 10−6 )
δ = Σ
Pst = σ st Ast
P = (120) (675) = 81000 N
= 2.50 × 10−3 m = 2.50 mm ◭
Pbr = σ br Abr
2P = (108) (900)
P = 48600 N
Pal = σ al Aal
2P = (72) (450)
P = 32400 N
PL
PL
PL
+
+
δ=
EA st
EA br
EA al
P (1000)
2P (2000)
2P (1500)
2.4=
−
+
3
3
(210 × 10 )(675) (84 × 10 )(900) (70 × 103 )(450)
P = 48600 N
2.13
(i) δ ≤ 3 mm = ΣP L/ (EA)
0.003 =
P = 39.0 × 103 N
P = (σA)st + (σA)cu
= (10000) (254.5) + (60) (309.3) = 2563.6 kN ◭
ǫ=
120 × 106
P (1.0)
(200 × 109 ) (480 × 10−6 )
2P (2)
−
9
(80 × 10 ) (650 × 10−6 )
0.003 = P (1.0417 − 7.6923) × 10−8
P = −45.1 × 103 N
|P | = 45.1 × 103 N
Comparing the above values, Pmax = 32400 N ◭ (governed
by the stress in the aluminum)
17
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2.18
2.16
Equilibrium:
Bronze: Pbr = 3P (C), L = 0.6 m, A = 450 mm2 ,
σ w = 120 MPa
Alum.: Pal = 2P (C), L = 1.0 m, A = 600 mm2 ,
σ w = 80 MPa
Steel: Pst = 2P (T), L = 0.8 m, A = 300 mm2 ,
σ w = 140 MPa
Pbr = σ br Abr
3P = 120 × 106 450 × 10−6
Pal = σ al Aal 2P = 80 × 106 600 × 10−6
ΣME = 0
PA x = PB (10 − x)
(1)
Compatibility: δ A = δ B
P = 18.0 × 103 N
PB = PA
P = 24.0 × 103 N
PB LB
PA LA
=
EAA
EAB
LA AB
(1) (180)
PA
= PA
=
LB AA
(1.5) (360)
3
Substituting (2) into (1):
Pst = σ st Ast
2P = 140 × 106 300 × 10−6
P = 21.0 × 103 N
PL
PL
PL
+
+
δ =
EA br
EA al
EA st
1
0.002 =
×
(109 ) (10−6 )
2P (1.0)
2P (0.8)
3P (0.6)
−
+
−
(83) (450) (70) (600) (200) (300)
PA x =
PA
(5 − x) which gives x = 1.25 m ◭
3
2.19
P = −28.93 × 103 N (negative means bar contracts)
Equilibrium ΣMA = 0: 2.5P = Pst gives Pst = 2.5P
Comparing the above values, Pmax = 18.0 kN ◭ (governed
by the stress in the bronze)
Compatibility
yC = 2.5δ st
σ st ≤ 35 MPa: Pst = σ st Ast
2.17
yC ≤ 3.6 mm:
yC = 2.5δst = 2.5
3.6 = 2.5
δ BD =
+
2.5P = (35) (450)
P = 6300 N
PL
EA
st
(2.5P ) (2000)
(200 × 103 ) (450)
which gives P = 25920 N
Comparing the above two values, Pmax = 6300 N ◭
(governed by the movement of C)
From the FBD:
ΣMA = 0
(2)
3PBD − 5(360) = 0
PBD = 600 N
PBD LBD
(600)(2500)
=
= 3.33 mm ◭
EA
(200 × 103 ) (2.25)
2.20
Equilibrium ΣMC = 0: 3.5Pal = 2.5Pst
ΣFy = 0: Pal + Pst = 50 kN
18
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Solving gives Pal = 20.83 kN and Pst = 29.17 kN. Using
δ = P L/(EA) gives
20.83 × 103 (3)
= 2.9757 × 10−3 m
δ al =
(70 × 109 ) (300 × 10−6 )
29.17 × 103 (4)
= 1.1668 × 10−3 m
δ st =
(200 × 109 ) (500 × 10−6 )
Compatibility
Equilibrium
AB: ΣMA = 0: 6Pst = 3Pal CD: ΣMC = 0: 6Pst = 3P
Solving yields: Pst = 0.5P and Pal = P.
Compatibility
yB = 2δ al
yD = yB + δ st = 2δal + δ st
yP = 0.5yD = δ al + 0.5δ st
PL
PL
+ 0.5
yP =
EA al
EA st
Using yP = 6 mm, the equation becomes
yC − δ st
δ al − δ st
=
2.5
6.0
2.5
(δ al − δ st ) + δ st
yC =
6.0
2.5
(2.9757 − 1.1668) + 1.1668
yC =
6.0
yC = 1.921 mm ◭
0.006 =
P (2)
(70 × 109 ) (300 × 10−6 )
0.5P (2)
+ 0.5
(200 × 109 ) (500 × 10−6 )
Solving gives P = 59.9 × 103 N = 59.9 kN ◭
2.21
2.23
Equilibrium:
Bar BD: ΣMD = 0 gives Pbr = 25 kN
ΣMD = 0
NC = 25 kN
Bar AB: ΣMA = 0 1Pal = 2.5Pbr Pal = 2.5 (25) = 62.5 kN
ΣMA = 0
4.5 (25)
3
= 37.5 kN
P =
Compatibilty:
37.5 × 103 (3)
PL
=
δ =
EA
(200 × 109 ) (300 × 10−6 )
= 1.875 × 10−3 m = 1.875 mm
From geometry of deformation: δ/3 = yC /4.5 which gives
Compatibility: ∆ =
4.5
4.5
yC =
δ=
(1.875) = 2.81 mm ◭
3
3
1
(2.5δal + δ br )
2
PL
PL
1
2.5
+
2
EA al
EA br
"
#
3
62.5 × 10 (1500)
25 × 103 (2000)
1
=
2.5
+
2
(70 × 103 ) (675)
(84 × 103 ) (225)
∆ =
2.22
= 3.8 mm ◭
19
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9.156 × 103 (2.5)
δ AC =
(200 × 109 ) (120 × 10−6 )
2.24
ΣMC = 0
= 0.9538 × 10−3 m = 0.9538 mm (elongation)
δ BC = 0.9538 mm (contraction)
Pal (4) = 22000 (1.5)
Pal = 8250 N
0.9538
δ AC
=
cos 35◦
cos 35◦
= 1.164 mm ◭
∆C =
PL
δ al =
EA al
8250(3000)
=
(70 × 103 ) (148.5)
= 2.38 mm
2.27
dh = 2.38 mm →
3
dv = rAB = δ al
4
3
(2.38) = 1.785 mm ↓ ◭
=
4
ΣFy = 0:
5
PAB = (60000) = 100000 N (T)
3
ΣFx = 0 :
4
4
PBC = PAB = (100000) = 80000 N (C)
5
5
80000 (4000)
PL
=
= 2.37 mm (contr.)
δ BC =
EA BC
(200 × 103 ) (675)
2.25
AAC = ABC = 120 mm2 ; LAC = LBC = 2.5 m;
E = 200 GPa
ΣFx = 0 :
PAB = PBC
ΣFy = 0:
◦
2PAB sin 35
(∆B )h = δ BC = 2.37 mm → ◭
= 15 kN
PAB = 13.076 kN
13.076 × 103 (2.5)
PL
δ = δ AB = δ BC =
=
EA
(200 × 109 ) (120 × 10−6 )
= 1.362 × 10−3 m = 1.362 mm
δ
1.362
=
sin 35◦
sin 35◦
= 2.37 mm ◭
∆C =
2.28
2.26
AAC = ABC = 120 mm2 ;
LAC = LBC = 2.5 m;
E = 200 GPa
ΣFy = 0: PAC = −PBC
ǫx =
σx
P
=
E
EA
=
(1)
ΣFx = 0: PBC cos 35◦ −PAC cos 35◦ + 15 = 0
ǫy =
(2)
Solving (1) and (2) gives PAC = 9.156 kN (T)
and PBC = −9.156 kN = 9.156 kN (C)
P
E (πd2 /4)
∆d
4P ν
which gives
= −νǫx = −
d
πEd2
4P ν
4P ν
=
(decrease) Q.E.D.
∆d = −
πEd
πEd
20
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2.29
2.33
P = τ w A = 60(60 × 120) = 432 kN ◭
LAB
3000
(σ x − νσ y ) =
[12 − 0.4(7.2)]
E
1800
= 15.2 mm ◭
LAC
2000
δ AC =
(σ y − νσ x ) =
[7.2 − 0.4(12)]
E
18000
= 2.67 mm ◭
δ AB =
δ = γt =
60
τw
(0.75) =
(22.5) = 0.28 mm ◭
G
4800
2.34
Hydrostatic pressure: σ x = σ y = σ z = −p
2.30
110 × 109
−3
9
[0.40 + 0.35(0.30)] 10−3
(ǫx + νǫy )E
=
2
1−ν
1 − 0.352
6
= 63.3 × 10 Pa = 63.3 MPa ◭
σx =
[0.30 + 0.35(0.40)] 10
(ǫy + νǫx )E
=
σy =
1 − ν2
1 − 0.352
6
= 55.2 × 10 Pa = 55.2 MPa ◭
Consider a volume element with sides of lengths a, b and c
that are parallel to the x-, y- and z-axes, respectively.
The initial volume of the element is V = abc. After
loading, the new volume is
V + ∆V = abc(1 + ǫx )(1 + ǫy )(1 + ǫz )
110 × 10
= abc [1 + (ǫx + ǫy + ǫz ) + (higher order terms)]
V + ∆V ≈ V [1 + (ǫx + ǫy + ǫz )]
Therefore, ∆V = V (ǫx + ǫy + ǫz ), from which we find
∆V
= ǫx + ǫy + ǫz
V
2.31
ǫx =
=
ǫy =
=
ǫz =
=
From Hooke’s law, Eqs.(2.12), the normal strains for
hydrostatic pressure are
1
[σ x − ν(σ y + σ z )]
E
48 − 0.3(−24 + 60)
= 0.186 × 10−3 ◭
200 × 103
1
[σ y − ν(σ z + σ x )]
E
−24 − 0.3 (60 + 48)
= −0.282 × 10−3 ◭
200 × 103
1
[σ z − ν(σ x + σ y )]
E
60 − 0.3 (48 + 24)
= 0.192 × 10−3 ◭
200 × 103
1
1
[σ x − ν(σ y + σ z )] = [−p − ν(−2p)]
E
E
p
= − (1 − 2ν)
E
p
ǫy = ǫz = − (1 − 2ν)
E
ǫx =
Therefore, the volumetric strain becomes
∆V
3p
= ǫx + ǫy + ǫz = − (1 − 2ν)
V
E
Q.E.D.
2.35
2.32
Assuming that the load acts in the negative z-direction,
σ z = −P/A, ǫx = ǫy = 0.
1
(σ y − νσ x ) = 0 which gives σ y = ν σ x
E
1
σx
1
(σ x − νσ y ) = [σ x − ν (ν σ x )] =
1 − ν2
ǫx =
E
E
E
P
2
ν −1
=
EA
PL 2
ν −1 ◭
δ = ǫx L =
EA
ǫy =
Using Hooke’s law,
ǫx = 0
ǫy = 0
gives σ x − ν(σ y + σ z ) = 0
gives σ y − ν(σ z + σ x ) = 0
Solving Eqs.(a) and (b) yields σ x = σ y =
(a)
(b)
ν
σz
1−ν
21
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2.39
The strain in the z-direction becomes
2ν 2
σz
1
1−
[σ z − ν (σ x + σ y )] =
ǫz =
E
E
1−ν
2
σ z 1 − ν − 2ν
σ z (1 + ν) (1 − 2ν)
=
=
E (1 − ν)
E
(1 − ν)
P (1 + ν) (1 − 2ν)
= −
EA
(1 − ν)
0.3
= 4.764◦
3.6
0.12
= 2.291◦
α = tan−1
3
θ = tan−1
γ = θ−α
= 4.764◦ − 2.291◦
The change in the thickness equals the deformation in the
z-direction:
P t (1 + ν) (1 − 2ν)
δ = ǫz t = −
EA
(1 − ν)
P t (1 + ν) (1 − 2ν)
(decrease) Q.E.D.
δ =
EA
(1 − ν)
= 2.473◦ = 0.0432 rad ◭
(angle increases)
2.40
2.36
AC = BD =
Solving Eq. (2.14) for Poisson’s ratio, and substituting the
given values for G and E, gives
ν=
√
2a
73.2
E
−1=
− 1 = 0.326 ◭
2G
2 (27.6)
Apply law of cosines to triangle AC ′ D:
π
2
AC ′
= a2 + a2 + 2a2 cos
+γ
2
π
π
2
= 2a 1 + cos cos γ + sin sin γ
2
2
2
2
= 2a (1 + sin γ) ≈ 2a (1 + γ)
√ p
√ γ
Therefore, AC ′ = 2a 1 + γ ≈ 2a 1 +
2
√
√
γ
2a 1 + 2 − 2a
γ
AC ′ − AC
√
=
◭
=
ǫAC =
2
AC
2a
2.37
0.1992 − 0.2
δ
=
= −4.0 × 10−3 ◭
L x
0.2
0.1515 − 0.15
δ
=
= 10 × 10−3 ◭
ǫy =
L y
0.15
ǫx =
Change in right angle is γ = 16.6◦ − 15.7◦ = 0.9◦
= 15.71 × 10−3 rad (increase) ◭
Applying a similar analysis to triangle AB ′ D:
π
2
− γ ≈ 2a2 (1 − γ)
B ′ D = a2 + a2 + 2a2 cos
2
√ p
√ γ
which gives B ′ D = 2a 1 − γ ≈ 2a 1 −
2
√
√
γ
2a 1 − 2 − 2a
γ
B ′ D − BD
√
=− ◭
=
ǫBD =
2
BD
2a
2.38
δ
2.015 − 2.0
ǫAC =
=
= 7.50 × 10−3 ◭
L AC
2.0
δ
1.995 − 2.0
ǫBD =
=
= −2.50 × 10−3 ◭
L BD
2.0
θ = tan−1
2.015
= 45.286◦
1.995
Change in right angle ABC is
γ = 90◦ − 2θ = 90◦ − 2 (45.286◦)
= −0.572◦ = −9.98 × 10−3 rad
γ = 9.98 × 10−3 rad (angle ABC increases) ◭
22
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2.41
Applying Hooke’s law, the axial displacement of the
element shown above is
τ
dδ = γ dr = dr (2)
G
Substituting τ from Eq.(1) into Eq.(2) gives
δ
5.4
γ =
=
L
1000
= 0.0054 rad
P
dr
2πrGL
The axial displacement of the shaft is found by integration:
Z (D/2)+t
P
dr
P
(D/2)+t
=
ln r|D/2
δ =
2πGL D/2
r
2πGL
P
P
(D/2) + t
D + 2t
=
=
Q.E.D.
ln
ln
2πGL
(D/2)
2πGL
D
τ = Gγ
6
dδ =
= 420 × 10 (0.0054)
= 2.268 MPa
P = V = τ A = τ t LAB = (2.268) (15)(1500) = 51030 N ◭
2.42
2.44
Equlibrium:
σ co Aco + σ st Ast = P
(a)
Compatibility:
εco = εst
P = V = τ A = τ (πDL)
P
(1)
τ =
πDL
δ = γ t (2)
Pt
τt
=
G
πDGL
σ co = σ st
Eco
Est
(b)
Substituting into Eq. (a), we get
Eco
σ st
Aco + Ast = P
Est
14 π(1802 ) π
2
2
+ (195 − 180 ) = 1200 × 103
σ st
200
4
4
σ st = 193.6 MPa ◭
Substituting Hooke’s law and Eq. (1) into Eq. (2) gives
δ =γt=
σ st
σ co
=
Eco
Est
Q.E.D.
From Eq. (b):
2.43
σ co = 193.6
Consider the equilibrium of the cylindrical surface of
radius r: [(D/2) ≤ r ≤ (D/2) + t]
2.45
P = V = τ A = τ (2πrL)
Compatibility: ǫst = ǫco gives (σ/E)st = (σ/E)co . If the
concrete governs, the stress in the steel would be
Therefore, the shear stress is
σ st =
τ=
P
2πrL
14
= 13.6 MPa ◭
200
(1)
200
Est
σ co =
(6) = 85.71 MPa
Eco
14
Because 85.71 MPa is less than (σ w )st = 120 MPa, the
concrete governs. Equilibrium gives Pst + Pco = 320 kN:
σ st Ast + σ co Aco = 320 kN
π(0.200)2
− Ast
85.71 × 106 Ast + 6 × 106
4
= 320 × 103 N
which gives Ast = 1.650 × 10−3 m2 = 1650 mm2 ◭
23
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2.46
Assuming that concrete governs, we let σ co = (σ w )co ,
which yields
−6
−3
6 200
(1250 × 10 ) + 61.25 × 10
P = 15 × 10
24
Compatibility: ǫst = ǫwood
gives (σ/E)st = (σ/E)wood .
= 1.075 × 106 N = 1.075 MN ◭
If the steel governs, the
stress in the wood would be
σ wood =
Check stress in steel using Eq. (b):
10
Ewood
σ st =
(20) = 1 MPa
Est
200
σ st =
Because 1 MPa is less than (σ w )wood = 7.2 MPa, the steel
governs. Equilibrium gives Pst + Pwood = 1500 kN.
σ st Ast + σ wood Awood = 1500 × 103 N
200
(15) = 125 MPa ≤ (σ w )st O.K.
24
2.49
(20) (4) (200t) + (1) (200 × 200) = 1500 × 103
From FBD:
which gives t = 91.3 mm◭
PA
b
2.47
PC
b
G
C
B
A
1.2b
Compatibility: δ st = δ cu gives (σL/E)st = (σL/E)cu . If
copper governs, the stress in the steel would be
σ st =
PB
W
Est Lcu
(200) (160)
σ cu =
(70) = 77.78 MPa
Ecu Lst
(120) (240)
ΣFy = 0
ΣMA = 0
Because 77.78 MPa is less than (σ w )st = 140 MPa, the
copper governs. Equilibrium gives:
+↑
PA + PB + PC − W = 0
+
bPB + 2bPC − 1.2bW = 0
(a)
(b)
Compatibility:
b
b
2Pcu + Pst = M g
δA
2σ cu Acu + σ st Ast = M g
2 70 × 106 900 × 10−6 + 77.78 × 106 1200 × 10−6
δB
δC
δB − δA = δC − δB
Because E,A and L are identical for the three wires, the
last equation is equivalent to
= M (9.81)
which yields M = 22.4 × 103 kg◭
−PA + 2PB − PC = 0
2.48
(c)
Solution of Eqs. (a)-(c) is
Ast = 1250 × 10−6 m2
Aco = 2502 − 1250 × 10−6 = 61.25 × 10−3 m2
PA =
7
W
30
PB =
1
W
3
PC =
13
W ◭
30
Equilibrium:
σ st Ast + σ co Aco = P
(a)
Compatibility:
εst = εco
σ co
σ st
=
Est
Eco
σ st =
Est
σ co
Eco
(b)
Substituting into Eq. (a):
Est
σ co
Ast + Aco = P
Eco
24
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2.50
2.52
Compatibility: δ st = δ al + δ 0 (δ 0 = 0.1 mm). Using
δ = σL/E, we have
σL
σL
=
+ δ0
E st
E al
Because B is the longest wire, σ A = σ C = σ w . Let the
stress in B be σ B .
which gives
σ st =
σL
E
al
+ δ0
E
L
Equilibrium: From above FBD PA = PC and
P = 2PA + PB , which becomes
P = A (2σ w + σ B )
(a)
Compatibility gives δ A = 6 mm + δ B . Using δ = σL/E:
st
σ al (0.250)
200 × 109
−4
σ st =
+ 1 × 10
70 × 109
0.250
6
(a)
σ st = 2.857σal + 80 × 10 Pa
σw L
σB L
= 6 mm +
E
E
Solving for σ B gives:
Equilibrium: 2Pst + Pal = P , or 2 (σA)st + (σA)al = P.
Solving for the stress in the aluminum gives:
400 × 103 − 2σ st 1400 × 10−6
P − 2σ st Ast
=
σ al =
Aal
2800 × 10−6
(b)
σ al = 142.86 × 106 − σ st
E
70 × 103
(6) = 82 MPa
(6) = (96) −
L
30 × 103
Substituting the values of σ B , σ w and A into Eq. (a) yields
σB = σw −
P = 270 [2 × 96 + 82]
= 73980 N ◭
Substituting σ st from Eq.(a) into Eq.(b) yields
σ al = 142.86 × 106 − 2.857σal + 80 × 106
which gives: σ al = 16.30 ×106 Pa = 16.30 MPa ◭
2.53
Equilibrium: ΣMO = 0 gives
2.51
1.5PB = 0.75PA
Middle bar is shorter by ∆ = 1.35 mm
1.5σ B AB = 0.75σA AA
AA
σA
σ B = 0.5
AB
(300)
σ B = 0.5
σA
(250)
σ B = 0.600σA (a)
Equilibrium: from FBD above: Pi = 2Po .
Compatibility:
Compatibility (∆ = 4 mm):
∆ − δ A = 0.5δB
σB L
σA L
= 0.5
∆−
E
E
∆ = δi + δo
L
3Po L
PL
PL
+
=
(Pi + Po ) =
=
EA i
EA o
EA
EA
Substituting Eq.(a) into Eq.(b) yields
from which the force in an outer bar becomes
(1.35) 200 × 103 (120 × 30)
∆EA
Po =
=
= 32400 N
3L
3 (10 × 1000)
∆−
(0.600σA ) L
σA L
= 0.5
E
E
σAL
[1 + 0.5 (0.600)] = ∆
E
200 × 109 (0.004)
E∆
=
σA =
1.300L
1.300 (2)
Inspection of the FBD shows that the shear force in the
pin is V = Po = 32400 N. The shear stress in the pin is
τ=
(b)
32400
V
= 59.9 MPa ◭
=
Apin
(π/4) (26.25)2
σ A = 308 × 106 Pa = 308 MPa ◭
25
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2.54
2.57
The equilibrium and compatibility analyses are the same
as in Prob. 2.53, with the results being:
Equilibrium:
Pal + Pst = P = 320 kN (a)
σ B = 0.600σA
(a) and ∆ −
σB L
σA L
= 0.5
E
E
Compatibility:
|δ al | = |δ st | gives
(b)
PL
PL
=
EA al
EA st
Pal (450)
Pst (360)
=
(1125) (70 × 103 )
(1800) (200 × 103 )
Pal = 0.175Pst (b)
Given ǫA = 1.5 × 10−3 :
1.5 × 10−3 = 300 × 106 Pa
σ B = 0.600σA = 0.600 300 × 106 = 180.0 × 106 Pa
σ A = EǫA = 200 × 109
Equation (b) becomes
Substituting Eq.(b) into Eq.(a): 0.175Pst + Pst = P =
320 kN
which gives: Pst = 272.34 kN and Pal = 47.66 kN
P
272.34 × 103
σ st =
=
= 151.3 MPa (C) ◭
A st
1800
47.66 × 103
P
=
σ al =
= 42.4 MPa (T) ◭
A al
1125
#
"
300 × 106 (2)
180.0 × 106 (2)
∆−
= 0.5
200 × 109
200 × 109
Solving yields ∆ = 3.90 × 10−3 m = 3.90 mm ◭
2.55
Using FBD in the text: Equilibrium: R1 + R2 = P
Compatibility: |δ a | = |δ b | which becomes
R1 a
R2 b
=
giving
EA
EA
Substituting Eq.(b) into Eq.(a):
a
R2 = R1
b
(a)
2.58
(b)
From the solution to Prob. 2.57:
Pal + Pst = P
a
R1 + R1 = P
b a+b
L
R1
= R1
=P
b
b
Pb
b
Pa
a
a
R1 =
P
=
◭
R2 = R1 =
◭
L
b
b
L
L
(a)
Pal = 0.175Pst
(b)
The working loads of the two segments are:
(Pal )w = σ al Aal = (80) (1125) = 90000 N = 90 kN
(Pst )w = σ st Ast = (144) (1800) = 259200 N = 259.2 kN
Using Eq. (b), if steel governs, Pal = 0.175(259.2) =
45.36 kN < 90 kN. Therefore, steel governs.
Using Eq.(a), P = Pal + Pst = 45.36 + 259.2 = 304.56 kN ◭
2.56
2.59
Equilibrium: PCD = RD , PBC = RD − P2 ,
PAB = RD − P1 − P2
P1 b1
P2 b2
20 (2.10) 60 (0.9)
+
=
+
= 35.56 kN
L
L
2.7
2.7
= RA − P1 = 35.56 − 20 = 15.56 kN
15.56 × 103
PBC
=
=
A
600 × 10−6
= 25.9 × 106 Pa = 25.9 MPa (T) ◭
RA =
PBC
σ BC
26
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.61
Compatibility: δ CD + δ BC + δ AB = ∆ = 0, which, using
δ = P L/(EA), becomes
π(0.015)2
4
= 176.72 × 10−6 m4
AAB =
(RD − P2 ) LBC
(RD − P1 − P2 ) LAB
RD LCD
+
+
=0
EACD
EABC
EAAB
Cancelling E and the common conversion factors gives
Equilibrium:
PAB + PBC = 25 × 103 N
RD (350) (RD − 90) (250) (RD − 150 − 90) (500)
+
+
=0
1200
2000
900
Solving yields
Compatibility:
|δ AB | = |δ BC | (b)
RD = 148.71 kN
Therefore, the forces in the segments are
PCD = RD = 148.7 kN (T) ◭
PBC = RD − P2 = 148.71 − 90 = 58.7 kN (T) ◭
δ AB =
PAB = RD − P1 − P2 = 148.71 − 150 − 90
= −91.3 kN = 91.3 kN (C) ◭
D
C
B
RD
P
x
RB + RD = P
(a)
δ BC + δ CD = 0
(b)
1131.8PAB
679.06PBC
=
E
E
Compatibility:
δ BC =
RB
RB dx
=
EA
Ebt
Z L/2
RB L
ln (1 + x/L)
Ebt
0
=
Z L
Z L
0
=
0
L/2
−RD dx
RD
=−
EA
Ebt
L/2
= −0.2877
or PAB = 0.600PBC
(e)
Substituting Eq.(e) into Eq.(a) yields
0.600PBC + PBC = 25 × 103 N
dx
1 + x/L
PBC = 15 625 N
[and PAB = 0.600 (15 625) = 9375 N]
RB L
RB L 3
ln = 0.4055
Ebt
2
Ebt
The maximum stress occurs in BC, just below B:
15 625
PBC
=
AB
176.72 × 10−6
= 88.4 × 106 Pa = 88.4 MPa (C) ◭
dx
1
+
x/L
L/2
L
3
RD L
RD L
ln(1 + x/L)
ln 2 − ln
=−
= −
Ebt
Ebt
2
L/2
δ CD =
(d)
Substituting Eqs.(c) and (d) into Eq.(b) gives
From the FBD:
Z L/2
PL
PAB (0.2)
1131.8PAB
=
=
m (c)
EA AB E (176.72 × 10−6 )
E
D = 0.015 + 0.05x m
Z P dx
δ BC =
EA BC
Z
4 PBC 0.2
dx
=
π E 0 (0.015 + 0.05x)2
679.06PBC
4 PBC
(533.33) =
=
π E
E
2.60
RB
(a)
σ max =
RD L
Ebt
Eq. (b) becomes
0.4055RB − 0.2877RD = 0
(c)
Solving Eqs. (a) and (c) simultaneously yields
RB = 0.415P
RD = 0.585P ◭
27
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.62
2.64
Equilibrium:
Pst = Pbr (= P )
Equilibrium: ΣMO = 0: 1.5Pbr + 6Pst = 4 400 × 103 (a)
Compatibility:
|δ st | + |δ br | = ∆ = 1 mm
PL
= 1 mm
EA br
st
P (1200)
P (1200)
+
= 1 mm
3
(200 × 10 ) (675) (83 × 103 ) (1350)
P = 51 kN
PL
EA
+
Compatibility: δ st = 4 δ br
(b)
δP =
8
δ br
3
(c)
PL
PL
= 4
EA st
EA br
Pbr (1.5 × 1000)
Pst (5 × 1000)
= 4
(200 × 103 ) (450)
(84 × 103 ) (1800)
Pst = 0.714Pbr (d)
Eq.(b) becomes
The increases in the stresses are
51 × 103
P
=
= 37.8 MPa (C)
Abr
1350
P
51 × 103
(σ st )increase =
=
= 75.6 MPa (T)
Ast
675
(σ br )increase =
Substituting Eq.(d) into Eq.(a):
Including the initial stresses [24 MPa (C) for the bronze
and 48 MPa (T) for the steel], the final stresses are
1.5Pbr + 6 (0.714Pbr ) = 1600×103 N·mm or Pbr = 276.6 kN
Then Eq.(c) becomes:
8
8 PL
δ P = δ br =
3
3 EA br
8 276.6 × 103 × 1.5 × 1000
= 7.32 mm ◭
=
3
84 × 103 × 1800
σ br = 37.8 + 24 = 61.8 MPa (C) ◭
σ st = 75.6 + 48 = 123.6 MPa (T) ◭
2.63
Equilibrium: ΣMO = 0
2PA + 4PB = 5 (66)
2.65
(a)
Compatibility:
Equilibrium: ΣMO = 0: 1.5Pst + 3.0Pbr = 2P
δ = 2δ A
B
PL
PL
= 2
EA B
EA A
PA (4)
PB (6)
= 2
EA
EA
4
PB = PA (b)
3
(a)
Compatibility: δ br = 2 δst , which becomes
PL
PL
= 2
EA br
EA st
Pbr (2)
Pst (1.5)
= 2
(83 × 109 ) (300 × 10−6 )
(200 × 109 ) (900 × 10−6 )
Pbr = 0.2075Pst (b)
Substituting Eq.(b) into Eq.(a):
4
2PA + 4
PA = 5 (66)
3
PA = 45 kN ◭
4
PB =
(45) = 60 kN ◭
3
28
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.67
Substituting Eq.(b) into Eq.(a):
Equilibrium: ΣMB = 0:
1.5Pst + 3.0 (0.2075Pst) = 2P
12 (PAC sin 45◦ ) + 16 (0.6PAD ) = 16W
which gives
Pst = 0.9423P
8.485PAC + 9.60PAD = 16W
and Pbr = 0.2075 (0.9423P ) = 0.1955P
(a)
If steel governs: Pst = (σ w )st Ast :
900 × 10−6 = 143.3 × 103 N
300 × 10−6 = 107.4 × 103 N
0.9423P = 150 × 106
If bronze governs: Pbr = (σ w )br Abr :
0.1955P = 70 × 106
We see that the bronze governs. The largest load that can
be applied is Pmax = 107.4 kN ◭
2.66
Equilibrium
ΣFy = 0: PA + PB + PC = 600 kN
√
2δ AC
(5/3) δ AD
Compatibility
=
12
16
which gives δ AC = 0.8839δAD
(a)
ΣMC = 0: 6PA + 2PB = 3(600) or PB = 900 − 3PA (b)
Substituting Eq.(b) into Eq.(a) gives:
PA + (900 − 3PA ) + PC = 600 or
Using δ = P L/ (EA), we get
PC = 2PA − 300 (c)
PAD (20)
PAC (16.971)
= 0.8839
EA
EA
PAC = 1.0417PAD (b)
Substituting Eq.(b) into Eq.(a):
8.485 (1.0417PAD ) + 9.60PAD = 16W
PAD = 0.8677W ◭
Compatibility
PAC = 1.0417 (0.8677W )
PAC = 0.9039W ◭
δC − δA
δB − δA
=
4
6
which simplifies to
δ A − 3δB + 2δC = 0
(d)
Substituting Eqs.(b) and (c) into Eq.(d), and using
δ = P L/ (EA), yields
PA (5) 3 (900 − 3PA ) (6)
(2PA − 300) (6)
=0
−
+2
EA
EA
EA
which gives PA = 238.6 kN ◭
PB = 900 − 3 (238.6) = 184.2 kN ◭
PC = 2 (238.6) − 300 = 177.2 kN ◭
29
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.68
Using δ = P L/(EA), Eq.(e) becomes
PL
PL
= 0.8152
EA AB
EA AD
PAD (5000/ cos 20◦ )
PAB (5000/ cos 40◦ )
= 0.8152
Est Ast
Est Ast
PAB = 0.6646PAD (f)
Equilibrium: ΣFx = 0 gives PAB = PAD
ΣFy = 0 gives 2PAB cos 25◦ + PAC = 7.5 kN (a)
Compatibility: δ AB = δ BD and δ AB = δ AC cos 25◦
Using δ = P L/(EA), Eq. (b) becomes
PL
PL
=
cos 25◦
EA AB
EA AC
LAC
PAB = PAC
cos 25◦
LAB
2.75
cos 25◦
PAB = PAC
2.75/ cos 25◦
PAB = 0.8214PAC (c)
Similarly, Eq.(d) becomes
PL
PL
=
cos 20◦
EA AD
EA AC
PAC (5000)
PAD (5000/ cos 20◦ )
=
cos 20◦
(200 × 103 ) (270)
(70 × 103 ) (540)
PAC = 0.7449PAD (g)
(b)
Substituting Eqs.(f) and (g) into Eq.(a):
(0.6646PAD ) cos 40◦ + 0.7449PAD + PAD cos 20◦ = 80 kN
Substituting Eq.(c ) into Eq.(a) gives
Solving gives PAD = 36.5 kN and
PAB = 0.6646 (36.5) = 24.3 kN. The force in the horizontal
strut is then found from Eq.(b):
2 (0.8214PAC ) cos 25◦ + PAC = 7.5 kN
PAE = 34.3 sin 40◦ − 36.5 sin 20◦
PAE = 9.56 kN (C) ◭
PAC = 3.013 kN ◭
PAB = 0.8214 (3.013) = 2.475 kN
Therefore, PAB = PAD = 2.48 kN ◭
2.70
Equilibrium: ΣMA = 0
2Pst + 4.5Pal = 3.5P
2.69
(1)
Equilibrium:
ΣFy = 0 PAB cos 40◦ + PAC + PAD cos 20◦ = 80 kN (a)
ΣFx = 0 PAE = PAB sin 40◦ − PAD sin 20◦
Compatibility: δ st /2 = δ al /4.5
9
δ al = δ st
4
PL
9 PL
=
EA al
4 EA st
9 Pst
Pal
=
Eal
4 Est
9 Eal
9 (70)
Pal =
Pst =
Pst = 0.7875Pst
4 Est
4 (200)
(b)
Compatibility:
δ AB = δ AC cos 40◦
(c) and δ AD = δ AC cos 20◦
(d)
Dividing Eq.(c) by Eq.(d) gives
(2)
Subsitute (2) into (1): 2Pst + 4.5 (0.7875Pst) = 3.5P
◦
δ AB
cos 40
=
= 0.8152 or δ AB = 0.8152δAD
δ AD
cos 20◦
Pst = 0.63134P = 0.63134 (80) = 50.5 kN ◭
(e)
Pal = 0.7875 (50.5) = 39.8 kN ◭
30
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.71
Compatibility:
√
3
13
5 δC
=
δB
2
which gives
Moment equilibriuim:
Equal forces in steel bars
Force equilibrium:
2Pst + Pal = 30 kN (1)
Compatibitlity (by symmetry): δ al = δ st
PL
PL
=
EA al
EA st
Pst
Pal
=
Eal
Est
Est
200
Pst =
Pal =
Pal (2)
Eal
70
6
√ δc
5 13
6 PC LC
PB LB
= √
EA
5 13 EA
√ 6
PB
13 = √ PC (2)
5 13
12
PB =
PC
(b)
65
δB =
Substitute (2) into (1): 2 (2.86Pal ) + Pal = 30 kN
30
= 4.46 N ◭
Solving gives Pal =
6.72
Solving (a) and (b) gives PC = 0.4711W ◭
PB = 0.0870W ◭
2.72
2.74
ΣMB = 0
PA = PC
Initial stress: σ i = P/A = 9600/225 = 42.67 MPa (T)
(1)
(a) Stress due to temperature change is given by:
σ T = α (∆T ) E
σ T = 12 × 10−6 (20) 200 × 103 = 96 MPa (T)
ΣF = 0
PA + PB + PC = 30 kN
(2)
Final stress at −20 ◦ C:
Compatibitlity: (δ B − δ A ) /1 = (δ C − δ A ) /2
σ F = σ i + σ T = 42.67 + 96 = 138.67 MPa (T) ◭
δ C − δ A = 2 (δ B − δ A )
(b) To remove initial stress:
δ − 2δ + δ = 0
C B A
PL
PL
PL
−2
+
= 0
EA C
EA B
EA A
∆T =
42.67
σi
=
= 17.8 ◦C
αE
(12 × 10−6 ) (200 × 103 )
Final temperature:
The lengths L and areas A cancel. Substituting the values
for E:
PA
PB
PC
+
−2
= 0 (3)
70
70
200
Tf = Ti + ∆T = 20 + 17.8 = 37.8 ◭
2.75
Solving (1), (2) and (3) gives:
PA = PC = 11.2 kN and PB = 7.6 kN ◭
P
+ α (∆T ) E
A
5000
130 × 106 =
+ 11.7 × 10−6 (40) 200 × 109
A
A = 137.36 × 10−6 m2
σ =
2.73
2
ΣMA = 0: √ PB (3) + PC (5) − W (2.5) = 0
13
Using A = πd2 /4, the required diameter is
r
r
4A
4 (137.36 × 10−6 )
d =
=
π
π
d = 13.22 × 10−3 m = 13.22 mm ◭
(a)
31
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.76
2.79
(a) Each rail must expand δ T = 3 mm.
Equilibrium: Pst = Pal (= P )
Compatibility: (δ T )al + (δ T )st = (δ P )al + (δ P )st
PL
PL
[α (∆T ) L]al + [α (∆T ) L]st =
+
EA al
EA st
23 × 10−6 (30) (300)+ 12 × 10−6 (30) (450)
δ T = α (∆T ) L
3 × 10−3 = 11.7 × 10
which gives ∆T = 25.64 ◦C
−6
(∆T ) (10)
Therefore, Tf = Ti + ∆T = 15 + 25.64 = 40.64 ◦C ◭
P (450)
P (300)
+
(70 × 103 ) (1350) (200 × 103 ) (1800)
Solving gives P = 83484 N
(b) If there is no clearance:
σ = Eǫ = Eα (∆T ) = 200 × 10
6
=
9
11.7 × 10
σ = 60.0 × 10 Pa = 60.0 MPa (C) ◭
−6
(25.64)
83484
P
=
= 61.8 MPa ◭
Aal
1350
P
83484
σ st =
=
= 46.4 MPa ◭
Ast
1800
σ al =
2.77
Initial stress: σ i = P/A = 9600/270 = 35.56 MPa (T)
2.80
(a) Final stress of 80 MPa:
σ T = 80 − 35.56 = 44.44 MPa (T)
Equilibrium: PA = PB
∆T =
σT
44.44
=
= 18.5 ◦ C (decrease)
Eα
(200 × 103 ) (12 × 10−6 )
Tf = Ti − ∆T = 25 − 18.5 = 6.5 ◦ C ◭
Note that ∆Tx = ∆TA +
(b) To reduce stress to zero, σ T = 35.56 MPa
∆T =
(∆TB − ∆TA ) x
L
Using dδ T = α (∆Tx ) dx, the thermal elongation becomes
Z L
Z L
(∆TB − ∆TA ) x
dx
∆TA +
δT = α
(∆Tx ) dx = α
L
0
0
Z L
Z
α (∆TB − ∆TA ) L
= α (∆TA )
dx +
x dx
L
0
0
αL
α (∆TB − ∆TA ) L
=
(∆TA + ∆TB )
= α (∆TA ) L +
2
2
35.56
σT
=
= 14.8 ◦ C (increase)
Eα
(200 × 103 ) (12 × 10−6 )
Tf = Ti + ∆T = 25 + 14.8 = 39.8 ◦ C ◭
2.78
Compatibility: δ T = δ P :
δT = δP + ∆
σL
+∆
α (∆T ) L =
E
σL
αL
(∆TA + ∆TB ) =
2
E
αE
(∆TA + ∆TB )
σ =
2
Solving for the temperature change:
∆
1 σ
+
∆T =
α E
L
1
30 × 106
2.2 × 10−3
∆T =
+
(18 × 10−6 ) 80 × 109
3
◦
∆T = 61.6 C (increase)
Q.E.D.
2.81
Equilibrium:
PA = 2PC
σ A AA = 2σ C AC
σ A (300) = 2σ C (1200)
σ A = 8σ C
Tf = Ti + ∆T = −20 + 61.6 = 41.6 ◦ C ◭
32
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Initial stresses are (σ br )i = 12 MPa (C)
and (σ st )i = 2 (12) = 24 MPa (T).
Compatibility:
Compatibility (relax support for left rod):
(δ A )T = (δ A )P + 0.5 (δ C )P
σ A LA (0.5) σ C LC
+
E
E
Using σ A = 8σ C , we obtain
(δ st )T + (δ st )P = (δ br )T − (δ br )P
σL
σL
= [α (∆T ) L]br −
[α (∆T ) L]st +
E st
E br
[α (∆T ) L]A =
8σ C (0.9)
(0.5) σ C (1.2)
11.7 × 10−6 (40) (0.9) =
+
9
200 × 10
200 × 109
6
σ C = 10.80 × 10 Pa
σ C = 10.80 MPa (T) ◭
Using σ st = 2σ br , and Lst = Lbr , we get
2σ br
12 × 10−6 (40)+
200 × 103
= 20 × 10−6 (40) −
σ A = 8 (10.80) = 86.4 MPa (T) ◭
σ br
84 × 103
The solution is σ br = 14.6 MPa (C) and
σ st = 2 (14.6) = 29.2 psi (T). Including the initial stresses,
the final stresses are
2.82
σ br = 14.6 + 12 = 26.6 MPa (C) ◭
Equilibrium:
σ st = 29.2 + 24 = 53.2 MPa (T) ◭
Pal = 2Pcu
σ al Aal = 2σ cu Acu
σ al (400) = 2σ cu (500)
σ al = 2.5σcu
2.84
Compatibility:
Equilibrium: ΣMO = 0
1.0Pst + 4.0Pbr = 2.5 80 × 103 = 200 × 103 N
Pbr = 50 × 103 − 0.25Pst
σ br Abr = 50 × 103 − 0.25σst Ast
σ br 1400 × 10−6 = 50 × 103
−0.25 50 × 106 300 × 10−6
(δ cu )T + (δ cu )P + ∆ = (δ al )T − (δ al )P
σL
σL
+ ∆ = [α (∆T ) L]al −
[α (∆T ) L]cu +
E cu
E al
Using σ al = 2.5σcu , we obtain
σ br = 33.04 × 106 Pa
σ cu (0.750)
16.8 × 10−6 (85) (0.750)+
+(0.00018)
120 × 109
(2.5σcu ) (0.750)
= 23.1 × 10−6 (85) (0.750) −
70 × 109
The solution of this equation gives
Compatibility:
6
σ cu = 6.709 × 10 Pa = 6.71 MPa (T) ◭
σ al = 2.5 (6.709) = 16.77 MPa (C) ◭
(δ br )T + (δ br )P = 4 [(δ st )T + (δ st )P ]
σL
σL
= 4 [α (∆T ) L]st +
[α (∆T ) L]br +
E br
E st
6
33.04 × 10 (3)
18.9 × 10−6 (∆T ) (3) +
83 × 109
"
#
6
50
×
10
(1.5)
= 4 11.7 × 10−6 (∆T ) (1.5) +
200 × 109
2.83
Equilibrium:
Pst = Pbr
σ st Ast = σ br Abr
Solving this equation gives
σ st (0.75) = σ br (1.5)
σ st = 2σ br
∆T = −22.7 ◦C (temperature decrease) ◭
33
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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2.85
[α (∆T ) L]al +[α (∆T ) L]st +
Equilibrium: ΣMO = 0
PL
EA
250 kN al
=
1.0Pst + 4Pbr = 2.5 120 × 103 = 300 × 103 N
Pst = 300 × 103 − 4Pbr
σ st Ast = 300 × 103 − 4σbr Abr
σ st 300 × 10−6 = 300 × 103 − 4σbr 1400 × 10−6
σ st = 1000 × 106 − 18.67σbr (1)
23.1 × 10
−6
PL
EA
al
+
PL
EA
st
(20) (450)
+ 12 × 10
=
−6
250 × 103 (450)
(20) (300) +
(70 × 103 ) (1800)
Pst (450)
Pst (300)
+
3
(70 × 10 ) (1800) (200 × 103 ) (2700)
The solution to this equation is Pst = 284.2 × 103 N
284.2 × 103
Pst
=
= 105.3 MPa (C) ◭
Ast
2700
Pal
(284.2 − 250) × 103
σ al =
=
Aal
1800
σ al = = 19 MPa (C) ◭
σ st =
Compatibility:
(δ br )T + (δ br )P = 4 [(δ st )T + (δ st )P ]
σL
σL
= 4 [α (∆T ) L]st +
[α (∆T ) L]br +
E br
E st
Substituting the numerical values and the expression for
σ st from Eq.(1), we obtain
2.87
σ br (3)
18.9 × 10−6 (20) (3) +
83 × 109
Equilibrium: Pal = Pst − 250 × 103 N
= 4 11.7 × 10−6 (20) (1.5)
#
"
1000 × 106 − 18.67σbr (1.5)
+4
200 × 109
For equal stresses:
σAal = σAst − 250 × 103
σ (1800) = σ (2700) − 250 × 103 which gives
The solution of this equation gives
σ br = 50.77 × 106 Pa = 50.8 MPa ◭
σ st = 1000 × 106 − 18.67 50.77 × 106
σ = 277.8 MPa
σ st = 52.12 × 106 Pa = 52.1 MPa ◭
Compatibility (relaxing right support):
(δ al )T + (δ st )T + (δ al )250 kN = (δ al )P + (δ st )P
PL
[α (∆T ) L]al +[α (∆T ) L]st +
EA 250 kN al
σL
σL
+
=
E al
E st
−6
−6
(∆T ) (300)
(∆T ) (450)+ 12 × 10
23.1 × 10
3
250 × 10 (450)
(277.8) (450) (277.8) (300)
+
=
+
3
(70 × 10 ) (1800)
(70 × 103 )
(200 × 103 )
Solving gives ∆T = 93.6 ◦C. The final temperature
becomes
2.86
Equilibrium: Pal = Pst − 250 × 103 N
Compatibility (relaxing right support):
Tf = Ti + ∆T = 20 + 93.6 = 113.6 ◦ C ◭
(δ al )T + (δ st )T + (δ al )250 kN = (δ al )P + (δ st )P
34
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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δD
δC
=
1.4δC = 0.8δD
0.8
1.4 PD L
+ α ∆T L
1.4α ∆T L = 0.8
EA
PD = 0.75α ∆T EA
2.88
Equilibrium:
ΣFx = 0: PAD = PCD
ΣFy = 0:
(PAD + PCD ) cos 60◦ + PBD = 0
= 0.75(12 × 10−6 )(80)(200 × 109 )(250 × 10−6 )
= 36.0 × 103 N
Solving yields PBD = −2PAD cos 60◦ = −PAD , which,
because of equal areas, gives σ BD = −σ AD .
P = 0.4667(36.0 × 103 ) = 16.80 × 103 N = 16.80 kN ◭
2.90
Equilibrium:
Pbr = Pal = Pst = P
Compatibility: δ AD = δ BD sin 30◦ = 0.5δBD , which
becomes
Compatibility:
L
EA
(25) 10−6 [(19.0) (0.8) + (23.0) (0.5) + (11.7) (0.4)]
P
0.8
0.5
0.4
=
+
+
(109 ) (10−6 ) (83) (2000) (70) (1400) (200) (800)
Solving gives 63.16 × 103 N.
∆T ΣαL = P Σ
(δ AD )T + (δ AD )P = 0.5 [(δ BD )T + (δ BD )P ]
σL
σL
α (∆T ) L +
= 0.5 α (∆T ) L +
E
E
AD
BD
Using σ BD = −σ AD from above, we obtain
σ AD (3)
11.7 × 10−6 (80) (3) +
200 × 109
(−σ AD ) (1.5)
= 0.5 11.7 × 10−6 (80) (1.5) +
200 × 109
P
63.16 × 103
=
Abr
2000 × 10−6
= 31.6 × 106 Pa = 31.6 MPa (T) ◭
63.16 × 103
P
=
σ al =
Aal
1400 × 10−6
= 45.1 × 106 Pa = 45.1MPa (T) ◭
P
63.16 × 103
σ st =
=
Ast
800 × 10−6
= 79.0 × 106 Pa = 79.0 MPa (T) ◭
σ br =
The solution yields:
σ AD = −112.3 × 106 Pa
σ AD = σ CD = −112.3 MPa = 112.3 MPa (C) ◭
σ BD = −σ AD = − (−112.3) = 112.3 MPa (T) ◭
2.91
2.89
ΣMO = 0
3Pal = 4Pst
Pst
3
=
(1)
Pal
4
Compatibility: For each bar, deformation due to
temperature equals deformation due to load.
From FBD:
0.8 m
3m
A
B
P
RB
C
PC = 0
ΣMB = 0
+
0.6 m
D
PD
1.4PD − 3P = 0
(δ T )al = (δ P )al
PL
(α ∆T L)al =
EA al
Pal = αal ∆T Eal Aal
Similarly, Pst = αst ∆T Est Ast
P = 0.4667PD
Compatibility:
0.8 m
B
0.6 m
δC
δD
C
D
(2)
(3)
Substituting (2) and (3) into (1) yields
35
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Substituting Eq. (a) into Eq. (b) and cancelling L, we get
1
1
+
= ∆T (αbr − αst )
Pst
Est Ast
Ebr Abr
1
1
= 90(19 − 12)×10−6
+
Pst
200×103 (675) (84×103 )(1350)
Pst = 38827.2 N
Pst
3
αst ∆T Est Ast
=
=
Pal
4
αal ∆T Eal Aal
(23) (70)
3 αal Eal
3
Ast
= 0.516 ◭
=
=
Aal
4 αst Est
4 (11.7) (200)
2.92
Pst
38827.2
=
= 57.5 MPa ◭
Ast
675
Pbr
−38827.2
σ br =
=
= −28.8 MPa ◭
Abr
1350
σ st =
Equilibrium:
Pal = Pbr = P = 2Pst
2.94
δ=
P (0.4L)
PL
P (0.6L)
+
= 1.6
◭
E(bt)
E(0.4bt)
Ebt
Compatibility:
2.95
(δ T )st + (δ P )st = (δ T )al + (δ T )br − (δ P )al − (δ P )br
PL
(α ∆T L)st +
EA st
PL
PL
−
= (α ∆T L)al + (α ∆T L)br −
EA al
EA br
δ=
P L/ (EA)
(30000) (1.5) − (5000) (2.5) + (20000) (2)
(70 × 109 ) (540 × 10−6 )
δ = 1.92 mm ◭
δ =
Lal = 90 mm Lbr = 120 mm Lst = 270 mm ∆T = 80◦
12 × 10−6 (80) (270) +
P
Pst (270)
(200 × 103 ) (675)
= 2.3 × 10−6 (80) (90) + 19 × 10−6 (80) (120)
−
2Pst (120)
2Pst (90)
−
(70 × 103 ) (1800) (84 × 103 ) (2700)
2.96
Solving gives Pst = 182425.4 N.
σ st =
182425.4
Pst
=
= 270.3 MPa (T) ◭
Ast
675
Equilibrium:
Pal (1) = Pst (2)
Pal /Pst = 2
Compatibility : δ al = δ st
2.93
Pal Lal
Pst Lst
=
Eal Aal
Est Ast
Aal
Est
Lal
Pal
=
Ast
Pst
Eal
Lst
2
200
Aal
= 3.8 ◭
= (2)
Ast
70
3
Equilibrium:
Pst + Pbr = 0
Pbr = −Pst
(a)
Compatibility:
δ st = δ br
Pst L
Pbr L
+ αst L ∆T =
+ αbr L ∆T (b)
Est Ast
Ebr Abr
36
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2.97
Lengths of the bars: LAC = LBC = 3500/ sin 60◦ =
4041.5 mm.
Elongation of bar AC:
(25000) (4041.5)
PL
h
i = 1.27 mm
=
δ AC =
2
EA AC
(200 × 103 ) π (22.5) /4
Equilibrium:
P = W = γ × (volume)
A = πx2 /4
Volume (cone):
(1/3)(area of base x height)
V =
1
3
2
πx
4
(L + y)−
1
3
πD
4
2
(a) Compatibility (using above sketch): δ AC = δ BC , which
gives
L
ABC =
2
1 πD2
1 πx
1
V
(L
+
y)
−
L
=
A
(πx2 /4) 3
4
3
4
1
D2 L
=
(L + y) − 2
3
x
=
BC
(25000) (4041.5)
= 6629.8 mm2 ◭
(1.27) (12 × 103 )
δ AC
1.27
=
= 2.54 mm ◭
◦
cos 60
cos 60◦
2.99
From the FBD:
L
L
Pb{Ps
A
ΣF = 0
Z L
γ
V
W dy
=
dy
EA
E
0 A
0
Z Lh
i
γ
−2
=
dy
(L + y) − L3 (L + y)
3E 0
"
#L
2
γ (L + y)
−1
3
=
+ L (L + y)
3E
2
"
# 0
2
3
2
L
L
γL2
γ (2L)
+
−
− L2 =
◭
=
3E
2
2L
2
3E
Compatibility:
P
B
+→
(δ s )AB = (δ b )AB
Ps L
Pb L
=
(EA)s
(EA)b
C
P's
Pb + Ps + Ps′ − P = 0
δ AB + δ BC = 0
P ′L
−Ps L
= 0
+ s
(EA)s
(EA)s
Pb =
(a)
Ps′ = Ps
(EA)b
1
Ps = Ps
(EA)s
3
From Eq. (a):
1
Ps + Ps + Ps − P = 0
3
3
1
Ps = P
Pb = P ◭
7
7
2.98
Equilibrium:
ΣFx = 0
ΣFy = 0
∆C =
Elongation (δ = P L/EA):
δ =
PL
δE
(b) The displacement of C (using above sketch):
Substituting x = D (L + y) /L:
"
#
D2 L
V
1
(L + y) −
=
2
A
3
[D (L + y) /L]
#
"
L3
1
(L + y) −
=
3
(L + y)2
Z L
PBC cos 30◦ − PAC cos 30◦ = 0
PBC sin 30◦ + PAC sin 30◦ = 25000 N
(a)
(b)
2.100
Solving Eqs.(a) and (b) gives: PBC = PAC = 25000 N
P
A
400 × 103
= −
π(0.04)2
σz = −
= −79.58 × 106 Pa
37
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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ν
∆
= − σz
r
E
rν
P
rν
−
∆ = − σz = −
E
E
A
(0.040) (1/3)
= −
−79.58 × 106
9
70 × 10
= 0.01516 × 10−3 m = 0.01516 mm ◭
the total load. For W = 4000 N, the load in the shortest
wire is
ǫx =
4000 − 2399
= 3199.5 N
2
3199.5
P1
=
= 71.1 (T) ◭
σ1 =
A
45
P1 = 2399 +
2.101
2.103
(a) Using Hooke’s law:
From FBD:
ǫx =
1
[σ x − ν (σ y + σ z )]
E
Solving: p = −
gives 0 =
1
[σ x − ν (−2p)]
E
Ps
σx
(−30)
=−
= 45.46 MPa ◭
2ν
2 (0.33)
ΣF = 0
(b) Hooke’s law (σ x = −30 MPa, σy = σ z = −45.46 MPa):
Compatibility:
1
ǫy =
[σ y − ν(σ z + σ x )]
E
1
{(−45.46) − (0.33) [(−45.46) + (−30)]}
=
70 × 103
= −293.7 × 10−6 ◭
Find P1 and P2 to bring all wires to the same length.
δAE
L
(0.01) (45) 200 × 103
=
37.51
= 2399 N
P2 =
′
=
(b)
Solution of Eqs. (a) and (b) is
Ps = 4660 N (T)
′
1500 − P1 − P2
12000 − 4798 − 2399
P3 =
=
= 1601 N
3
3
1601
P3
=
= 35.6 MPa (T) ◭
σ3 =
A
45
Pa = 9310 N (C) ◭
2.104
(b) The force required to bring wires 1 and 2 to the same
length is
′
8.842Ps + 216 × 103 = −2.398Pa + 279.5 × 103
8.842Ps + 2.398Pa = 63.5 × 103
(a) When all three wires have the same length, each
carries one-third of the load. When W = 12000 N, the load
in the longest wire is
′
(a)
(αL)a ∆T = (23 × 10−6 )(0.135)(90) = 279.5 × 10−6 m
P1 = 2P2 = 2 (2399) = 4798 N
′
Pa − 2Ps = 0
δs = δa
−P L
+ (αL)s ∆T =
+ (αL)a ∆T
EA a
s
PL
EA
L
EA
+↑
0.2
= 8.842 × 10−9 m/N
π
9
2
(200 × 10 ) (0.012)
s
4
L
0.135
=
π
EA a
(70 × 109 ) (0.0682 − 0.0602)
4
= 2.398 × 10−9 m/N
(αL)s ∆T = 12 × 10−6 (0.2)(90) = 216 × 10−6 m
2.102
′
Ps
Pa
From FBD:
0.8 m
3m
′
0.6 m
P = P1 + P2 = 4798 + 2399 = 7179 N
A
B
C
D
Therefore, wire 3 remains slack for W = 4000 N. The force
required to bring wire 1 to the same length as wire 2 is
2399 N. After that, wires 1 and 2 each carry one-half of
P
RB P
C
PD
ΣMB = 0
+
0.8PC + 1.4PD − 3P = 0
(a)
38
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Compatibility:
0.8 m
0.6 m
B
C
δC
δD
D
PC L
PD L
=
0.8EA
1.4EA
δD
δC
=
0.8
1.4
1.4PC = 0.8PD
(b)
Compatibility: δ A = 2δC :
Solution of Eqs. (a) and (b) is
PC = 0.923P
σ A LA
σ C LC
= 2
E
E
1.5
LC
σC
σC = 2
σA = 2
LA
2
σ A = 1.5σC (2)
PD = 1.615P ◭
2.105
Substituting Eq.(2) into Eq.(1) and solving gives
σ C = (711.1) − 3 (1.5σ C ) = 129.3 MPa (T) ◭
(a) Compatibility:
σ A = 1.5σ C = 1.5 (129.3) = 194 MPa (T) ◭
2δ BC = δ DE
PBC LBC
PDE LDE
2
=
EA
EA
PDE (2LBC )
=
EA
gives PBC = PDE
2.107
The equilibrium equation, Eq.(1) from the solution to
Prob. 2.106, is still valid:
Equilibrium: ΣMA = 0
PDE (3)+PBC (1.5)
σ C = (711.1) − 3σ A
= 200 (3 tan 35◦ )
The stresses are known to be equal. Letting σ C = σ A = σ
yields
σ = (711.1) − 3σ or σ = 177.8 MPa
PDE (3)+PDE (1.5)
= 200 (3 tan 35◦ )
The compatibility equation, δ A = 2δ C , from the solution
to Prob. 2.106,
is also still valid:
σ C LC
σ A LA
= 2 α (∆T ) LC +
α (∆T ) LA +
E
E
1
α (∆T ) (LA − 2LC ) =
(2σ C LC − σ A LA )
E
σ
(2LC − LA )
α (∆T ) (LA − 2LC ) =
E
PDE = 93.36 kN ◭
(b)
PDE LDE
tan 35◦
δ v = δ DE tan 35◦ =
EA
93.36 × 103 (3 tan 35◦ )
=
tan 35◦
(70 × 109 ) (50 × 10−6 )
= 0.0392 m = 39.2 mm ◭
Solving for ∆T gives
2.106
∆T = −
Equilibrium: ΣMB = 0
σ A AA (2) + σ C AC (1) =
σ C AC =
σ C (450) =
σ
177.8
=−
= −74.1 ◦ C ◭
Eα
(200 × 103 ) (12 × 10−6 )
160 × 103 (2)
320 × 103 − 2σ A AA
320 × 103 − 2σ A (675)
σ C = 711.1 − 3σ A
(1)
39
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2.108
Use Simpson’s rule to evaluate the integral:
1
1
1
1
1
∆x
+
+4·
+
+
·
I :=
3 A1
A
A2
A4
A6
7
1
1
+2 ·
+
A3
A5
P
·I
δ = 0.1400 mm
δ :=
E
Equilibrium: ΣMA = 0 (P = σA)
3σ cu Acu = 5σ br Abr
5 (90) (1200)
5 σ br Abr
=
= 120 MPa
σ cu =
3
Acu
3
(1500)
C2.2 MathCad worksheet
Given:




 10 






 20 




 24 
d := 24 · mm






 23 






 20 


16
Compatibility: (Relax support for bronze rod; assume
temperature decrease. Apply force P to reattach bronze
rod.)
5
5
(δ cu )T + (δ br )T = (δ br )P + (δ cu )P
3
3 5 σL
5
σL
+
[α (∆T ) L]cu + [α (∆T ) L]br =
3
E br 3 E cu
5
16.8 × 10−6 (∆T ) (3) + 18.7 × 10−6 (∆T ) (2)
3
120 × 106 (3)
90 × 106 (2)
5
+
=
(100 × 109 )
3
(120 × 109 )
t := 10 · mm
∆x := 80 · mm
E := 70 · 109 · Pa
P := 6 · 103 · N
Computations:
A := t · d
Cross-sectional areas
Z L
P
1
Formula for elongation (for information
δ :=
·
dx
only; its evaluation is suppressed)
E 0 A
Evaluate the integral by Simpson’s rule:
1
1
1
1
1
∆x
+
+4·
+
+
·
I :=
3 A1
A7
A2
A4
A6
1
1
+2 ·
+
A3
A5
P
δ :=
·I
δ = 0.2056 mm
E
∆T = 56.0 ◦ C (decrease) ◭
C2.1 MathCad worksheet
C2.3 MathCad worksheet for
Part (a)
Given:


10 






20 







 24 

d := 24 · mm



 23 






 20 





16
Given:
∆x := 80 · mm
E := 70 · 109 · Pa
P := 6 · 103 · N
L := 500 mm
E := I (Value of E not given).
x
x2
d(x) := (25 · mm) · 1 + 3.8 · − 3.6 · 2
L
L
Computations:
Computations:
πd2
A :=
4
P
δ :=
·
E
Z L
0
A(x) :=
Cross-sectional areas
k(E) := R
1
dx Formula for elongation (for information
A
only; its evaluation is suppressed)
E
1
L
0 A(x) dx
π · d(x)2
4
k(E)
= 2.5737 × 10−3 m
E
40
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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C2.3 MathCad worksheet for
Part (b)
Compatibility (due to symmetry, displacement of D is
vertical):
δ CD = δ AD = δ BD sin α
PCD (1.5/ sin α)
PAD (1.5/ sin α)
PBD (1.5)
=
=
sin α
EA
EA
EA
PCD = PAD = PBD sin2 α
Given:
L := 200 mm
E := I
(24 · mm − 0.05 · x) if x ≤ 120 · mm
(18 · mm) otherwise
d(x) :=
The solution is
Computations:
PAD = PCD
π · d(x)2
4
k(E)
= 1.4969 × 10−3 m
E
E
1
L
dx
0 A(x)
PBD =
P cos θ
1 + 2 sin3 α
Superposition
A(x) :=
k(E) := R
P cos θ sin2 α
1 + 2 sin3 α
P sin θ P cos θ sin2 α
+
2 cos α
1 + 2 sin3 α
P sin θ P cos θ sin2 α
+
PCD =
2 cos α
1 + 2 sin3 α
P cos θ
PBD =
1 + 2 sin3 α
PAD =
C2.4
Forces caused by horizontal component of P
C2.4 MathCad worksheet for
Part (a)
Given:
Equilibrium:
ΣFx = 0
ΣFy = 0
P := 10000 · N
(PCD − PAD ) cos α − P sin θ = 0
Computations:
(PCD + PAD ) sin α + PBD = 0
P · sin(θ) P · cos(θ) · sin(α)2
+
2 · cos(α)
1 + 2 · sin(α)3
−P · sin(θ) P · cos(θ) · sin(α)2
Axial forces
+
PCD (θ) :=
2 · cos(α)
1 + 2 · sin(α)3
P · cos(θ)
PBD (θ) :=
1 + 2 · sin(α)3
Compatibility (due to skew-symmetry, displacement of D
is horizontal):
δ BD = 0
PAD (θ) :=
PBD (1.5)
=0
EA
The solution is
PAD =
P sin θ
2 cos α
PCD =
P sin θ
2 cos α
α := 30 · deg
PBD = 0
θ := −90·deg, −88·deg..90·deg
Forces caused by vertical component of P
Plotting range and interval
1.104
Axial Force (N)
PBD
Equilibrium:
ΣFx = 0
ΣFy = 0
(PCD − PAD ) cos α = 0
(PCD + PAD ) sin α + PBD − P cos θ = 0
5000
0
PAD
–5000
–1.104
–90
–60
–30
PCD
0
30
Theta (deg)
60
90
41
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C2.4 MathCad worksheet for Part
(b)
C2.5 MathCad worksheet for Part
(b)
Given:
f (θ) :=
P := 10000 · N
α := 60 · deg
Computations:
θ := 20 · deg, 21 · deg..85 · deg
2
P · sin(θ) P · cos(θ) · sin(α)
+
2 · cos(α)
1 + 2 · sin(α)3
−P · sin(θ) P · cos(θ) · sin(α)2
Axial forces
+
PCD (θ) :=
2 · cos(α)
1 + 2 · sin(α)3
P · cos(θ)
PBD (θ) :=
1 + 2 · sin(α)3
PAD (θ) :=
θ := −90·deg, −88·deg..90·deg
10
F (theta)
8
6
4
Plotting range and interval
2
20
Axial Force (N)
PCD
30
40
PAD
5000
Plotting range and interval
12
1.5·104
1.104
1
sin(θ)2 · cos(θ)
θ := 55 · deg
PBD
–5000
70
80
90
Initial value used in the solution
d
f (θ) = 0
dθ
Given:
0
50
60
Theta (deg)
θopt := Find(θ)
θ opt := 54.74 deg
–1.104
–1.5·104
–90
–60
–30
0
30
Theta (deg)
60
90
C2.6
Equilibrium:
Pal + Pst = 0
C2.5
Pst = P
Pal = −P
Compatibility:
δ al = δ st
−P L
P (L − b)
Pb
+ αal L∆T =
+
+ αst L∆T
Eal A0
Est A0
Est A
Equilibrium:
ΣFy = 0 PAC sin θ − P = 0
PAC =
P
sin θ
b
L
L−b
= (αal − αst )L∆T
+
+
Est A0
Est A Eal A0
P 1 − b/L A
b/L
1 A
= (αal − αst )∆T
+
+
A
Est A0
Est
Eal A0
Compatibility:
P
PAC LAC
(P/ sin θ)(b/ cos θ)
δ AC
=
=
sin θ
EA sin θ
EA sin θ
1
Pb
=
EA sin2 θ cos θ
1
f (θ) =
◭
sin2 θ cos θ
∆C =
42
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Letting P/A = σ yp and solving for A/A0 :
Est
b/L
(αal − αst )∆T − b/L
(αal − αst )∆T −
σ yp
A
σ
E
yp
st
= =
Est
1
1 − b/L
A0
1 − b/L +
σ yp
+
Eal
Est
Eal
C2.6 MathCad worksheet
Given:
αal := 23 · 10−6
αst := 12 · 10−6
9
Eal = 70 · 10 · Pa Est := 200 · 109 · Pa
∆T := 90
σ yp := 210 · 106 · Pa
Computations:
Let x = b/L and y = A/A0
Est
(αal − αst ) · ∆T − x
σ yp
y(x) :=
Est
1−x+
Eal
x : = 0, 0.02.. 1
Plotting range and increment
0.3
0.25
A/A0
0.2
0.15
0.1
0.05
0
–0.05
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
b/L
1
43
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Chapter 3
3.5
3.1
16 45 × 106
16T
τ max =
= 229.2 MPa ◭
=
3
πd3
π (100)
45 × 106 (1 × 1000)
TL
180
h
i
θ=
= 3.28◦ ◭
=
4
GJ
π
3
(80 × 10 ) π (100) /32
For θ = 4(π/180) rad,
θ =
TL
τJ L
τL
=
=
JG
r JG
rG
(48) (8 × 1000)
4π
= d
3
180
2 (80 × 10 )
3.2
d = 137.5 mm ◭
(a) J = πd4 /32 = T L/ (Gθ) becomes
s
r
32 (12 × 103 ) (6)
4 32T L
d =
= 4
π
πGθ
π (83 × 109 ) 3 × 180
3.6
= 0.1140 m = 114.0 mm ◭
16 12 × 103
16T
=
(b) τ max =
3
πd3
π (0.1140)
Assuming that stress governs:
τw =
120 × 106 =
16(0.3P )
π(0.015)3
P = 265 N
Assuming that twist governs:
= 41.3 × 106 Pa = 41.3 MPa ◭
θ =
3.3
T L 32
TL
=
GJ
G πd4
5
32
π
(0.3P )(0.5)
=
180
80 × 109 π(0.015)4
P = 231 N
i
h
π
π
4
D4 − d4 = D4 1 − (2/3) = 7.878 × 10−2 D4
J=
32
32
Tr
(a)
J =
τ
4
250 × 106 (D/2)
−2
D =
7.878 × 10
48
The maximum allowable force is P = 231 N, governed by
the angle of twist ◭
3.7
For shaft BC:
D = 321 mm ◭
6
(b) θ =
16T
πd3
J=
250 × 10 (8 × 1000)
TL
=
4
JG
(7.878 × 10−2 ) (321) (80 × 103 )
= 0.02989 rad = 1.71
◦
B
θ
360 m
m
◭
A
60 mm
The angle of twist of the shaft is
3.4
60
= 0.167 45 rad
360
The torque that produces this twist is
θ = sin−1
From T = τ max J/r, we see that the ratio of the maximum
torques that can be carried by the hollow and solid shafts
is given by
Jhollow
Thollow
=
Tsolid
Jsolid
h
i
4
π D4 − (D/2) /32
1
15
= 1−
=
=
πD4 /32
16
16
π(37.5)4
πd4
=
= 194144.4 mm4
32
32
T =
(80 × 103 )(194144.4)
GJ
θ=
(0.16745) = 1156 N · m
L
2250
∴ P =
Q.E.D.
1156
= 3211 N
0.360
τ max =
Td
(1156 × 103 )(37.5)
=
= 111.6 MPa ◭
2J
2(194144.4)
45
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3.8
3.11
i
π h
Jhollow =
(0.100)4 − (0.070)4 = 7.460 × 10−6 m4
32
π
4
(0.070) = 2.357 × 10−6 m4
Jsolid =
32
Hollow shaft:
70 × 106 7.460 × 10−6
τJ
T =
=
= 10.444 × 103 N · m
r
0.050
Solid shaft:
70 × 106 2.357 × 10−6
τJ
T =
=
= 4.714 × 103 N · m
r
0.035
Angle of twist: θ = 2.5◦ = 2.5(π/180) rad
Equilibrium: TBC = 1200 N·m
TAB = 1200 − 750 = 450 N·m
If maximum stress governs:
d=
16T
πτ max
1/3
16 (1200)
=
π (60 × 106 )
1/3
= 0.0467 m
If angle of rotation governs: J = πd4 /32 = ΣT L/ (Gθ)
which gives
1/4
32 ΣT L
d =
π Gθ
1/4
32 (1200) (2.5) + (450) (2.5)
= 0.0519 m
=
π
(83 × 109 ) (4π/180)
T L
Σ
G J
2
1.5
T
2.5π
+
=
180
83 × 109 7.460 × 10−6 2.357 × 10−6
θ =
Comparing the above results, the required diameter is
d = 0.0519 m = 51.9 mm ◭
T = 4.00 × 103 N · m
3.12
Comparing the above three values, Tmax = 4.00 kN·m ◭
π
4
(0.100) = 9.817 × 10−6 m4
32
π
4
Jbr = Jst =
(0.075) = 3.106 × 10−6 m4
32
Jal =
3.9
π (0.120)3 100 × 106
πd3 τ
=
= 33 929 N · m
T =
16
16
T XL
θ =
J
G
4.5
3.5
180
33 929
+
θ =
4
83 × 109
π
π (0.120) /32 35 × 109
(a) Shear stresses: Tal = 5.5 kN·m; Tbr = Tst = 1.5 kN·m;
τ = 16T /πd3
16 5.5 × 103
= 28.0×106 Pa = 28.0 MPa ◭
τ al =
3
π (0.100)
16 1.5 × 103
τ br = τ st =
= 18.11 × 106 Pa = 18.11 MPa ◭
3
π (0.075)
θ = 16.30◦ ◭
3.10
(b) Angle of rotation:
Maximum shear stress:
The maximum safe torque for τ = 14 GPa is determined
by the smaller diameter:
3
π (0.060) 12 × 106
πd3 τ
T =
=
= 509 N · m
16
16
Angle of twist:
T 32
L
TL
T
L
=
Σ
θ=Σ
= Σ
GJ
G
J
G π
d4
θ = Σ
θ =
TL
GJ
(1500) (2)
(5500) (3)
(1500) (1.5)
+
+
(3.106) (35) (3.106) (83) (9.817) (28)
1
180
×
(10−6 ) (109 )
π
θ = 5.29◦ ◭
For θ = 4◦ = 4 (π/180) rad,
#
"
4π
3
4
32
T
=
4 +
4
180
83 × 109 π
(0.060)
(0.080)
3.13
Jst =
T = 1490 N · m
Comparing the above two values, Tmax = 509 N·m ◭
π (60)4
π (45)4
= 1272345 mm4 Jal =
= 402578 mm4
32
32
Tst = −T
Tal = T
46
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Stress in steel:
3.16
(84) (1272345)
τJ
=
= 3562.6 N · m
r
30
Stress in aluminum:
(45) (402578)
τJ
=
= 805.2 N · m
T =
r
22.5
◦
Angle of rotation: θ = 8 = 8(π/180) rad
X TL
θ =
GJ
8π
T (1.5 × 1000)
T (1 × 1000)
= −
+
3
180
(80 × 10 ) (1272345) (28 × 103 ) (402578)
Torque per unit length: T0 /L
Torque at distance x from free end: T = T0 x/L
Z L
T
θ=
GJ
0
Z L
T0
=
x dx
GJL 0
Comparing the above three values, Tmax = 805.2 N · m ◭
3.17
3.14
Equilibrium: Z
|T | =
=
T = 1887.4 N · m
T0 L
◭
2GJ
Tx = TA −
x
tA
= TA −
L
= TA −
The torques shown above are in N·m, and the lengths are
in meters. Using the right-hand rule, the torques are:
TDC = −800 N·m, TCB = 300 N·m, TBA = −600 N·m.
Z L
(L − x) dx
3.18
RL
0
Z L
0
TA =
3.15
[Tx / (GJ)] dx = 0, which gives
x2
dx = 0
T A − tA x −
2L
tA L 2
tA L 2
+
=0
2
6
tA L
(200) (1.5)
=
= 100 N · m
3
3
16 (100)
16TA
6
=
3 = 32.6 × 10 Pa = 32.6 MPa ◭
πd3
π (0.025)
Using T = τ max J/r, the angle of twist of a shaft can be
expressed as follows:
τ max J L
τ max L
TL
=
=
θ=
GJ
r GJ
rG
Using this expression, the compatibility condition,
θ br = θst , becomes
τ max L
τ max L
=
rG
rG
br
st
60 − 24
d − 24
=
2000
x
d = 0.018 x+24 mm
πd4
J =
32
4
π (0.018 x + 24)
mm4
32
0
Tx dx =
TA L −
τ max =
Z 60
0
x
tA
x2
x2
Lx −
= T A − tA x −
L
2 0
2L
0
The negative sign indicates that the rotation is clockwise
when viewed from D toward A.
TL
32T
θB =
=
GJ
πG
Z x
Compatibility: θBA =
ΣT L
θDA =
GJ
(−800) (2) + (300) (3) + (−600) (2)
=
4
(28 × 109 ) π (0.050) /32
180
= −6.34◦ ◭
= −0.11059 rad = −0.11059
π
=
t dx
0
(τ max )br (2)
(τ max )st (1.5)
=
(0.075/2) (35)
(0.050/2) (83)
dx
4
(0.018 x + 24)
(τ max )br = 0.4744 (τ max )st
3
32 30 × 10 (0.00125)
=
= 0.00545 rad = 0.3◦ ◭
π (70 × 103 )
If steel governs, (τ max )br = 0.4744 (τ max )st = 0.4744 (80) =
37.95 MPa < 60 MPa. Therefore steel governs.
47
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3.21
Equilibrium gives: T = Tst + Tbr , which yields
π X 3
d τ max
T =
16
i
π h
3
3
(0.075) 37.95 × 106 + (0.050) 80 × 106
=
16
= 5110 N · m = 5.11 kN · m ◭
Relax support at A. Let θ1 be the angle of rotation caused
by TA = T /2, and let θ2 be the angle of rotation caused by
T.
3.19
As in the solution for Prob. 3.18, the compatibility
condition,, θbr =θ st , becomes
τ max L
τ max L
=
rG
rG
br
st
Subtituting the values of the maximum shear stresses, and
Lbr = a and Lst = b
(60)(a)
(80) (b)
=
(0.075/2) (35)
(0.050/2) (83)
Solving for the ratio b/a gives
b
= 1.186 ◭
a
Equilibrium gives: T = Tst + Tbr , from which the required
torque is
π X 3
T =
d τ max
16
h
i
π
=
(0.075)3 60 × 106 + (0.050)3 80 × 106
16
= 6930 N · m = 6.93 kN · m ◭
"
#
T 32 L/2
L/2
TA 32
L
TL
=
+
=
Σ
θ1 = Σ
4
JG
G π
d4
2 πG d4
(0.75d)
"
#
1
TL
8T L
1+
θ1 =
4 = (10.595) Gd4
πGd4
(0.75)
θ2 =
Compatibility: θ1 = θ2
(10.595)
a = 0.671L ◭
3.22
Shear stress: τ max = 16T / πd3 can be rewritten as
T = πτ max d3 /16
π (84) d3st
Tst =
= 16.5 d3st (1)
16
π (60) d3br
Tbr =
= 11.8 d3br (2)
16
Compatibility: θst = θbr
TL
TL
=
GJ st
GJ br
3
16.5 dst (2 × 1000)
11.8 d3br (3 × 1000)
=
(80 × 103 ) (d4st )
(35 × 103 ) (d4br )
Compatibility: θB/A = 0
32 T L
TL
=0
=
Σ
GJ
π Gd4
Cancelling common factors gives (torques in N·m)
θ B/A = Σ
TB (1.5 × 1000)
4
(30)
(3)
Equilibrium:
= 40 × 10
(35 × 103 )
+
+
Tst + Tbr = 40 × 106 N · mm (4)
Substituting Eqs.(1)-(3) into Eq.(4) gives
16.5 d3st + 11.8 (2.452dst)3
TL
T (L − a)
= (32.193)
Gd4
Gd4
10.595L = 32.193 (L − a)
3.20
which gives dbr = 2.452 dst
TL
T (L − a)
T (L − a)
=
= (32.193)
4
JG
Gd4
Gπ (0.75d) /32
(TB − 1250) (2 × 1000)
4
(60) (30 × 103 )
(TB − 1875) (3 × 1000)
(30)4 (80 × 103 )
=0
Solving gives TB = 893.5
6
Equilibrium: TA = 7 − TB = 1875 − 893.5 = 981.5 N·m
dst = 59.4 mm ◭
Tbr = TB = 893.5 N · m
dbr = 2.452 (59.4)
Tal = TB − 1250 = −356.5 N · m
= 145.6 mm ◭
Tst = TA = 981.5 N · m
48
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Shear stress: τ = 16T /πd3
Initial twist: θ0 =
3
τ br =
16 893.5 × 10
τ al =
16 356.5 × 103
= 8.4 MPa ◭
τ st =
16 981.5 × 103
= 185.1 MPa ◭
3
π (30)
3
π (60)
3
π (30)
= 168.5 MPa ◭
T0 L
(750) (3)
=
= 0.11188 rad
Gst Jst
20.11 × 103
Equilibrium: Tst = Tal (= T )
Compatibility
θst + θ al = θ0
TL
TL
+
= θ0
Gst Jst
Gal Jal
3
3
T
= 0.11188
+
20.11 × 103
10.143 × 103
Solving gives T = Tst = Tal = 251 N · m ◭
3.25
3.23
π(60)4
= 1272345 mm4 ; Jbr
32
i
π h
4
4
(90) − (60) = 5168902 mm4
=
32
Jst =
Compatibility:
θ AB = θBC
TL
TL
=
GJ AB
GJ BC
Compatibility: θst = θbr becomes
Tst L
Tbr L
=
Gst Jst
Gbr Jbr
TAB (1.2)
TBC (0.8)
=
4
Gπ(0.06) /32
Gπd4BC /32
TAB
2
=
TBC
3
0.06
dBC
4
Tst =
(a)
Tst = 0.5626 Tbr
If bronze governs,
τ max J
(60) (5168902)
Tbr =
=
= 6.89 kN · m
r
(45)
Equating maximum shear stress:
(τ AB )max = (τ BC )max
16T
16T
=
πd3 AB
πd3 BC
Tst = 0.5626 Tbr = 0.5626 (6.89) = 3.88 kN · m
If steel governs,
τ max J
(100) (1272345)
Tst =
=
= 4.24 kN · m
r
(30)
16TBC
16TAB
=
3
π(0.06)
πd3BC
TAB
=
TBC
0.06
dBC
3
(80) (1272345)
Gst Jst
Tbr =
Tbr
Gbr Jbr
(35) (5168902)
Because 4.24 kN·m is greater 3.88 kN·m, bronze governs.
Equilibrium T = Tbr + Tst = 6.89 + 3.88
(b)
= 10.77 kN · m ◭
Equating Eqs (a) and (b):
2 0.06
=1
dBC = 0.04 m = 40 mm ◭
3 dBC
3.26
π(60)4
= 1272345 mm4 ;
32
i
π h
4
4
Jbr =
(90) − (60) = 5168902 mm4
32
Jst =
3.24
Compatibility: θst = θbr becomes
Gst Jst =
Gal Jal =
Tst L
Tbr L
=
Gst Jst
Gbr Jbr
π
4
(0.040) = 20.11 × 103 N · m2
32
i
π h
(0.050)4 − (0.040)4
28 × 109
32
80 × 109
Tst =
= 10.143 × 103 N · m2
Gst Jst
(80) (1272345)
Tbr =
Tbr
Gbr Jbr
(35) (5168902)
Tst = 0.5626 Tbr
(1)
49
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Equilibrium: Tst + Tbr = 8000 N · m
3.28
(2)
Substituting (1) into (2):
Equilibrium:
0.5626 Tbr + Tbr = 8000
Tbr = 5119.7 N · m
Tst = 0.5626 Tbr = 0.5626 (5119.7) = 2880.3 N · m
2880.3 × 103 (30)
Tr
τ st =
=
= 67.9 MPa
J st
1272345
5119.7 × 103 (45)
Tr
=
= 44.6 MPa ◭
τ br =
J br
5168902
FBD of gear B: ΣMB = 0
T −TA −F R = 0
(1)
FBD of gear D: ΣMD = 0
2R
= 0 (2)
TC −F
3
Eqs.(1) and (2) yield
3
T = TA + TC
2
3.27
(3)
Compatibility: The distance s is given by
Equilibrium
F1
s = RθA = (2R/3) θ C :
3
θA
θC =
2
TC L
3 TA L
=
GJ
2 GJ
which gives
T1
A
20
F1
500 N .m
F2
B
40
2
TC (4)
3
Substituting TA from Eq.(4) into Eq.(3):
3
13
6
2
TC + TC =
TC or TC =
T ◭
T =
3
2
6
13
3
3 6
4
TA = T − TC = T −
T =
T ◭
2
2 13
13
F2
T2
C
TA =
30
From the FBDs of the gears:
ΣMA = 0
T1 − 0.02F1 = 0
ΣMB = 0
500 − 0.04(F1 + F2 ) = 0
ΣMC = 0
T2 − 0.03F2 = 0
3.29
Eliminating F1 and F2 yields
4
2T1 + T2 = 500
3
(a)
π(60)4
π(45)4
= 1272345 mm4 J2 =
= 402578 mm4
32
32
Equilibrium: T1 = T2 (= T )
Compatibility (relax right support):
θ 1 + θ 2 = θ0 = 6◦ = 6(π/180) rad
Compatibility When gear B rotates through angle θ,
the corresponding rotations of gears A and C are
θA =
40θ
= 2θ
20
θC =
J1 =
40θ
4
= θ
30
3
which yields
3
θC
2
Since θA and θ C represent the angles of twist in shafts 1
and 2, the last equation is equivalent to
T L2
T L1
+
= θ0
G1 J1
G2 J2
L2
T L1
= θ0
+
G J1
J2
θA =
3
θ2
2
3
T1 = T2
2
θ1 =
T1 L
3 T2 L
=
GJ
2 GJ
which gives
T =
(b)
Solution of Eqs. (a) and (b) is
T1 = 173.1 N · m
Gθ0
(L1 /J1 ) + (L2 /J2 )
80 × 103 [(6) (π/180)]
=
[(3.5 × 1000) / (1272345) + (1.75 × 1000)/(402578)]
T2 = 115.4 N · m ◭
= 1180305 N · mm
50
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252626.9 × 103 (420/2)
T (D/2)
(a) τ max =
=
J
2259.68 × 106
= 23.5 MPa ◭
252626.9 × 103 (9 × 1000) 180
TL
= 0.72◦ ◭
(b) θ =
=
GJ
(80 × 103 ) (2259.68 × 106 )
π
The maximum stresses are
(τ max )1 =
(1180305) (30)
T r1
=
= 27.8 MPa ◭
J1
(1272345)
(τ max )2 =
T r2
(1180305) (45)
=
= 131.9 MPa ◭
J2
402578
3.30
3.33
T =
60P
60 × 20 × 1000
T =
=
2πN
2(π)(120)
τw =
T = 1591.5 N · m
If shear stress governs:
s
r
16T
16 (1591.5 × 103 )
= 3
= 54.5 mm
d= 3
πτ max
π (50)
60(P )
60(200 × 103 )
=
= 545.7 N · m
2πN
2π(3500)
16T
16(545.7 × 1000)
80 =
3
πd
πd3
T L T L 32
=
θ =
GJ
G πd4
=
(545.7 × 103 )(4 × 1000)
= 7.68 × 10−3 rad
(80 × 103 )[π(32.6)4 ]
= 0.44◦ ◭
If angle of twist governs [θ = 9 (π/180) = π/20 rad]:
TL
Gθ
4
1591.5 × 103 (5 × 1000)
πd
=
32
(80 × 103 ) (π/20)
3.34
J =
d = 50.4 mm
TAB =
35 × 103
P
=
= 1392.6 N · m
2πf
2π (4)
TBC =
P
55 × 103
=
= 2188 N · m
2πf
2π (4)
Comparing the above values, the smallest safe diameter is
d = 54.5 mm ◭
3.31
(a) AB
τ max =
16T
16 (1392.6)
6
=
3 = 42.6 × 10 Pa
πd3
π (0.055)
BC
τ max =
16 (2188)
16T
6
=
3 = 40.6 × 10 Pa
πd3
π (0.065)
Comparing the above values, the maximum shear stress is
42.6 MPa (occurs in segment AB)◭
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4
4
J = π (90) − (45) /32 = 6038669 mm4
T =
60(P )
60(200 × 103 )
=
2πN
2π(120)
(b) θAC =
T = 15915.5 N · m
3.35
3.32
TAB =
= 0.10479 rad = 6.00◦ ◭
60(P )
60(5000 × 103 )
=
2πN
2π(189)
60(P )
60(3000 × 103 )
=
= 19098.6 N·m
2πN
2π(1500)
TBC =
T = 252626.9 N · m
J=
1 X TL
G
J
"
#
(1392.6) (4) (2188) (2)
1
=
+
4
4
(83 × 109 ) (π/32)
(0.055)
(0.065)
15915.5 × 103 (45)
Tr
=
= 118.6 MPa
(a) τ max =
J
6038.669
T L (15915.5 × 103 )(3 × 1000) 180
= 5.66◦ ◭
=
(b) θ =
GJ
(80 × 103 )(6038669)
π
T =
d = 32.6 mm ◭
60(2000 × 103 )
= 12732.4 N · m
2π(1500)
i
π
π h
4
4
D4 − d4 =
(420) − (300) = 2259.68×106
32
32
51
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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(a) AB
BC
16T 16 19098.6 × 103
=
τ max =
= 56.3 MPa
3
πd3
π (120)
16T 16 12732.4 × 103
= 56 MPa
=
τ max =
3
πd3
π (10.5)
3.38
Cylindrical tube: thickness t, r = 10 t, R = 11 t, r = 10.5 t.
(a) Using Eq. (3.5d):
τ max =
Comparing the above values, the maximum shear stress is
56.3 MPa (occurs in segment AB)◭
1 X TL
(b) θDA =
G
J
19098.6 × 103 (3 × 1000)
=
π
(120)4
(80 × 103 ) 32
12732.4 × 103 (2 × 1000)
+
π
(80 × 103 ) 32
(105)4
τ max =
τ max =
1.444 × 10−3 T
i =
◭
τ max = h
2
t3
2 π (10.5t) t
% difference =
(a) Approximate area of the tube
is A ∼
= 2πrt and distance from
center is r. Therefore, the polar
moment of inertia can be
approximated by
J = (2πrt) r = 2πr t
q
T
T
=
=
t
2A0 t
2 πr2 t
T
3.36
3
1.509 × 10−3 T
◭
t3
(b) Using Eq. (3.8b) (q = T /2A0 ):
= 0.06185 rad = 3.54◦ ◭
2
2T (11 t)
2T R
i
= h
4
4
4
4
π (R − r )
π (11 t) − (10 t)
1.509 − 1.444
× 100% = 4.31% ◭
1.509
3.39
T = 800 N·m, A0 = (30 × 80) = 2400 mm2 , τ max = 90 MPa
Q.E.D.
Using Eq. (3.8b) [q = T /(2A0 )]:
τ max = q/t = T /(2A0 t)
2
(b) From Eq. (3.8b), q = T /2A0 , where A0 = πr ,
Tr
q
T
T
Tr
=
=
τ= =
=
3
2
t
2A0 t
J
2πr t
2 πr t
t =
Q.E.D
t = 1.852 × 10−3 m = 1.852 mm ◭
Using Eq. (3.9b),
θ=
TL
T L (2πr)
TL
T LS
=
=
=
3
2 2
4GA20 t
GJ
G
2πr
t
4G πr
t
800
T
=
2A0 τ max
2 (2400 × 10−6 ) (90 × 106 )
3.40
Q.E.D.
Using Eq. (3.8b) [q = T /(2A0 )]: τ max = q/t = T /(2A0 t)
π
T = τ max 2A0 t = (36) 2
× 180 × 90 (3.6)
4
3.37
A0 = πr2 = π (150)2 = 22500π mm2
S = 2πr = 2π (150) = 300π mm
T = 3297918 N · mm = 3297.9 N · m ◭
If shear stress governs [from Eq.(3.8b)]:
q =
3.41
18000 × 103
T
=
= 127.32 N/m
2A0
2 (22500π)
T = 600 N·m, t = 2 mm = 2 × 10−3 m
2
A0 = 30 (20) − π (10) = 285.8 mm2 = 285.8 × 10−6 m2
q
127.32
=
= 1.77 mm
τ
72
If angle of twist governs [from Eq.(3.9b)]:
18000 × 103 (7 × 1000) (300π)
T LS
t=
=
= 2.1 mm
4GA20 θ
4 (80 × 103 ) (22500π)2 2 × π
t =
S = 30 (2) + 2π (10) = 122. 83 mm = 122. 83 × 10−3 m
(a) Using Eq. (3.8b) [q = T / (2A0 )]:
τ max =
180
Comparing the above values, the minimum thickness is:
T
600
q
=
=
t
2A0 t
2 (285.8 × 10−6 ) (2 × 10−3 )
τ max = 524.8 × 106 Pa = 524.8 MPa ◭
tmin = 2.1 mm ◭
52
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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(b) From Eq.(3.9b):
3.44
TS
θ
=
L
4GA20 t
The area consist of 6 equilateral triangles of base 45 mm.
and height 45 sin 60◦ .
−3
(600) 122. 83 × 10
θ
=
2
L
4 (80 × 109 ) (285.8 × 10−6 ) (2 × 10−3 )
A0 = 6(1/2) (45) (45 sin 60◦ ) = 5261 mm2
S = 6(45) = 270 mm
θ
= 1.410 rad/m = 80.8 deg./m ◭
L
(a) τ = q/t = T /(2A0 t)
Using t = 2.25 mm. and τ max = 40 MPa,
3.42
T = 2A0 τ t = 2 (5261) (40) (2.25) = 946980 N · mm ◭
(b) Using L = 1000 mm,
Circle: A0 = πr2 , S = 2πr
Square: A0 = (πr/2)2 = π 2 r2 /4, S = 4 (πr/2) = 2πr
θ=
(a) τ = q/t = T / (2A0 t) . Because T, 2, and t cancel:
(A0 )square
π 2 r2 /4
τ circle
π
=
= = 0.785 ◭
=
τ square
(A0 )circle
πr2
4
T LS
(946980) (1000) (270) 180
=
= 2.1◦ ◭
2
4GA20 t
4 (28 × 103 ) (5261) (2.25) π
3.45
Area consists of a semi-circle of radius 120 mm and
a 120 by 240 mm rectangle:
TS
θ
=
Because T, S, 4, G, and t cancel:
L
4GA20 t
h
i2
2
(A
)
0
π 2 r2 /4
square
(θ/L)circle
π2
=
=
=
= 0.617 ◭
2
2
(θ/L)square
16
[(A0 )circle ]
(πr2 )
(b)
2
A0 =
π (120)
+ (120 × 240) = 51419.5 mm2
2
Given τ = τ max = 60 MPa and t = tmin = 1.2 mm.,
τ = q/t = T / (2A0 t) gives
T = 2A0 τ t = 2 (51419.5) (60) (1.2) = 7404408 N · mm ◭
3.43
T = 7.5 kN·m, G = 80 GPa
3.46
2
A0 =
π (120)
+(80 × 120) = 20909.7 mm2
4
A0 = 10800 mm2
T
= τ all
2A0 t
T = 2A0 t τ all = 2(10800)(3.6)(48) = 3732480 N · mm ◭
(b) θ =
(a) τ = q/t = T / (2A0 t) For τ max , use 5 mm.
7.5 × 106
T
=
= 1535.9 MPa ◭
τ max =
2A0 t
2 (20909.7) (5)
(3732480)(3000)(480)
T LS
=
2
4GA0 t
4(80 × 103 )(10800)2 (3.6)
= 0.04 rad = 2.29 deg. ◭
3.47
(b) Angle of twist per foot of length:
I
7.5 × 106 (81.9)
θ
T
ds
=
=
2
L
4GA20 S t
4 (80 × 103 ) (20909.7)
=
q
= τ all
t
(a)
s2
s1
ds
+
=
t
t
t2
1
S
I
ds
200 + 120 (120π/2) + 80
=
+
= 81.9
(5)
(15)
S t
I
= 4.39 × 10
S = 480 mm
Cylindrical tube AB: Median line is circle of radius 39 mm
Square tube BC: Median line is a 97 mm by 97 mm square
2
(A0 )AB = π (39) = 4778.4 mm2
−6
rad/mm
4.39 × 10−6 rad 1000 mm. 180 deg.
×
×
mm
m
π rad
SAB = 2π (39) = 245.04 mm
tAB = 2 mm
(A0 )BC = (97)2 = 9409 mm2
(S)BC = 4 (97) = 388 mm
= 0.252 deg./m ◭
tBC = 3 mm
53
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Integrating from x = 0 to x = L gives
Z L
x −3
4T
dx
1
+
θ =
3
2L
πGt (dA ) 0
#
" Z
L
4T
x −3 dx
θ =
1+
3 2L
2L
2L
πGt (dA )
0
L
4T
1
x −2
θ =
(2L)
−
1
+
3
2
2L
πGt (dA )
0
L
−2
x
4T L
1+
θ = −
3
2L
πGt (dA )
0
"
#
−2
1
4T L
−2
1+
− (1)
θ = −
3
2
πGt (dA )
5
4
4T L
4T L
−
−
1
=
−
θ = −
3
3
9
9
πGt (dA )
πGt (dA )
Equilibrium: TA + TC = T (1)
Compatibility: θAB = θAC becomes
T LS
T LS
=
4GA20 t AB
4GA20 t BC
Cancelling common factors and solving for TC gives
2
TC =
SAB [(A0 )BC ] tBC
2
SBC [(A0 )AB ] tAB
TC = 3.673TA
2
TA =
(245.04) (9409) (3)
2
(388) (4778.4) (2)
TA
(2)
Solving (1) and (2) gives
T = 4.673TA
(3) and T = 1.2723TC
(4)
If the shear stress in AB governs: TA = 2 (A0 τ t)AB
TA = 2 4778.4 × 10−6 60 × 106 (0.002) = 1146.8 N · m
θ =
T = 4.673 (1146.8) = 5360 N · m
If the shear stress in BC governs: TC = 2 (A0 τ t)BC
TC = 2 9409 × 10−6 60 × 106 (0.003) = 3387.2 N · m
The aspect ratio of the cross section is
a
20
=
= 3.333
b
6
The corresponding values of C1 and C2 can be determined
from Table 3.1 by linear interpolation:
Comparing the above two values, the maximum torque is
4310 N·m. From Fig.P3.47, the torque is T = 0.3P N·m,
from which the largest value of P is
4310
T
=
= 14 370 N = 14.37 kN ◭
0.3
0.3
a/b
3.0
3.333
4.0
3.48
d = dA [1 + x/ (2L)]
(a)
S = πd = πdA [1 + x/ (2L)]
h
i
2
2
A0 = πd2 /4 = π (dA ) /4 [1 + x/ (2L)]
dθ =
dθ =
T S dx
2
(b)
T πdA [1 + x/ (2L)] dx
h
i
4Gt π 2 (dA )4 /16 [1 + x/ (2L)]4
3
πGt (dA )
1+
C2
0.263
C2
0.281
C1 = 0.272
C2 = 0.269
T
C1 ab2
∴ T = C1 ab2 τ max = 0.272(0.02)(0.006)2(120 × 106 )
4G (A0 ) t
4T
C1
0.267
C1
0.282
0.282 − 0.267
C1 − 0.267
=
3.333 − 3.0
4.0 − 3.0
0.281 − 0.263
C2 − 0.263
=
3.333 − 3.0
4.0 − 3.0
τ max =
Modifying Eq. (3.9b):
dθ =
Q.E.D.
3.49
T = 1.2723 (3387.2) = 4310 N · m
Pmax =
20 T L
9π Gt (dA )3
= 23.50 N · m ◭
θ =
x −3
dx
2L
23.50(0.8)
TL
=
C2 ab3 G
0.269(0.02)(0.006)3(80 × 109 )
= 0.02022 rad = 11.59◦ ◭
54
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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3.50
Bar (b) Here a/b = 10 which yields C1 = 0.312 and
C2 = 0.312.
The aspect ratio of the cross section is
a
150
=
= 2.143
b
70
The corresponding value of C2 can be determined from
Table 3.1 by linear interpolation:
a/b
C2
2.0
0.229
2.143
C2
2.5
0.249
C2 − 0.229
0.249 − 0.229
=
2.143 − 2.0
2.5 − 2.0
TL
θ =
C2 ab3 G
∴T =
τ max =
θ =
TL
(10 × 106 )(4000)
=
C2 ab3 G
0.312(300)(30)3(80 × 103 )
= 0.1978 = 11.3◦ ◭
3.53
C2 = 0.235
The maximum shear stress in the circular bar caused by a
torque T1 is
16T1
T1
τ max =
= 5.093 3
πd3
d
The height of the hexagonal bar is
√
3
0
d
c = d sin 60 =
2
C2 ab3 G
θ
L
0.235(150)(70)3(6 × 103 ) π(5)
4000
180
= 1582675.5 N · mm ◭
=
so that the maximum shear stress due to a torque T2 is
3.51
(a)
10 × 106
T
=
= 118.7 MPa ◭
2
C1 ab
0.312(300)(30)2
5.7T2
T2
τ max = √
= 8.776 3
d
( 3/2)3 d3
√
2 πT
2T
2T
=
(τ max )a =
=
√ 3
πr3
a3
π (a/ π)
(τ max )b =
Equating τ max of the two bars, we get
T
T
=
3
C1 a
0.208a3
T1
T2
5.093 3 = 8.776 3
d
d
√
(τ max )a
= 2 π(0.208) = 0.737 ◭
(τ max )b
5.093
T2
=
= 0.580
T1
8.776
(b)
The percentage loss of strength is
T1 − T2
T2
× 100% = 1 −
× 100% = 42.0% ◭
T1
T1
2πT L
TL
TL
√ 4 =
θa =
=
GJ
Ga4
Gπ(a/ π) /2
θb =
TL
TL
=
C2 Ga4
0.141Ga4
3.54
θa
= 2π(0.141) = 0.886 ◭
θb
For a/b = 4 Table 3.1 yields C1 = 0.282 and C2 = 0.281.
3.52
τ max =
Bar (a) We have a/b = 2.5 for which Table 3.1 gives
C1 = 0.258 and C2 = 0.249.
τ max =
θ =
T
10 × 106
=
= 71.8 MPa ◭
2
C1 ab
0.258(150)(60)2
T
C1 ab2
∴ T = C1 ab2 τ max
TL
C1 ab2 τ max L
C1 τ max L
=
=
3
3
C2 ab G
C2 ab G
C2 G b
L
0.281(80 × 109 ) π
C2 G
θ=
=
= 835 ◭
b
C1 τ max
0.282(150 × 106 ) 2
TL
(10 × 106 )(4000)
θ =
=
C2 ab3 G
0.249(150)(60)3(80 × 103 )
= 0.06197 rad = 3.55◦ ◭
55
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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3.55
3.57
.
2
(τ max )tube = (τ max )solid
r2 r1
.
16T
16T D
=
π(D4 − d4 )
πd3
1
D4 − Dd3 − d4 = 0
3 4
d
d
1−
−
= 0
D
D
The inner and outer portions of the shaft (denoted by 1
and 2) have the same angle of twist, Hence
T1 L
T2 L
=
GJ1
GJ2
The solution is
Letting T1 = T2 , we get
π r24 − r14
πr4
= 1
2
2
J1 = J2
r1 =
d
= 0.8192
D
∴D=
r2
= 0.841r2 ◭
21/4
40
d
=
= 48.8 mm ◭
0.8192
0.8192
3.58
L = 4 m, τ = 70 MPa, θ = 3◦ , G = 83 GPa, f = 18 Hz
3.56
(a)
(a) Shear stress τ =
The torques in the segments of the shaft are
TAB = 600 N · m
TBC = 600 − 360 = 240 N · m
16TAB
πd3
16TAB
16(600 × 103 )
d3 =
=
= 33953 mm3
πτ max
π(90)
2 (4) 70 × 106
2Lτ
d =
=
Gθ
(83 × 109 ) [3 (π/180)]
d = 0.1289 m = 128.9 mm ◭
d = 32.4 mm ◭
(b)
(b)
π(32.4)4
πd4
=
= 108188.2 mm4
32
32
P = 2πf T = 2πf
πd3 τ
16
3
P = 2π (18)
θ AD
= θAB + θBC + θ CD
π (0.1289) 70 × 106
16
!
P = 3.33 × 106 W = 3.33 MW ◭
TAB LAB + TBC LBC + TCD LCD
=
GJ
3.59
(600 × 103 )(1080) + (240 × 103 )(900) + (480 × 103 )(450)
(28 × 103 )(108188.2)
= 0.3565 rad = 20.4
πd3 τ
16
Gθπd4
πd3 τ
=
16
32L
τ max =
◦
T =
TL
GθJ
Gθπd4
or T =
=
GJ
L
32L
Equating the above two expressions for T,
Note that all torques have the same sense.
=
or
Angle of rotation θ =
TCD = 240 + 240 = 480 N · m
J=
16T
πd3
do = 100 mm, di = 80 mm, τ = 70 MPa, θ/L = 0.4◦ /m,
◭
G = 83 GPa
J=
i
π h
4
4
(0.1) − (0.080) = 5.796 × 10−6 m4
32
56
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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If shear stress governs:
Angle of twist: θ = 6◦ = 6(π/180) rad
X TL TL
TL
+
=
θ =
GJ
GJ al
GJ st
6π
1
T (0.6)
3T (0.9)
1
=
+
180
109
10−6
(28) (0.2513) (83) (0.6136)
70 × 106 5.796 × 10−6
τJ
T =
=
= 8110 N · m
r
0.050
If angle of twist governs: θ = T L/ (GJ) gives
T = GJ
T =
θ
L
Solving yields
83 × 10
9
5.796 × 10
−6
(0.4)
π 180
= 3360 N · m
T = 757 N · m
Comparing the above three values, Tmax = 679 N·m ◭
Comparing above values: Tmax = 3360 N·m = 3.36 kN·m ◭
3.62
3.60
d = 60 mm, f = 240 rev/min, τ = 72 MPa
T =
P × 60
2πN
Substituting T = πd3 τ /16 gives:
πd3 τ
=
16
63.0 × 103
3
The applied torque at B is oppositely directed from the
applied torques at A, C, and D. f = 2 Hz.
P [hp]
f [rev/min]
P × 60
π (60) (72)
=
16
2π(240)
Solving gives: P = 76.75 mm ◭
3.61
Steel: Tst = 3T , d = 50 mm, L = 900 mm, τ = 83 MPa,
G = 83 GPa; θ = 6◦
TBC =
50 × 103
PBC
=
= 3979 N · m
2πf
2π (2)
TCD =
30 × 103
PCD
=
= 2387 N · m
2πf
2π (2)
(b) Shaft diameter d = 100 mm, G = 83 GPa.
4
π (0.040)
= 0.2513 × 10−6 m4
32
4
J =
4
Jst =
PAB
20 × 103
=
= 1592 N · m
2πf
2π (2)
(a) The maximum shear stress (60 MPa) is governed by
the torque in shaft BC.
s
r
16 (3979)
3 16TBC
=3
= 0.06964 m = 69.6 mm ◭
d=
πτ max
π (60 × 106 )
Aluminum: Tal = T, d = 40 mm, L = 600 mm, τ = 55
MPa, G = 28 GPa
Jal =
TAB =
π (0.050)
= 0.6136 × 10−6 m4
32
π (0.100)
= 9.8175 × 10−6 m4
32
θ D/A = Σ
Stress in aluminum:
1
TL
=
ΣT L
GJ
GJ
(2387) (1.5) + (3979) (1.5) − (1592) (2)
θ D/A =
(83 × 109 ) (9.8175 × 10−6 )
55 × 106 0.2513 × 10−6
τJ
=
= 691 N · m
T =
r
0.020
180
π
θ D/A = 0.448◦ (gear A lags gear D) ◭
Stress in steel:
83 × 106 0.6136 × 10−6
τJ
3T =
=
r
0.025
T = 679 N · m
57
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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3.63
3.65
T /L = 2.5 N·m/m
τ = 120 MPa
G = 80 × 103 MPa
d = 6 mm
Aluminum: do = 90 mm, di = 70 mm, Gal = 26 GPa
Copper: d = 70 mm, Gcu = 47 GPa, T = 2 kN·m
i
π h
4
4
Jal =
(0.090) − (0.070) = 4.084 × 10−6 m4
32
π
4
(0.070) = 2.357 × 10−6 m4
Jcu =
32
Compatibility: θcu = θal gives [T L/ (GJ)]cu = [T L/ (GJ)]al
(a) T = 2.5L = πd3 τ /16 which gives
2.5L =
L = 2035.8 mm ◭
(GJ)cu
(47) (2.357)
Tal =
Tal = 1.0433 Tal
(GJ)al
(26) (4.084)
Tcu =
Equilibrium
π (6)3 (120)
16
(b) Apply θ = T L/ (GJ) to dx and integrate over L.
Use T (x) = 0.5x N·m
Tcu + Tal = 2000 N · m
Tal (1 + 1.0433) = 2000
θ =
Tal = 978.8 N · m
Tcu = 1.0433 ( 978.8) = 1021.2 N · m
Z L
0
θ =
Maximum shear stresses:
Tr
(978.8) (0.045)
τ al =
=
J al
4.084 × 10−6
1
T dx
=
GJ
GJ
Z L
0.5x dx
0
0.5L2
2GJ
0.5(2035.8)2
θ =
2(80 × 103 )(π(6)4 )/32
τ al = 10.79 × 106 Pa = 10.79 MPa ◭
Tr
(1021.2) (0.035)
τ cu =
=
J cu
2.357 × 10−6
180
π
= 5.8◦ ◭
3.66
For the distributed torque: T = 2T0 x/L for 0 ≤ x ≤ L/2.
τ cu = 15.16 × 106 Pa = 15.16 MPa ◭
3.64
(a) Compatibility
θAB = θBC
TA a
TC b
=
GJ
GJ
a
TA (1)
TC =
b
Equilibrium
Compatibility: Relax support at C. θ C = 0.
TC L
=
GJ
Tb
Tb
=
a+b
L
T dx
1 2T0
=
GJ
GJ L
Z L/2
x dx
0
L/2
2T0 L2
T0 L
2T0 x2
=
=
TC L =
L
2 0
L 8
4
TA + TC = T (2)
Substituting (1) into (2) gives
a
TA + TA = T
b
TA (b + a) = T b
TA =
Z
TC =
T0
◭
4
Equilibrium: TA + TC = T0 which gives
Q.E.D.
TA = T0 − TC = T0 −
a
a Tb
Ta
TA =
=
Q.E.D.
b
b L
L
(b) The results of Part (a) would not change if the shaft
were hollow because J cancels out in the analysis.
TC =
T0
3T0
=
◭
4
4
3.67
Square tube: a mm by a mm A0 = a2 mm2 , t = 3 mm,
T = 1000 N · m, τ = 640 MPa
58
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Cancelling common factors:
Using Eq. (3.8b) [q = T / (2A0 )]: τ max = q/t = T / (2A0 t) ,
which gives
1000 × 103
T
2
A0 = a =
=
2τ max t
2 (40) (3)
Tsq Ssq
2
(A0 )sq
=
Tcirc Scirc
2
(A0 )circ
2
Tsq =
a2 = 4166.7 mm ◭
a = 64.5 mm
Tsq =
3.68
(A0 )sq Scirc
2
(A0 )circ Ssq
Tcirc
(21.025)2 (424.12)
2
(14.314) (580.0)
Tcirc = 1.5777 Tcirc
Equilibrium: Tsq + Tcirc = T, which becomes
1.5777 Tcirc + Tcirc = T
Cross section: Equilateral triangle with sides b = 45 mm.
and wall thickness t = 1.8 mm.
τ = 48 MPa, θ/ () L = 2◦ /m = (2/1000) /mm, G = 40 MPa
Tcirc = 0.3879 T
Tsq = 0.6121 T
A0 = (1/2) (45) (45 sin 60◦ ) = 876.85 mm2
S = 3b = 3 (45) = 135 mm
Therefore, circular tube: 38.8% and square tube: 61.2% ◭
If shear stress governs: τ = T / (2A0 t) gives
3.70
T = 2A0 τ t = 2 (876.85) (48) (1.8) = 151519.68 N · mm
2
If angle of twist governs θ = T LS/ 4G (A0 ) t
Median line is the 50-mm by 100-mm rectangle.
L = 3 m, T = 200 N·m, G = 28 GPa
A0 = (100) (50)
4G (A0 )2 t
T =
S
A0 = 5000 mm2
θ
L
50 + 100 + 50 100
ds
=
+
= 166.67
2.4
1.2
S t
I
2
4 40 × 103 (876.85) (1.8)
2
π T =
135
1000
180
(a) Maximum shear stress:
τ max =
T = 57255.4 N · mm
Comparing above results, Tmax = 57255 N·mm ◭
200
T
=
−6
2A0 tmin
2 (5000 × 10 ) (1.2 × 10−3 )
τ max = 16.67 × 106 Pa = 16.67 MPa ◭
(b) Relative angle of rotation:
I
TL
(200) (3) (166.67)
ds
180
θ =
=
2
4GA20 S t
π
4 (28 × 109 ) (5000 × 10−6 )
3.69
Circular tube:
rinner = 75 − 10 = 65 mm, r = 65 + 0.5 (5) = 67.5 mm
A0 = πr2 = π (67.5)2 = 14.314 × 103 mm2
S = 2πr = 2π (67.5) = 424.12 mm
θ = 2.05◦ ◭
C3.1 MathCad Worksheet
Square tube:
(median line is a square with sides d = 150 − 5 = 145 mm)
2
A0 = d × d = (145) = 21.025 × 103 mm2
S = 4d = 4 (145) = 580.0 mm
Given:


10 






20 







 24 

d := 24 · mm





 23 






 20 



16
Compatibility: θsq = θ circ which, using Eq. (3.9b), becomes
!
!
T LS
T LS
=
2
2
4G (A0 ) t sq
4G (A0 ) t circ
∆x := 80 · mm
G := 30.109 · Pa
T := 15 · N · m
59
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Computation:
J :=
C3.3 MathCad Worksheet for
Part (a)
πd4
32
Given:
T
θ :=
·
G
Z L
0
G := 80 · 109 · Pa L := 500 · mm
x2
x
d(x) := (25 · mm) · 1 + 3.8 · − 3.6 · 2
L
L
Formula for the angle of twist
1
dx (for information onlyl;
J
not to be evaluated)
Use Simpon’s rule to compute the integral
1
1
1
1
1
T · ∆x
+
+4·
+
+
·
θ :=
3·G
J1
J7
J2
J4
J6
1
1
+ 2·
+
J3
J5
Computations:
J(x) :=
v
u
D := u
u
4
t
θ := 1.482 deg
π
4
· d(x)
32
32 · L
Z L
J(x)−1 dx
π·
0
L
· ·L
100
x := 0,
C3.2 MathCad Worksheet
D = 38.9 mm
0.06
d


2.0 






2.0 







1.9 







 1.7 

d := 1.4
·m



 1.2 






 1.1 






 1.0 





1.0
d (m)
Given:
k := Z L
J −1 dx
0
k
Nm
:= 4.828 × 108
rad
∆x
·
3
0.1
0.2
0.3
0.4
0.5
C3.3 MathCad worksheet for
Part (b)
Given:
G := 80 · 109 · Pa L := 650 · mm
Formula for torsional stiffness
(for information only;
not to be evaluated)
(20 · mm) if x ≤ 200 · mm
(35 · mm) if x ≤ 350 · mm
d(x) := x − 200 · mm
otherwise
(20) · mm +
10
G
P4
P3
1
1
1
1
+
+ 4 · i=1
+ 2 · i=1
J1
J9
J2·i
J2·i+1
:=
0
x (m)
Evaluate the integral by Simpson’s rule:
k
D
0.02
∆x := 5 · m
G := 80 × 109 Pa
π
· d4
J :=
32
Computation:
G
0.04
Computations:
J(x) :=
v
u
D := u
u
4
t
x := 0,
π
4
· d(x)
32
32 · L
Z L
J(x)−1 dx
π·
D = 24.5519 mm
0
L
· ·L
100
Plotting range and increment
60
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0.04
d (m)
d (m)
0.04
0.03
0.02
0.01
0.02
0
D
0
0.05
0.1
x (m)
d
0
0.1
0.2
0.3
0.4
x (m)
0.5
0.15
0.6
C3.4
C3.4 Mathcad worksheet for
Part (b)
Equilibrim:
Given:
TAB + TBC = T
(a)
L := 400 · mm
Compatibility:
b := 275 · mm
(25 · mm) if x ≤ 200 · mm
θAB = θBC
Z
Z
TAB b 1
TBC L 1
dx =
dx
G 0 J
G b J
Z L
1
dx
TAB
J
= Zb b
TBC
1
dx
0 J
(35 · mm) if x ≥ 250 · mm
d(x) := (x − 200 · mm)2
otherwise
(25) · mm +
250 · mm
Computations:
π
J(x) :=
· d(x)4
32
Letting TAB /TBC = α, Eq. (a) yields
T
TBC =
1+α
0.2
αT
TAB =
1+α
Z L
J(x)−1 dx
b
α := Z b
J(x)−1 dx
0
α·T
TAB = 0.1194T
1+α
T
TBC :=
TBC = 0.8806T
1+α
L
x := 0,
· ·L
Plotting range and interval
100
TAB :=
C3.4 mathcad worksheet for
Part (a)
0.04
d (m)
Given:
L := 200 · mm
b := 110 · mm
π · x
d(x) := 30 · mm − (20 · mm) · sin
L
Computations:
π
· d(x)4
J(x) :=
32
Z L
0.02
0
0
0.1
0.2
x (m)
0.3
J(x)−1 dx
C3.5 MathCad worksheet
J(x)−1 dx
Given:
b
α := Z b
0
α·T
1+α
TAB = 0.3478T
T
TBC :=
1+α
TBC = 0.6522T
TAB :=
x := 0,
L := 1.8 · m
r := 75 · mm
t1 := 2 · mm
t2 := 4 · mm
G := 40 · 109 · Pa
τ max := 110 · 106 · Pa
α
t(α) := 0, t1 + (t2 − t1 ) · sin
2
L
· ·L Plotting range and interval
100
61
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Computations:
A0 := πr2
Area enclosed by the median line
T := 2 · t1 · A0 · τ max
Torque required to produce τ max
Z 2·π
r
T ·L
·
dα
θ :=
2
4A0 · G 0
t(α)
θ = 2.4074 deg
C3.6 MathCad worksheet
Given:
L := 5 m
t := 6 mm
r1 := 90 mm
r2 := 360 mm
G := 80 · 109 Pa
x
x
r(x) := r1 + (r2 − r1 ) ·
T (x) := T0 ·
L
L
T0 := 150000 mm
Computations for Part (a):
A0 (x) := 0, πr(x)2
τ (x) :=
T (x)
2 · A0 (x) · t
L
· ·L
Plotting range and interval
100
Computations for Part (b):
x := 0,
s(n) = 2πr(n)
Z L
T (n) · s(n)
2G[A0 (n)]2 t
0.5
1
θ :=
0
θ = 0.77 deg
45
40
Shear Stress (MPa)
35
30
25
20
15
10
5
0
0
1.5
2
2.5 3
x (m)
3.5
4
4.5
5
62
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Chapter 4
w0 L w0 x2
−
◭
6
2L
w0 Lx w0 x2 x −
M =
6
2L
3
w0 Lx w0 x3
−
◭
M =
6
6L
Find where shear is zero, which is also where M is
maximum.
w0 L w0 x2
L
V =
−
= 0 gives x = √
6
2L
3
3
L
w0 L2
w0
w0 L L
√
√
= √
−
Mmax =
6
6L
3
3
9 3
4.1
V =
Derive expressions for V and M ; draw V - and
M -diagrams.
w0 L
− w0 x ◭
2
w0 Lx w0 x2
−
◭
M =
2
2
4.4
V =
Derive expressions for V and M ; draw V - and
M -diagrams.
4.2
Derive expressions for V and M ; draw V - and
M -diagrams.
V =
w0 x2
V = −w0 x +
◭
2L
x w x2 x 0
+
M = −w0 x
2
2L
3
w0 x3
w0 x2
+
◭
M = −
2
6L
C0
◭
L
M = −C0 +
C0 x
◭
L
4.5
4.3
Derive expressions for V and M ; draw V - and
M -diagrams.
Derive expressions for V and M ; draw V - and
M -diagrams.
63
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30x2
kN ◭
9
x 30x2 x +
M = −510 + 270x − 60x
2
9
3
4.7
V = 270 − 60x +
M = −510 + 270x − 30x2 +
Derive expressions for V and M ; draw V - and
M -diagrams.
10x3
kN · m ◭
9
AB
V = 400 kN ◭
M = 400x kN · m ◭
4.6
BC
Derive expressions for V and M ; draw V - and
M -diagrams.
V = 400 − 400 (x − 4) = 2000 − 400x kN ◭
x−4
M = 400x − 400 (x − 4)
2
M = −200x2 + 2000x − 3200 kN · m ◭
4.8
Derive expressions for V and M ; draw V - and
M -diagrams.
AB
BC
V =
Pb
◭
a+b
M =
P bx
◭
a+b
V = −
M =
Pa
◭
a+b
P a (a + b − x)
a+b
AB
P ax
P a (a + b)
−
a+b
a+b
x
◭
M = Pa 1 −
a+b
M =
BC
V = −120x kN ◭
x M = −120x
2
2
M = −60x kN · m ◭
V = −960 kN ◭
M = −960 (x − 4) = −960x + 3840 kN · m ◭
64
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.9
Find where shear is zero, which is also where M is
maximum.
Derive expressions for V and M ; draw V - and
M -diagrams.
V = 810 − 120x = 0
gives x = 6.75 m
2
Mmax = −60 (6.75) + 810 (6.75) − 2430 = 303.8 kN · m
4.11
Derive expressions for V and M ; draw V - and
M -diagrams.
AB
V = −4 kN ◭
M = −4x kN · m ◭
BC
V = −14 kN ◭
M = 14 (5 − x)
M = 70 − 14x kN · m ◭
4.10
Derive expressions for V and M ; draw V - and
M -diagrams.
AB
y = 40x kN/m; P1 =
1
1
xy = x (40x) = 20x2 kN
2
2
V = 80 − 20x2 kN ◭
x
M = 80x − 20x2
3
M = 80x − 6.667x3 kN · m ◭
3
At x = 2m: M = 80 (2) − 6.667 (2) = 106.7 kN · m
AB
V = −120x kN ◭
x
M = −120x
2
M = −60x2 kN · m ◭
BC
y = 40 (4 − x) kN/m
P2 =
1
(4 − x) y = 20 (4 − x)2 kN
2
2
BC
V = 810 − 120x kN ◭
M = 810 (x − 3) − 120x
V = 20 (4 − x) − 80 = 20x2 − 160x + 240 kN ◭
4−x
2
M = 80 (4 − x) − 20 (4 − x)
3
M = 6.667x3 − 80x2 + 240x − 106.7 kN · m ◭
x
2
M = −60x2 + 810x − 2430 kN · m ◭
65
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4.12
AB
V = −2x kN ◭
x
= −x2 kN · m ◭
M = −2x
2
Derive expressions for V and M ; draw V - and
M -diagrams.
BC
P1 = 2x kN
P2 =
AB
V = 29 − 8x kN ◭
x
M = 29x − 8x
2
M = 29x − 4x2 kN · m ◭
2
(x − 2)
x2
2x 4
=− −
− kN ◭
3
3
3
3
"
#
2 x
(x − 2)
x−2
−
M = −2x
2
3
3
V = −2x −
Find where shear is zero, which is also where M is
maximum.
M =
V = 29 − 8x = 0 gives x = 3.625 m
2
Mmax = 29 (3.625) − 4 (3.625) = 52.56 kN · m
BC
2
(x − 2)
(x − 2) 2
(x − 2) =
kN
2
3
3
−
x3
x2
4x 8
−
−
+ kN · m ◭
9
3
3
9
4.14
V = −11 kN ◭
M = 11 (8 − x)
Derive expressions for V and M ; draw V - and
M -diagrams.
= 88 − 11x kN · m ◭
4.13
Derive expressions for V and M ; draw V - and
M -diagrams.
66
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AB
BC
CD
P
◭
3
Px
M =
◭
3
4.16
V =
Derive expressions for V and M ; draw V - and
M -diagrams.
2P
P
−P =−
◭
3
3 Px
L
M =
−P x−
3
3
P
M =
(L − 2x) ◭
3
V =
P
◭
3
P
M = − (L − x) ◭
3
V =
AB
V = 12 kN ◭
M = 12x kN · m ◭
BC
V = 12 − 12 = 0 ◭
M = 12x − 12 (x − 4)
4.15
Derive expressions for V and M ; draw V - and
M -diagrams.
M = 48 kN · m ◭
CD
V = 0◭
M = 0◭
4.17
Derive expressions for V and M ; draw V - and
M -diagrams.
AB
V = −2 kN ◭
M = −2x kN · m ◭
BC
V = −2 + 4 = 2 kN ◭
M = −2x + 4 (x − 2)
M = 2x − 8 kN · m ◭
CD
V = −2 kN ◭
M = 2 (12 − x)
M = 24 − 2x kN · m ◭
67
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AB
V = 800 N ◭
M = 800x N · m ◭
BC
V = 800 − 800 (x − 2)
V = −800x + 2400 N ◭
2
800 (x − 2)
M = 800x −
2
M = −400x2 + 2400x − 1600 N · m ◭
AB
V = 670 − 60x N ◭
x
M = 670x − 60x
2
M = 670x − 30x2 N · m ◭
BC
V = 670 − 60x − 900
V = −230 − 60x N ◭
M = 670x − 30x2 − 900 (x − 4)
M = −30x2 − 230x + 3600 N · m ◭
Find where shear is zero, which is also where M is
maximum.
V = −800x + 2400 = 0 gives x = 3.00 m
Mmax = −400 (3.00)2 + 2400 (3.00) − 1600 = 2000 N · m
CD
V = 800 (8 − x)
V = 6400 − 800x N ◭
2
800 (8 − x)
M = −
2
M = −400x2 + 6400x − 25 600 N · m ◭
CD
V = 60 (18 − x) + 400
V = 1480 − 60x N ◭
2
60 (18 − x)
M = −400 (18 − x) −
2
2
M = −30x + 1480x − 16 920 N · m ◭
4.18
4.19
Derive expressions for V and M ; draw V - and
M -diagrams.
Derive expressions for V and M ; draw V - and
M -diagrams,
y
A
B 120 kN/m C
2.2 m
144 kN
D
x
2.2 m
144 kN
2.4 m
144
V
(kN)
–144
317 403.2 317
M
(kN. m)
Segment AB:
A
144 kN
V = 144 kN
M
x
V
M = 144x kN · m ◭
68
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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BC :
Segment BC:
P1 P2
120(x – 2.2) kN
B
M
A
2.2 m
x
144 kN
A
V
B
144 kN
V = 540 − P1 − P2
= 940 − 100x2 N ◭
M
V
x−2
x−2
− P2
2
3
x−2
= 540x − 400(x − 2)
2
x−2
2
− 100x − 400x + 400
3
100 3
1600
= −
x + 940x −
N·m ◭
3
3
√
Mmax occurs where V = 0, i.e. at x = 9.4 = 3.066 m
4.20
w = 200x N/m
600 N/m
540
V
(N)
B
2m
Mmax = −
400 N/m
C x
A
540 N
3m
1600
100
(3.066)3 + 940(3.066) −
= 1388.0 N · m
3
3
1560 N
4.21
3.066 m
1080
6 kN/m
A
72 kN . m
12 kN
–1560
1388
6m
B
B
540 N
12 kN
12
V
(kN)
M
A
C
4m
12 kN 12 kN
12
M
(N . m)
AB :
M = 540x − P1
V = 144 − 288 = −144 kN ◭
M = 144x − 288(x − 3.4) = 979.2 − 144x kN · m ◭
y
x
= 540 − 400(x − 2) − 100x2 − 400x + 400
C
2.2 m 1.2 m
x
M
V
x–2m
P1 = 400(x − 2) N
1
P2 = (200x − 400)(x − 2) = 100x2 − 400x + 400 N
2
288 kN
A
B
2m
540 N
V = 144 − 120(x − 2.2) = 408 − 120x kN ◭
(x − 2.2)2
M = 144x − 120
2
= −290.4 + 408x − 60x2 kN · m ◭
Segment CD:
200x – 400 N/m
400 N/m
–12
12
x
M
(kN. m)
V
–72
V = 540N ◭
M = 540x N · m ◭
6(x – 6) kN
A
x
72 kN . m
12 kN
M
V
B
x–6 V
12 kN
M
69
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Section BC (Note: α = π − θ)
Segment AB:
V = 12 kN
ΣFx = 0
M = 12x − 72 kN · M ◭
P
sin α
2
P
V = − sin (π − θ)
2
P
V = − sin θ ◭
2
ΣMQ = 0
PR
(1 − cos α)
M =
2
PR
PR
M =
[1 − cos (π − θ)] =
(1 + cos θ) ◭
2
2
V = −
Segment BC:
V = 12 − 6(x − 6) = 48 − 6x kN ◭
(x − 6)2
= −180 + 48x − 3x2 kN · m ◭
M = 12(x − 6) − 6
2
4.22
Derive V and M as functions of θ. Neglect weight of arch.
y
M
A
M0
2R
y F
F x
P
Q
V
R
R
θ
α
R(1 – cosα)
R(1 – cosθ)
x
V
M
4.24
Draw the V - and M -diagrams by the area method.
Area of w-diagram:kips; area of V -diagram: kN·m
C
M0
2R
6 kN(+) 4 kN(+) 2 kN(+)
B
C
D
Section AB :
ΣFy = 0
ΣMP = 0
A
M0
sin θ ◭
2R
M0
(1 − cos θ) ◭
M=
2
V =
4m
7 kN(–)
ΣFx = 0
ΣMQ = 0
V
(kN)
M0
sin α
2R
M0
M =−
(1 − cos α)
2
–12
M0
sin θ ◭
2R
M =−
–20
–3
V =
28
32
–5
20
But α = π − θ, so that sin α = sin θ and cos α = − cos θ.
Therefore,
V =
E
4m
5 kN(–)
+4
1
+28
Section BC :
4m
4m
M
(kN . m)
M0
(1 + cos θ) ◭
2
Vmax = 7000 kN; Mmax = 32 000 kN · m ◭
4.23
Derive V and M as functions of θ. Neglect weight of arch.
Section AB
ΣFx = 0
P
V =
sin θ ◭
2
ΣMO = 0
PR
(1 − cos θ) ◭
M =
2
70
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4.25
4.27
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V-diagram: kN·m
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V-diagram: kN·m
20 kN
40 kN . m
600 kN 800 kN 1500 kN 400 kN
3500 kN . m
V= M = 0
D
B
A
A
C
2m
4m
V
(kN)
2m
20 kN
300 kN
–20
1500
–300
E
5m
5m
400
300
V
(kN)
–40
5m
5m
D
C
B
2000
–1500
–5500
40
–1100
M
(kN . m)
5000
3500
Vmax = −20 kN; Mmax = 40 kN · m ◭
3500
M
(kN . m)
–2000
4.26
Vmax = −1100 kN; Mmax = 5000 kN · m ◭
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V-diagram: kN·m
D
C
B
2m
1.5 m 1.5 m 1.5 m
118.89 kN
31.11 kN
A
50
31.11
46.67
V
(kN) –28.89 –43.34 –103.34
4.28
50 kN
60 kN 40 kN
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V-diagram: kN·M
E
2000 kN
–68.89
8m
46.67
M
(kN . m)
B
A
C
16 m
4000 kN
3.33
2000 kN
250 kN/m
4000
100
D
8m
4000 kN
2000
V
(kN)
–100
2000
16000
8000
–8000
–16000
–2000
–2000
M
(kN . m)
Vmax = −68.9 kN; Mmax = −100 kN · m ◭
–16000
–8000
–16 000
Vmax = ±2000 kN; Mmax = −16 000 kN · m ◭
71
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4.29
4.31
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V-diagram: kN·m
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
250 kN/m
2000
4000
2000
A
C
B
8m
4000 kN
16 m
8m
4000 kN
2000
2000
16000
V
(kN) –16000
–16000
–2000
D
16000
–2000
M
(kN . m)
–16000
–16000
Vmax = ±2000 kN; Mmax = −16 000 kN · m ◭
Vmax = ±20 kN; Mmax = −20 kN · m ◭
4.30
4.32
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
Vmax = −1320 kN; Mmax = 4320 kN · m ◭
Vmax = 240 N; Mmax = −240 kN · m ◭
72
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.33
4.35
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
16 kN (+) 4 kN/m 20 kN (+)
+16 B
C
A
4m
8m
52 kN (–)
1100 kN (+) 400 kN/m
2400 (+)
2 kN/m
+16
D
8m
16 kN (–)
A
V
(kN)
–16
B
6m
2400 kN (–)
2m
20
1300
160
–64
–96
2113
V
(kN) –2200
–16
–32
6000 kN . m (+)
.
E
D
C
2m 2m
1100 kN (–)
–1512
3.25 m
–2200 –2200
–1100
64
–1100
2200
M
(kN . m)
M
(kN . m)
–2200
–96
–87
–1600
–3800
Vmax = −32 kN; Mmax = −96 kN · m ◭
Vmax = 1300 kN; Mmax = −3800 kN · m ◭
4.34
4.36
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
30 kN (+)
140 kN . m (–)
.
A
C
2m
3m
16 kN (–)
B
D
2m
14 kN (–)
30
16
60
32
48
V
(kN)
32
M
(kN . m)
60
–108
Vmax = 30 kN; Mmax = −108 kN · m ◭
Vmax = −49 kN; Mmax = 54.25 kN · m ◭
73
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.37
4.39
Draw the V - and M -diagrams by the area method.
Draw the V - and M -diagrams.
Area of w-diagram: kN; area of V-diagram: kN·m
Area of w-diagram: kN; area of V -diagram: kN·m
24 kN/m
120
A
C
C
B
2m
5m
36 kN 36 kN
168 kN
84
36 kN
36
48
1.5 m
V 27
(kN)
M
(kN . m)
120 36
24 kN/m
72
D
3m
36 kN
36
–27
27
–147
27
–36
–84
27
–120
Vmax = ±84 kN; Mmax = −120 kN · m ◭
Vmax = 60 kN; Mmax = 80 kN · m ◭
4.40
4.38
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·M
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
Vmax = −2100 kN; Mmax = −8200 kN · M ◭
Vmax = 60 kN; Mmax = 90 kN · m ◭
74
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.41
4.43
Draw the V - and M -diagrams by the area method.
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
Vmax = −700 kN; Mmax = −1000 kN · m ◭
Vmax = −
4.42
4.44
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·M
2000 kN (+)
A
w0 L2
w0 L
; Mmax = √ ◭
3
9 3
Draw the V - and M -diagrams by the area method.
400 kN/m
4000 (+)
E
C
B
D
10 m
4m 4m 4m
3400 kN (–)
1000 kN (–)
1600 kN (–)
2400
1000
7200
4000
V
–4000 –4000
(kN)
6m
–1000
–3200
–1600
3200
4000
M
(kN . m)
Vmax = ±
–4000
w0 L
w0 L2
; Mmax =
◭
4
12
Vmax = 2400 kN; Mmax = ±4000 kN · m ◭
75
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Find where shear is zero, which is also where M is
maximum.
4.45
Draw the V - and M -diagrams by the area method
2
84 = 20x + 10 (x − 1)
gives x = 2.72 m
Find maximum moment using equilibrium:
2.72
1.72
Mmax = 84 (2.72) − 20 (2.72)
− 10 (1.72)2
2
3
Mmax = 137.5 kN · m
Vmax = −86 kN; Mmax = 137.5 kN · m ◭
Vmax = ±
4.47
w0 L
w0 L2
; Mmax =
◭
4
24
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN; area of V -diagram: kN·m
4.46
Draw the V - and M -diagrams by the area method.
Area of w-diagram: kN
Vmax = ±40 kN ◭
Mmax = 35 kN · m ◭
76
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.48
4.50
Draw the w- and M -diagrams for the given V -diagram.
Area of V -diagram: kN·m
Draw the w- and M -diagrams for the given V -diagram.
Area of V -diagram: kN·m
4000 kN
4m
4800 kN
8800 kN
6m
2000 kN
4m
4m
10000 kN
4800
19200
V
(kN)
4800
800
8000
2000
–32000
–8000
19200
24000
M
(kN . m)
–8000
Vmax = −780 kN; Mmax = −1800 kN · m ◭
Vmax = −8000 kN; Mmax = 24 000 kN · m ◭
4.51
4.49
Draw the w- and M -diagrams for the given V -diagram.
Area of V -diagram: kN·m
Draw the w- and M -diagrams for the given V -diagram.
Area of V -diagram: kN·m
Vmax = −1380 kN; Mmax = 3600 kN · m ◭
Vmax = −4000 kN; Mmax = −16 000 kN · m ◭
77
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4.52
4.53
Draw the w- and M -diagrams for the given V -diagram.
Area of V -diagram: kN·m
Draw the V - and M -diagrams for the beam shown.
Area of w-diagram: kN; area of V -diagram: kN·m
600 kN (+) 800 kN (+)
600 kN (+)
E
A
C
B
6m
6m
200 kN (–)
D
2m
1200
600
3600
200
V
(kN)
–400
6m
1800 kN (–)
–2400
–2400
–1200
1200
M
(kN . m)
–1200
–3600
Find where shear is zero,
which is also where M is
maximum.
1
8x
10 = x
gives x = 2.74 m
2
3
2 2.74
8 (2.74)
= 18.3 kN · m
Mmax = 10 (2.74) −
6
3
Vmax = −1200 kN; Mmax = −3600 kN · m ◭
4.54
Draw the V - and M -diagrams for the beam shown.
Area of w-diagram: kN; area of V -diagram: kN·m
Vmax = 10 kN; Mmax = 18.3 kN · m ◭
Vmax = −1500 kN; Mmax = 8000 kN · m ◭
78
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4.55
4.57
Draw the V - and M -diagrams
Draw the V - and M -diagrams
4P
P
B
A
L/4
P
D
C
L/4
A
E
L/4
L/4
w0L/3
3P
3P
2P
w0
w0L/3
P
PL/2
V
−PL/4
−P
B
L/3
w0L/3
w0
w0L/3
C
L/3
D
L/3
w0L/3
w0L2/18
V
PL/4
−w0L2/18
−PL/2
−w0L/3
w0L2/18
−2P
PL/4
M
M
−PL/4
−PL/4
Vmax = ±
Vmax = ±2P ; Mmax = ±P L/4 ◭
w0 L
w0 L2
; Mmax =
◭
3
18
4.58
4.56
Draw the V - and M -diagrams.
Draw the V - and M -diagrams.
Area of w-diagram: kN; area of V-diagram: kN·m
500 kN/m
5000 kN
8000 kN . m
4100 kN
P1
A
A
B
10 m
2w0
B
w0
MB
VB
8000 kN . m
V
900 kN
−2w0L
4100
V
(kN)
L
P2
M
–810
16810
8.2 m
1.8 m
–900
− 5 w0L2
6
8810
8000
P1 = w0 L
M
(kN .m)
–8000
P2 =
1
(2w0 L) = w0 L
2
VB = − (P1 + P2 ) = −2w0 L
1 1
5
L
L
= −w0 L2
= − w0 L2
MB = − P1 + P2
+
2
3
2 3
6
Vmax = 4100 kN; Mmax = 8810 kN · m ◭
5
Vmax = −2w0 L; Mmax = − w0 L2 ◭
6
79
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4.59
4.61
Draw the V - and M -diagrams for the beam shown.
Draw the V - and M -diagrams for the beam shown.
Area of w-diagram: kN; area of V -diagram: kN·m
Area of w-diagram: kN; area of V -diagram: kN·m
2 kN/m
8 kN (+)
A
B
4m
2 kN . m (+)
2m
5 kN (–)
C
3 kN (–)
5
6.25
V
(kN)
–2.25
2.5 m
–6
–3
6.25
4
M
(kN . m)
Vmax = ±960 kN; Mmax = ±1920 kN · m ◭
–2
Vmax = 5 kN; Mmax = 6.25 kN · m ◭
4.60
4.62
Draw the V - and M -diagrams for the beam shown.
Draw the V - and M -diagrams for the beam shown.
Area of w-diagram: kN; area of V -diagram: kN·m
Area of w-diagram: kN; area of V -diagram: kN·m
Vmax = 1300 kN; Mmax = 2667 kN · m ◭
Vmax = −1500 kN; Mmax = −4500 kN · m ◭
80
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.63
4.65
Draw the V - and M -diagrams for the beam shown.
Draw the V - and M -diagrams for the beam shown.
Area of w-diagram: kN; area of V -diagram: kN·m
Area of w-diagram: kN; area of V -diagram: kN·m
2.5 kN/m
10 kN (+)
10 kN (+)
A
B
4m
4m
15 kN (–)
5 kN (–)
5
V
(kN)
C 40 kN . m (+)
5
2m
–5
–5
5
–40
–15
M
(kN . m)
–40
Vmax = −15 kN; Mmax = −40 kN · m ◭
Vmax = −8 kN; Mmax = −10 kN · m ◭
4.64
Draw the V - and M -diagrams for the beam shown.
Area of w-diagram: kN; area of V -diagram: kN·m
Vmax = ±2.25 kN; Mmax = −1.0 kN · m ◭
81
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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4.66
4.67
Draw the V - and M -diagrams for the beam shown.
Draw the V - and M -diagrams for the beam shown.
Section AB (0 ≤ x ≤ 5 m)
Area of w-diagram: kN; area of V -diagram: kN·m.
V = 0.49 − 0.15x2 kN
At x = 5 m: V = 0.49 − 0.15 (5)2 = −3.26 kN
At x = 5 m:
M = 0.49x − 0.05x3 kN · m
M = 0.49 (5) − 0.05 (5)3 = −3.80 kN · m
V = 0.49 − 0.15x2 = 0 gives x = 1.8074 m
3
Mmax = 0.49 (1.8074) − 0.05 (1.8074) = 0.590 kN · m
Section BC (5 m ≤ x ≤ 7 m)
V = 7.35 − 0.15x2 kN
M = 0.49x − 0.05x3 + 6.86 (x − 5)
M = 7.35x − 0.05x3 − 34.3 kN · m
Vmax = ±800 kN; Mmax = 1600 kN · m ◭
4.68
Draw the w- and M -diagrams for the given V -diagram.
Area of V -diagram: kN·m
Vmax = 3.60 kN; Mmax = −3.8 kN · m ◭
82
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C4.1 MathCad worksheet
Given:

70
 191 


 271 


 293 



w := 
 230  · N/m ∆x := 12 · m
 162 


 96 


 55 
0

Mmax = 3200 kN · m ◭
Computations:
4.69
VB :=
Draw the w- and M -diagrams for the given V -diagram.
Z L
w dx
0
Area of V -diagram: kN·m
MB :=
Z L
0
w · x dx
Formulas for shear force and bending
moment at B (for information only;
not to be evaluated)
Evaluate the integrals by Simpson’s rule:
i := 1, 2.. 9
xi := (i − 1) · ∆x
j := 2, 4.. 8
k := 3, 5..7


X
X
∆x 
VB =
wj + 2 ·
wk 
· w1 + w9 + 4 ·
3
j
k
4
VB = 1.6272 × 10 N

X
∆x 
MB =
wj · xj
· w1 · x1 + w9 · x9 + 4 ·
3
j
!
X
+2·
wk · xk
k
MB = 6.3053 × 105 N · m
Mmax = 64 kN · m ◭
C4.2
83
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From FBD of beam:
x := 0.75 · L Trial value for optimal x
w0 x2
w0 (L − x)2
+
+ P (L − x) = 0
2
2
P (L − x)
w0 x w0 (L − x)2
−
−
RA =
2
2x
2x
ΣMB = 0RA x −
Given MB (x) = MC (x)
xopt = 13.73 m
Maximum bending moment occurs at B or at C. The
corresponding moments are
MB =
C4.2 MathCad worksheet for
Part (b)
w0 c2
w0 (L − x)2
+ P (L − x) MC = RAC −
2
2
Given:
where c is the coordinate of C, obtained from
VC = 0
P := 0 · N
RA
RA − w0 c = 0 c =
w0
w0 := 200 · Nm−1
RA (x) :=
C4.2 MathCad worksheet for
Part (a)
w0 · x w0 · (L − x)2
P · (L − x)
−
−
2
2·x
x
c(x) :=
RA (x)
w0
Given:
MB (x) := P · (L − x) +
L := 16 m
Computations:
RA (x) :=
RA (x)
w0
Mmax (x) :=
w0 · (L − x)2
MB (x) := P · (L − x) +
2
2
w0 · c(x)
if RA (x) > 0
RA (x) · (x) −
MC (x) :=
2
0
otherwise
8000
MB (x) if MB (x) > MC (x)
MC (x) otherwise
x := 0.5 · L, 0.505 · L..L
Plotting range and interval
2·104
Max. M (N·m)
MB (x) if MB (x) > MC (x)
MC (x) otherwise
x := 0.5 · L, 0.505 · L..L Plotting range and interval
Max. M (N·m)
Mmax (x) :=
w0 · (L − x)2
2
w0 · c(x)2
RA (x) · c(x) −
MC (x) :=
if RA (x) > 0
2
0 otherwise
w0 · x w0 · (L − x)2
P · (L − x)
−
−
2
2·x
x
c(x) :=
L := 16 m
Computations:
Note that c is positive only when RA > 0.
P := 1200 · N · m w0 := 200 · Nm−1
xopt := Findx
6000
4000
2000
1.5·104
8
10
12
x (m)
14
16
x := 0.75 · L Trial value for optimal x
1·104
Given MB (x) = MC (x)
5000
xopt := Find(x)
xopt = 11.31 m
0
8
10
12
x (m)
14
16
84
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C4.3 MathCad worksheet for
Part (a)
C4.3 MathCad worksheet for
Part (b)
Given:
Given:
a := 9 · m b := 18 · m L := 44 · m
a := 5 · m b := 28 · m L := 80 · m
P1 := 4000 · N · m P2 := 8000 · N · m P3 := 6000 · N · m
P1 := 8000 · N · m P2 := 4000 · N · m P3 := 6000 · N · m
Computations:
Computations:
P3 · (x + a + b) + P2 · (x + a) + P1 · x
L
RA (x) := P1 + P2 + P3 − RB (x)
RB (x) :=
M1 (x) := RA (x) · x
M2 (x) := RA (x) · (x + a) − P1 · a
M3 (x) := RA (x) · (x + a + b)
−P1 · (a + b) − P2 · b
x := 0,
L−a−b
..L − a − b
100
Support
reactions
P3 · (x + a + b) + P2 · (x + a) + P1 · x
L
RA (x) := P1 + P2 + P3 − RB (x)
RB (x) :=
Bending
moments
under the
loads
M1 (x) := RA (x) · x
M2 (x) := RA (x) · (x + a) − P1 · a
M3 (x) := RA (x) · (x + a + b)
−P1 · (a + b) − P2 · b
Plotting range and increment
1.5·105
x := 0,
M2
L−a−b
..L − a − b
100
Support
reactions
Bending
moments
under the
loads
Plotting range and increment
3·105
1·105
M (N·m)
M3
M1
M (N·m)
2·105
5·105
M3
M2
1·105
0
0
5
10
M1
16
x (m)
0
0
10
20
x (m)
30
40
85
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C4.4 MathCad worksheet for
Part (a)
C4.4 MathCad worksheet for
Part (b)
Given:
Given:
L := 3 · m
3
−1
w(x) := (50 · 10 · N · m
) · sin
Computations:
V (x) :=
Z x
w(ξ)dξ
M (x) :=
0
Z x
0
L := 5 · m
π · x
(20 · 103 · N · m−1 ) x ≤ 1.0 · m
2·L
w(x) := 0 if x ≥ 4 · m
h
(20 · 103 · N · m−1 ) ·
Computations:
(x − ξ) · w(ξ)dξ
x := 0, 0.01 · L..L Plotting range and interval
V (x) :=
M (x) :=
Z x
0
(x − ξ) · w(ξ)dξ
x := 0, 0.01 · L..L Plotting range and interval
5·104
0
2·105
0
0.5
1
1.5
x (m)
2
2.5
3
1·105
1.5·105
0
0
1
2
3
4
3
4
5
x (m)
1·105
6·105
5·104
0
0
0.5
1
1.5
x (m)
2
2.5
M (N·m)
M (N·m)
w(ξ)dξ
0
V (N)
V (N)
1·105
Z x
x i
otherwise
1.0 · m
3
4·105
2·105
0
0
1
3
5
x (m)
86
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4·104
M (N·m)
C4.5
2·104
0
ΣMB = 0
RA L −
0
1
RA =
L
(L − ξ)w(ξ)dξ = 0
Z L
0
(L − ξ)w(ξ)dξ
V = RA −
w(ξ)dξ
0
M = RA x −
Z x
0
w(x) := 0 if x ≥ 4 · m
h
(20 · 103 · N · m−1 ) ·
(x − ξ)w(ξ)dξ
Computations:
1
RA := ·
L
0
Z x
w(ξ)dξ
0
2·105
V (N)
Z x
V (x) := RA −
w(ξ)dξ
0
Z x
M (x) := RA · x −
(x − ξ) · w(ξ)dξ
0
–1·105
0
1
2
3
4
5
3
4
5
x (m)
x := 0, 0.01 · L..L Plotting range and interval
2·105
M (N·m)
5·104
V (N)
x i
otherwise
1.0 · m
Z L
0
0
–5·104
–1·105
3
x := 0, 0.01 · L..L Plotting range and interval
Computations:
(L − ξ) · w(ξ)dξ
2.5
0
L := 3 · m
π · x
w(x) := (50 · 103 · N · m−1 ) · sin
2·L
0
2
(L − ξ) · w(ξ)dξ V (x) := RA −
Z x
(x − ξ) · w(ξ)dξ
M (x) := RA · x −
Given:
Z L
1.5
x (m)
L := 5 · m
(20 · 103 · N · m−1 ) if x ≤ 1.0 · m
C4.5 MathCard worksheet for
Part (a)
1
RA :=
L
1
Given:
From FBD of segment:
Z 2
0.5
C4.5 MathCard worksheet for
Part (b)
From FBD of beam:
Z L
0
1·105
0
0
1
3
x (m)
0
0.5
1
1.5
x (m)
2
2.5
3
87
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Chapter 5
5.1
(b)
Web: ymax = ± [(387/2) − 33.3] = ±160.2 mm
M |ymax |
(460 × 103 )(160.2 × 10−3 )
=
I
894 × 10−6
6
= 82.4 × 10 Pa = 82.4 MPa ◭
Cross section: 50 mm by 250 mm rectangle removed from
200 mm by 300 mm rectangle. M = 24000 N·m
3
3
3
bh
250 (300) (50) (250)
I=Σ
=
−
= 497.4×106 mm4
12
12
12
σ max =
Board A: ymax = ± 150 mm
5.4
24000 × 103 (150)
M |ymax |
(σ A )max =
= 7.24 MPa ◭
=
I
497.4 × 106
M = 3.6 kN·m
bh3
I =
36
3
(60) (120)
I =
36
I = 2.88 × 106 mm4
Board B: ymax = ± 100 mm
(σ B )max =
24000 × 103 (100)
M |ymax |
= 4.83 MPa ◭
=
I
497.4 × 106
5.2
3.6 × 103 80 × 10−3
M ctop
=−
σ top = −
I
2.88 × 10−6
6
σ top = −100.0 × 10 Pa
3.6 × 103 40 × 10−3
M cbot
σ bot =
=
I
2.88 × 10−6
6
σ bot = 50.0 × 10 Pa
Circular cross section D = 200 mm, d = 150 mm
M = −90000 N·m
π
π
D4 − d4 =
2004 − 1504 = 53.69 × 106 mm4
I=
64
64
(a) yA = + 100 mm
−90000 × 103 (100)
M yA
σA = −
=−
I
53.69 × 106
= 167.6 MPa (T) ◭
Therefore, the maximum tensile and compressive bending
stresses are
(σ C )max = 100.0 MPa and (σ T )max = 50.0 MPa ◭
(b) yB = −75 mm
−90000 × 103 (−75)
M yB
=−
I
53.69 × 106
= −125.7 MPa (C) ◭
σB = −
5.5
Find σ max .
180 kN/m
2160 kN
(c) yD = 100 sin 60◦ mm
3
A
◦
−90000 × 10 (100 sin 60 )
M yD
=−
I
53.69 × 106
= 145.2 MPa (T) ◭
σD = −
B
12 m
1200 kN
960 kN
1200
4000 kN . m
5.3
V(N)
6
1440 kN . m
Location of Mmax
6.667 m
4
From Appendix B-2 for W360×262: I = 894 × 10 mm
M = 460 kN·m
(a)
–960
Flange: ymax = ±(387/2) = ±193.5 mm
Mmax = 4000 kN · m
S=
bh2
120(240)2
=
6
6
= 1152000 mm3
4000 × 106
Mmax
=
= 3472.2 MPa ◭
σ max =
S
1152000
(460 × 103 )(193.5 × 10−3 )
M |ymax |
=
I
894 × 10−6
6
= 99.6 × 10 Pa = 99.6 MPa ◭
σ max =
89
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5.6
5.8
Orientation (a)
Cross section is rectangle 60 mm wide and 120 mm high.
3
bh3
150(150)3
=
= 42.2 × 106 mm4
12
12
(24000 × 103 )(75)
M ymax
= 42.7 MPa ◭
=
σ max =
I
42.2 × 106
(60) (120)
bh3
=
= 8.64 × 106 mm4
12
12
(a) Mmax = 1500(3) = 4500 kN·m
I =
I=
|M |max c
I
4500 × 106 (60)
σ max =
8.64 × 106
σ max = 31250 MPa ◭
σ max =
Orientation (b) Consider the cross-section as an
assembly of two triangles.
150 mm
2
150 2 mm
0
15
m
m
(b) At section D (midspan): M = 500 (6) = 3000 kN·m
y = 60 − 15 = 45 mm
3000 × 106 (45)
My
=−
σD = −
I
8.64 × 106
σ D = −15625 MPa ◭
m
50
m
1
√
√
(150 2)(150/ 2)3
bh3
=2
= 42.2 × 106 mm4
I = 2
12
12
√
M ymax
(24000 × 103 )(150/ 2)
σ max =
= 60.3 MPa ◭
=
I
42.2 × 106
5.9
5.7
Steel band saw: 20 mm wide by 0.6 mm thick.
E = 200 GPa. Diameter of pulleys is d.
Cross section is rectangle 50 mm wide and 150 mm high.
3
(50) (150)
bh3
=
= 14.063 × 106 mm4
12
12
(a) Maximum moment occurs at the wall.
(a) Maximum bending stress given d = 500 mm.
(ρ = 500/2 = 250 mm; c = 0.6/2 = 0.3 mm)
I=
M
σI/c
σ
1
=
=
=
which gives:
ρ
EI
EI
Ec
200 × 109 0.3 × 10−3
Ec
=
σ =
ρ
0.250
6
σ = 240 × 10 Pa = 240 MPa ◭
Mmax = − (1/2) (6) (1000) (6/3) = −6000 N·m
|M |max c
(6000) (0.075)
=
I
14.063 × 10−6
σ max = 32.0 × 106 N · m = 32.0 MPa ◭
σ max =
(b) Smallest value of d if σ = 400 MPa.
200 × 109 0.3 × 10−3
Ec
ρ =
= 0.150 m
=
σ
400 × 106
d = 2ρ = 2 (0.150) = 0.300 m = 300 mm ◭
(Occurs at wall: tension in top, compression in bottom)
(b) At 2 m from the free end,
5.10
1000
2
1
(2)
Mmax = −
2
3
3
= − 222.2 N · m
W250 × 28.4 is used as cantilever beam, L = 6 m. Find
maximum uniformly distributed load over entire length (in
addition to weight of the beam) if σ max = 120 MPa.
At y = 75 − 20 = 55 mm,
From Appendix B-2: Sx = 308 × 103 mm3 , mass/length
= 28.4 kg/m
My
(− 222.2) (0.055)
σ = −
=−
I
14.063 × 10−6
3
σ = 869 × 10 Pa = −869 kPa ◭
Mmax =
wtotal L2
wtotal (6)2
=
= 18 wtotal N · m
2
2
90
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Mmax = σ max S gives 18 wtotal = 120 × 106
wtotal = 2053 N/m
308 × 10−6
5.13
Find W if σ max = 120 MPa for an S380 × 74 beam.
wnet = wtotal − wbeam
wnet = 2053 − (28.4 × 9.81)
wnet = 1774 N/m = 1.774 kN/m ◭
5.11
From above shear force diagram:
Mmax = (1/2) (2.30W ) (2.3) = 2.645W N·m
Repeat Prob. 5.10 using W150 × 29.8 with L = 4 m.
For S380 × 74 section, S = 1060 × 103 mm3 .
M = σS gives: 2.645W = 120 × 106 1060 × 10−6
From Appendix B-2: Sx = 220 × 103 mm3 ,
mass/length = 29.8 kg/m
2
Mmax =
wtotal L2
wtotal (4)
=
= 8 wtotal N · m
2
2
Mmax = σ max S gives 8 wtotal = 120 × 106
wtotal = 3300 N/m
220 × 10−6
W = 48.1 × 103 N = 48.1 kN ◭
5.14
Find P for σ max = 10 MPa, b = 120 mm, h = 400 mm.
2
10 × 106 (0.120) (0.4)
σbh2
=
= 32 000 N · m
Mmax =
6
6
wnet = wtotal − wbeam
wnet = 3300 − (29.8 × 9.81)
wnet = 3010 N/m = 3.01 kN/m ◭
First find critcal value of w0 that will cause V = 0 under
the load P . The shear force FBD and shear force diagram
for this condition are shown below.
5.12
Find σ max given b = 50 mm and h = 80 mm.
From the above shear force diagram,
Mmax = (1/2) (4) [4 (w0 )cr ] = 8 (w0 )cr , which gives
8 (w0 )cr = 32 000 N · m
(w0 )cr = 4000 N/m
Therefore, for w0 = 3000 N/m, the FBD and shear force
diagram are as shown below.
3
(50) (80)
bh3
=
= 2.133 × 106 mm4
12
12
From the above bending moment diagram, Mmax = 4
kN·m.
I=
Mmax c
(4000) (0.040)
= 75.0 × 106 Pa
=
I
2.133 × 10−6
σ max = 75.0 MPa ◭ (top or bottom, just to right of B)
σ max =
91
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1
P
P
Mmax = 4
+ 7500 +
− 4500
2
5
5
4P
=
+ 6000 N · m
5
which gives
4P
+ 6000 = 32 000 or P = 32 500 N = 32.5 kN ◭
5
From the above shear force diagram,
Mmax = (1/2) (9) [9 (w0 )cr ] = 40.5 (w0 )cr , which gives
5.15
40.5 (w0 )cr = 21464.8 N · m
80(9.81) = 784.8 N
B
A
(w0 )cr = 530 N/m
Therefore, for w0 = 530 N/m, the shear force diagram is as
shown below.
C
x
12 m
784.8(1 − x/12)
784.8(x/12)
784.8(1 − x/12)
V
(N)
784.8(x/12)
784.8x(1 − x/12) N. m
Mmax =
Location of Mmax
x
Mmax = 784.8x 1 −
N·m
12
360(22.5)2
bh2
=
= 30.38 × 103 mm3
S =
6
6
2P
9P
1
(9) 900 +
= 4050 +
N·m
2
4
4
which gives
4050 +
784.8x(1 − x/12) × 103
Mmax
50 =
S
30.38 × 103
x = 2.4 m ◭
9P
= 21464.8 or P = 7740 N ◭
4
5.17
σ ult =
Find σ max .
400 N
200 N/m
1000 N
400 N
800 N
A
5.16
B
5m
(a) Consider the cross section to be an 200 mm by 100
mm rectangle from which a 100 mm by 200 mm rectangle
has been removed.
3
3
3
(200) (300)
(100) (200)
bh
=
−
I = Σ
12
12
12
2m
2014 N
586 N
858.5 N . m
586
V (N)
= 383.3 × 106 mm4 Q.E.D.
D
C
4m
1600 N . m
800
2.07 m
2.93 m
Location
of Mmax
-414
(b) Find P for σ max = 8400 N/mm2 given that w0 = 3000
N/m
-814
Mmax = 1600 N · m
(8.4) 383.3 × 106
σI
=
= 21464.8 N · m
Mmax =
c
150
S=
πd3
π(50)3
=
32
32
= 12.27 × 103 mm3
1600 × 103
Mmax
= 130.4 N/m2
=
σ max =
S
12.27 × 103
= 130.4 MPa ◭
First find critcal value of w0 that will cause V = 0 under
the load P . The FBD and shear force diagram for this
condition are shown below.
92
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5.18
5.20
Find the smallest allowable width b if σ w = 10 MPa.
Find the largest w0 if σ w = 120 MPa.
Two C380 × 50.4 sections with webs vertical
From Appendix B, section modulus: Sx = 2 688 × 103
mm3
w0
A
1m B
8m
RB = 5w0
4w0
D
C 1m
RC = 5w0
Mmax w0
V
–w0
Equilibrium analysis gives
at B
at D
–4w0
1
2
(1) (w0 ) +
Mmax = −
Mmax = σ w S becomes
MB = −2 (1) (0.5) = −1 kN · m
MD = 6 (1) − 2 (1) (0.5) = 5 kN · m
7.5 w0 =
Therefore, Mmax = 5000 N·m.
6M
σ= 2
bh
becomes
10 × 106 =
1
2
(4) (4w0 ) = 7.5 w0 N·m
120 × 106 (2) 688 × 10−6
w0 = 22.0 × 103 N/m = 22.0 kN/m ◭
6 (5000)
2
b (0.200)
Solving gives: b = 0.075 m = 75.0 mm ◭
5.21
Rectangular section: width 80 mm, height h.
Find minimum h if σ w = 20 MPa.
5.19
Find the largest w0 if σ w = 60 MPa; 40-mm diameter
shaft.
1
Mmax =
2
1
MB = − (1) (2.5) = −1.25 kN · m
2
1
MD =
(2) (5) = 5.00 kN · m = Mmax
2
Using σ = 6M/ bh2 gives
πr3
S=
4
#
"
π (0.020)3
9w0
6
= 60 × 10
32
4
9w0
3
3w0
=
4
4
32
Mmax = σS
becomes
Solving gives: w0 = 1340 N/m ◭
h =
r
6M
=
bσ w
s
6 (5.00 × 103 )
(0.080) (20 × 106 )
h = 0.1369 m = 136.9 mm ◭
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5.22
Mmax =
Find w0 if σ w = 60 MPa. For each tube: A = 80 mm2 .
I = ΣAy 2 = 80 2 × 1202 + 4 × 602 = 3.456 × 106 mm2
σw =
6Mmax
bh2
200L2
wL2
=
= 25L2 N · m
8
8
6 2.5L2 × 103
becomes 18 =
2
40 (40)
Solving gives L = 2.8 m ◭
5.25
Circular bar 40 mm diameter. Determine σ max at B.
(Assume flexure formula applies.)
1
Mmax =
2
Mmax =
w0 L2
L
w0 L
=
2
2
8
w0 (12)2
= 18 w0 N · m = 18 w0 × 103 N · mm
8
Using σ w = Mmax c/I gives
σw I
c
(60) 3.456 × 106
18 w0 × 103 =
120
w0 = 96 N/mm = 96000 N/mm ◭
Mmax =
ΣMD = 0:
4RA = 2 (2000 sin 30◦ ) + 2 (4000 sin 60◦ )
RA = 2232 N
Summing moments about B for left half of arch gives:
MB = 2 (2232) − 2 (4000 cos 60◦ ) = 464 N · m
5.23
The maximum bending stress is σ = M/S = M/ πr3 /4
which yields:
4 464 × 103
4MB
=
= 9.2 MPa ◭
(σ B )max =
πr3
π (40)3
Find the maximum w0 if σ w = 50 MPa.
Beam diameter: 50-mm S = πr3 /4 .
5.26
1
Mmax =
2
4w0
3
Two C200 × 28 back to back, cantilever beam, L = 4 m.
Find uniformly distributed load w, in addition to the
weight, if σ w = 120 MPa. For the beam, if load is w0 N/m,
maximum bending moment is
8w0
4
=
3
9
Mmax = σ w S gives:
8w0
= 50 × 106
9
2
3
π (0.025)
4
w0 (4)
w0 L2
=
= 8 w0 N · m
2
2
Properties from Appendix B:
Mmax =
!
y
14.4 mm
x
Solving yields w0 = 690 N/m ◭
Iy = 0.820 × 106 mm4
Sx = 180 × 103 mm3
A = 3550 mm2 m
= 27.9 kg/m
64.3 mm
L
The load due to the dead weight of the beam is
5.24
Loading is w = 200 N/m over entire span of length L.
Find maximum length L if cross section is 80 mm × 80 mm
and σ w = 18 MPa Letting L be measured in meter,
wd = 2 (27.9) (9.81) = 547.4 N/m
94
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From Appendix B for S310 × 74:
Sx = 829 × 103 mm3 , m/L = 74.0 kg/m.
(a) Vertical webs
= 360 × ×103 mm3
Mmax = σ w Sx = 120 × 106 360 × 10−6 = 8 w0
Sx = 2 180 × 10
3
Mmax = σ w Sx = 120 × 106
Solving gives w0 = 5400 N/m. Therefore, the intensity of
the applied load is
Solving gives: w = 22 110 N/m.
w (6)2
829 × 10−6 =
8
w0 = w − wd = 22 110 − (74.0) (9.81) = 21 500 N/m
w = w0 − wd = 5400 − 547.4 = 4850 N/m ◭
w0 = 21.5 kN/m ◭
(b) Horizontal webs
i
h
2
I = 2 Iy + Ad2 = 2 0.820 × 106 + 3550 (14.4)
= 3.126 × 106 mm4 = 3.112 × 10−6 m4
120 × 106 3.112 × 10−6
σw I
=
= 8 w0
Mmax =
y
0.0643
Solving gives w0 = 726.0 N/m. Therefore, the intensity of
the applied load is
5.29
Find largest value of P if σ w = 180 MPa.
w = w0 − wd = 726.0 − 547.4 = 178.6 N/m ◭
5.27
Two S250 × 52 welded as shown in Fig. P5.27. Uniformly
distributed load w0 in addition to wd , the dead weight of
the beam. Find maximum w0 , if σ w = 240 MPa. Let
w = w0 + wd .
At B
πd3
π (30)3
=
= 2650.7 mm3
32
32
P (600)
PL
=
= 75P N · mm
M =
8
8
M = 75P = σ w S = (180) (2650.7)
S =
Solving gives: P = 6361.7 N
At C
1
1
(9) (9w0 ) − (6) (6w0 ) = 22.5w N · m
2
2
From Appendix B-8 for S250 × 52: Sx = 55.1 × 103 mm3 .
Mmax = σ w Sx = (240) 55.1 × 103 = (22.5 × 103 w)
w = 587.7 N/m
Mmax =
πd3
π (37.5)3
=
= 5177.2 mm3
32
32
P (600)
PL
=
= 150P N · mm
M =
4
4
M = 150P = σ w S = (180) (5177.2)
S =
Solving gives: P = 6212.6 N
Comparing above values, Pmax = 6212.6 N ◭
5.30
w0 = w − wd = 587.7 − (52) (9.81) = 77.58 N/m ◭
Find largest w0 if σ w = 140 MPa.
5.28
An S310 × 74, simply supported with L = 6 m, uniformly
distributed load w0 in addition to wd , the dead weight of
the beam. Find maximum w0 , if σ w = 120 MPa.
Let w = w0 + wd .
2
Mmax =
w (6)
wL2
=
8
8
95
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At B
M = −w0 (0.75)
π R4 − r 4
S =
h 4R
0.75
2
4
= −0.2813 w0 N · m
4
π (0.025) − (0.0175)
i
= 9. 3254 × 10−6 m3
|M | = 0.2813 w0 = σ w S = 140 × 106 9.3254 × 10−6
=
4 (0.025)
MB = 6RA N · m and MD =
Solving gives: w0 = 4640 N/m
At D
At D
M = −w0 (1.0)
1.0
2
MD =
σ w bh2
6
1
(6) (6 w0 ) = 18 w0 N · m
2
(12) (180) (360)2
6
Solving, w0 = 2592
(18 w0 × 1000) =
N/m ◭
= −0.500 w0 N · m
For simultaneous maximums of w0 and P ,
3
π (0.05)
πd3
=
= 12.272 × 10−6 m3
32
32
|M | = 0.500 w0 = σ w S = 140 × 106 12.272 × 10−6
S =
MB = MD = 18 (2592) = 46656 = 6RA N · m
which gives RA = 7776 N
Using free-body diagram of entire beam (above),
Solving gives: w0 = 3440 N/m
ΣMD = 0
Comparing values, (w0 )max = 3440 N/m= 3.44 kN/m ◭
− 7776 (12) + P (6) − (2592) (6) (3) = 0
Solving, P = 23328 N ◭
5.31
5.33
Find smallest b for the square timber beam if σ w = 8 MPa.
Find maximum values of P and overhang b that can be
applied simultaneously if σ w = 14 MPa
MB = 6RA N · m and MD =
1
Mmax =
(0.6) (20) = 6.00 kN · m
2
r
6M
6M
6M
= 3
gives
b= 3
σw =
bh2
b
σw
r
3
3 6 (6.00 × 10 )
b =
= 0.1651 m = 165.1 mm ◭
8 × 106
At D
MD =
σ w bh2
6
1
(b) (400 b) = 200 b2 N · m
2
2
(14) (180) (360)
200 b2 (1000) =
6
Solving, x = 16.5 m ◭
For simultaneous maximums of w0 and P ,
2
MB = MD = 200 (16.5) = 6RA N · m
5.32
which gives RA = 9075 N
Using free-body diagram of entire beam (above),
Find maximum values of P and w0 that can be applied
simultaneously if σ w = 12 MPa
ΣMD = 0
2
− 9075 (12) + P (6) − (200) (16.5) = 0
Solving, P = 27225 N ◭
96
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At x = 6 m
5.34
6
M = −12 kN·m (tension in top, compression in bottom)
4
Find (σ T )max and (σ C )max : I = 30 × 10 mm ,
ctop = 200 mm, cbot = 80 mm.
M ctop
I
−12 × 103 (0.130)
= −
= 15.60 × 106 Pa
100 × 10−6
−12 × 103 (0.2)
M cbot
σ bot =
= −24.0 × 106 Pa
=
I
100 × 10−6
σ top = −
Above results give
(σ T )max = 25.0 MPa ◭
(σ C )max = 24.0 MPa ◭
Mmax = +18 kN·m (compression in top, tension in bottom)
5.36
3
18 × 10 (0.2)
M ctop
= −120.0 × 106 Pa
=−
I
30 × 10−6
18 × 103 (0.080)
M cbot
= 48.0 × 106 Pa
=
σ bot =
I
30 × 10−6
σ top = −
Above results give
Find (σ T )max and (σ C )max . I = 64 × 106 mm4 ,
ctop = 180 mm, cbot = 60 mm
(σ T )max = 48.0 MPa ◭
(σ C )max = 120.0 MPa ◭
5.35
Find (σ T )max and (σ C )max . I = 100 × 106 mm4 ,
ctop = 130 mm, cbot = 200 mm.
At section just to left of B
M = −24 kN·m (tension in top, compression in bottom)
−24 × 106 (180)
M ctop
= 67.5 MPa
=−
σ top = −
I
64 × 106
−24 × 106 (60)
M cbot
= −22.5 MPa
=
σ bot =
I
64 × 106
At section just to right of B
M = +14 kN·m (compression in top, tension in bottom)
14 × 106 (180)
M ctop
σ top = −
= −39.4 MPa
=−
I
64 × 106
14 × 106 (60)
M cbot
= 13.1 MPa
=
σ bot =
I
64 × 106
At x = 2.5 m
M = +12.5 kN·m (compression in top, tension in
bottom)
M ctop
I
12.5 × 103 (0.130)
= −
= −16.25 × 106 Pa
100 × 10−6
12.5 × 103 (0.2)
M cbot
= 25.0 × 106 Pa
=
σ bot =
I
100 × 10−6
σ top = −
Above results give
(σ T )max = 67.5 MPa ◭
(σ C )max = 39.4 MPa ◭
97
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ΣAi ȳi
(90 × 120)(60) + (40 × 120)(40)
=
ΣAi
90 × 120 + 40 × 120
= 53.85 mm = 0, 053 85 m
40(120)3
90(120)3
+2
I = ΣI¯i + ΣAi (ȳi − ȳ)2 =
12
36
+(90 × 100)(60 − 53.85)2 + (40 × 120)(40 − 53.85)2
5.37
ȳ =
Find (σ T )max and (σ C )max . I = 78 × 106 mm4 ,
ctop = 75 mm, cbot = 240 mm
= 18.061 × 106 mm4 = 18.061 × 10−6 m2
Stresses at B
(−2 × 103 )(0.12 − 0.053 85)
MB ytop
=−
I
18.061 × 10−6
6
= 7.325 × 10 Pa
MB ybot
(−2 × 103 )(−0.053 85)
σ bot = −
=−
I
18.061 × 10−6
6
= −5.963 × 10 Pa
σ top = −
At B
M = +3600 N·m (compression in top, tension in
bottom)
3600 × 103 (75)
M ctop
= −3.46 MPa
=−
σ top = −
I
78 × 106
3600 × 103 (240)
M cbot
= 11.1 MPa
=
σ bot =
I
78 × 106
Stresses at C
2.8 × 103 (0.12 − 0.053 85)
MC ytop
σ top = −
=−
I
18.061 × 10−6
= −10.255 × 106 Pa
2.8 × 103 (−0.053 85)
MC ybot
σ bott = −
=−
I
18.061 × 10−6
6
= 8.348 × 10 Pa
At D
M = −1800 N·m (tension in top, compression in
bottom)
−1800 × 103 (75)
M ctop
= 1.73 MPa
=−
σ top = −
I
78 × 106
−1800 × 103 (240)
M cbot
σ bot =
= −5.54 MPa
=
I
78 × 106
Above results give
∴ (σ T )max = 8.35 MPa ◭
(σ C )max = 10.26 MPa ◭
5.39
Find maximum value of W if (σ T )max = 30 MPa and
(σ C )max = 100 MPa I = 85 × 106 mm4 , ctop = 210 mm,
cbot = 90 mm
(σ T )max = 11.1 MPa ◭
(σ C )max = 5.54 MPa ◭
5.38
6 kN
2 kN
2.8 kN . m
B
A
1.0 m
1.2 m
C
4 kN
4
4.8 kN . m
V (kN)
–2
MB = −2 kN·m
At B and C
–2 kN . m
M = −4W kN·m (tension in top, compression in
bottom)
−4W × 106 (210)
M ctop
30 = −
W = 3 kN
=−
I
85 × 106
−4W × 106 (90)
M cbot
W = 23.6 kN
=
−100 =
I
85 × 106
MC = 2.8 kN·m
120
n.a. –
y
40
90
40
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At midspan
5.41
M = +12W kN·m (compression in top, tension in
bottom)
12W × 106 (210)
M ctop
W = 3.37 kN
=−
−100 = −
I
85 × 106
12W × 106 (90)
M cbot
30 =
W = 2.36 kN
=
I
85 × 106
For the inverted T-beam, find maximum P if
(σ T )max = 21 MPa and (σ C )max = 48 MPa
Letting y be measured from bottom of cross section:
y = ΣAy/ΣA = [(120 × 15) + (240 × 150)] / (240 + 120) =
105 mm
from which: ctop = 270 − 105 = 165 mm; cbot = y = 105
mm
Comparing above results, Wmax = 2.36 kN ◭
bh3
+ Ad2
12
3
3
(30) (240)
(120) (30)
2
2
+ 3600 (90) +
+ 7200 (45)
=
12
12
= 78.57 × 106 mm4
I = Σ
5.40
Find maximum value of W if (σ T )max = 60 MPa and
(σ C )max = 100 MPa. I = 75 × 106 mm4 , ctop = 84 mm,
cbot = 116 mm.
At B and E
M = −2P N·m (tension in top, compression in bottom)
−2P × 103 (165)
M ctop
=−
21 = −
I
78.57 × 106
P = 5000 N
−2P × 103 (105)
M cbot
−48 =
=
I
78.57 × 106
P = 17958.9 N
At B
M = −2W N·m (tension in top, compression in bottom)
M ctop
(−2 W ) (0.084)
=−
I
75 × 10−6
W = 26 790 N
M cbot
(−2 W ) (0.116)
−100 × 106 =
=
I
75 × 10−6
W = 32 330 N
60 × 106 = −
At D
M = +4 P N·m (compression in top, tension in bottom)
4 P × 103 (165)
M ctop
−48 = −
=−
I
78.57 × 106
P = 5714.2 N
4 P × 103 (105)
M cbot
=
21 =
I
78.57 × 106
P = 3928.5 N
At E
M = + 1.125 W N·m (comp. in top, tension in bottom)
(+ 1.125 W ) (0.084)
M ctop
=−
I
75 × 10−6
W = 79 370 N
M cbot
(+ 1.125 W ) (0.116)
60 × 106 =
=
I
75 × 10−6
W = 34 480 N
−100 × 106 = −
Comparing above results, Pmax = 3928.5 N ◭
Comparing above results, Wmax = 26.8 kN ◭
99
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5.42
5.44
Loading on cantilever beam, L = 10 m, varies linearly
from zero at free end to w0 at the support. Find maximum
value of w0 if (σ T )max = 40 MPa and (σ C )max = 100 MPa.
I = 38 × 106 mm4 , ctop = 90 mm, cbot = 180 mm
The bending moment is always negative (tension in top,
compression in bottom)and its maximum value occurs at
the support. Let w0 be measured in kN/m.
1
−50
10
Mmax = − (10) (w0 )
=
w0 N · m
2
3
3
50
× 106 × w0 (90)
M ctop
3
40 = −
w0 = 1.01 kN/m
=−
I
38 × 106
50
− × 106 × w0 (180)
M cbot
3
w0 = 1.27 kN/m
=
−100 =
I
38 × 106
Find the ratio b/h that maximizes the section modulus.
Letting the diameter
of the log be D, the
constraint is
b 2 + h2 = D 2
or h2 = D2 − b2
(1)
For a rectangle the section
modulus is S = bh2 /6, which gives
b D 2 − b2
bD2
b3
S=
=
−
6
6
6
To maximum the section modulus,
d bD2
b3
dS
=
−
db
db
6
6
2
2
dS
D
b
=
−
= 0 or D2 = 3b2
db
6
2
Comparing above results, (w0 )max = 1.01 kN/m ◭
5.43
Find the ratio a/b that maximimizes the section modulus.
Substituting Eq. (2) into Eq. (1) yields
Consider the area to be composed of four triangles (1-4)
and the rectangle (5).
The moment of inertia for one triangle about its base is
h2 = D2 − b2 = 3b2 − b2 = 2b2
√
from which, b/h = 1/ 2 = 0.707 ◭
bh3
a4
=
12
12
The moment of intertia
of the rectangle about
its centriodal axis is
5.45
Simply supported beam, L = 12 m, uniformly distibuted
load over entire length w = 17.5 kN/m. Find the lightest
W-shape if σ w = 120 MPa. What is the actual maximum
bending stress for the beam selected?
bh3
2 (b − a) (2a)3
=
12
12
4a3 b 4a4
−
=
3
3
Therefore, the moment of inertia for the total area
becomes
4 3
4a3 b
a
4a b 4a4
I=4
+
=
−
− a4
12
3
3
3
2
wL2
17.5 (10)
=
= 315 kN · m
8
8
315 × 103
Mmax
=
Smin =
σw
120 × 106
= 2.625 × 10−3 m3 = 2625 × 103 mm3
Mmax =
Referring to Appendix B, try W610 × 113:
Sx = 2880 × 103 mm3 , and m/L = 113 kg/m. [weight per
unit length: 113 (9.81) = 1109 N/m]. The total distributed
load intensity, including the weight of the beam:
Because c = a, the section moduls (S = I/c) becomes:
4a2 b
1 4a3 b
I
− a4 =
− a3
S= =
a
a
3
3
w0 = (17.5000 + 1.109) × 103 = 18.609 × 103 N/m
2
18.609 × 103 (12)
w0 L2
Mmax =
=
= 335.0 × 103 N · m
8
8
335.0 × 103
Mmax
=
σ max =
S
2880 × 10−6
= 116.3 × 106 Pa = 116.3 MPa
To maximize the section modulus:
dS
8ab
d 4a2 b
=
− a3 =
− 3a2 = 0
da
da
3
3
3a2 =
8ab
3
(2)
which gives
a
8
=
◭
b
9
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Because 116.3 MPa is less than the working stress (120
MPa) the lightest W-shape is
W610 × 113; actual maximum bending stress is
116.3 MPa ◭
Because 102.3 MPa is less than the working stress (108
MPa) the lightest S-shape is
S460 × 81.4; actual maximum bending stress is
102.3 MPa ◭
5.46
5.48
Repeat Prob. 5.45 if L = 8 m and w = 12 kN/m.
Select the lightest W-shape if σ w = 120 MPa. What is the
actual maximum bending stress for the beam selected?
2
wL2
12 (8)
Mmax =
=
= 96.0 kN · m
8
8
96.0 × 103
Mmax
=
Smin =
σw
120 × 106
= 0.800 × 10−3 m3 = 800 × 103 mm3
Referring to Appendix B, try W460 × 52:
Sx = 944 × 103 mm3 , and m/L = 52.0 kg/m. [weight per
unit length: 52.0 (9.81) = 510 N/m]. The total distibuted
load intensity, including the weight of the beam:
Mmax = (2) (2 w0 ) +
Using w0 = 8000 N/m,
w0 = (12.000 + 0.510) × 103 = 12.510 × 103 N/m
12.510 × 103 (8)2
w0 L2
Mmax =
=
= 100.1 × 103 N · m
8
8
100.1 × 103
Mmax
=
σ max =
S
944 × 10−6
= 106.0 × 106 Pa = 106.0 MPa
Mmax = 6 (8000) = 48000 N · m
48000 × 103
Mmax
Smin =
=
= 400 × 103 mm3
σw
120
Referring to Appendix B, try W200 × 46.1: Sx = 451 × 103
mm3 .
Because the maximum moments for both the applied load
and the dead weight of the beam (wd = 46.1 kg/m) both
occur at midspan, they can be added.
Because 106.0 MPa is less than the working stress (120
MPa) the lightest W-shape is
W460 × 52; actual maximum bending stress is
106.0 MPa ◭
wd =(46.1)(9.81) = 452.24 N/m
#
"
(452.24) 103 (8)2
wd L2
× 106
= 48 +
Mmax =6w0 +
8
8
5.47
Simply supported beam, L = 12 m, load P = 45000 N at
its midpoint. Find the lightest S-shape if σ w = 108 MPa.
What is the actual maximum bending stress for the beam
selected?
PL
45000 (12)
Mmax =
=
= 135000 N · m
4
4
Mmax
135000 × 103
Smin =
=
= 1.25 × 106 mm3
σw
180
=(48 + 3.6) × 103 = 51.6 × 106 N · mm
51.6 × 106
Mmax
σ max =
=
= 114.4 MPa
S
451 × 103
Because 114.4 MPa is less than the working stress (120
MPa) the lightest W-shape is
W200 × 46.1; actual maximum bending stress is
114.4 MPa ◭
Referring to Appendix B, try S460 × 81.4 Sx = 1.46 × 106
mm3 .
Because the maximum moments for both the applied load
and the dead weight of the beam (wd = 81.4 kg/m) both
occur at midspan, they can be added.
5.49
Repeat Prob. 5.48 if w0 = 20000 N/mm
2
wd = 81.4 kg/m × (9.81m/s ) = 798.5N/m
2
1
(2) (2 w0 ) = 6w0
2
From the solution to Prob. 5.48,using w0 = 20000 N/m,
2
(798.5) (12)
P L wd L
+
= 135000 +
4
8
8
= 135000 + 14373 = 149373 N · m
Mmax
149373 × 103
σ max =
= 102.3 MPa
=
S
1.46 × 106
Mmax = 6w0 = 6 (20000) = 120000 N · m
= 120 × 106 N · mm
120 × 106
Mmax
Smin =
=
= 1000 × 103 mm3
σw
120
Mmax =
101
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1
(13.3) (39.88) = 265.2 kN · m
2
265.2 × 106
Mmax
= 105.2 MPa
σ max =
=
S
2520 × 103
Because 105.2 MPa is less than the working stress (108
MPa) the lightest W-shape is
Referring to Appendix B, try W410 × 60 Sx = 1060 × 103
mm3 .
Because the maximum moments for both the applied load
and the dead weight of the beam (wd = 60 kg/m) both
occur at midspan, they can be added.
Mmax =
wd = (60)(9.81) = 588.6 N/m
wd L2
Mmax = 121.5w0 +
= 120 × 106
8
(588.6) (10−3 ) (8000)2
+
8 = (120 + 4.7) 106 = 124.7 × 106 N · mm
124.7 × 106
Mmax
σ max =
= 117.6 MPa
=
S
1060 × 103
Because 117.6 MPa is less than the working stress (120
MPa) the lightest W-shape is
W610×101; actual maximum bending stress is 105.2 MPa ◭
5.51
40 kN/m
160 kN(+)
96 kN(+)
C
A
B
4m
2.4 m
204.8 kN(–)
51.2 kN(–)
96
W410×60; actual maximum bending stress is 117.6 MPa ◭
115.2 kN . m
32.77 kN . m
51.2
V
(kN) 1.28 m
5.50
Select the lightest W-shape if σ w = 120 MPa. What is the
actual maximum bending stress for the beam selected?
–108.8
Mmax = 115.2 kN·m occurs at B.
S=
Mmax 115.2 × 103
=
= 960 × 10−6 m3 = 960 × 103 mm3
σw
120 × 106
From Appendix B: S380×64 has S = 973 × 103 mm3 ,
w = 64.0(9.81) = 627. 8
The actual loading is 40 × 103 + 627.8 = 40.63 kN/m.
40.63
= 117.01 kN · m
40
117.01 × 103
Mmax
= 120.3 × 106 Pa < σ w
=
σ max =
S
973 × 10−6
∴ Choose S380×64, resulting in σ max = 120.3 MPa ◭
∴ Mmax = 115.2
1
(14) (28) = 196.0 kN · m = 196.0 × 103 N · m
2
196 × 106
Mmax
=
= 1633.3 × 103 mm3
Smin =
σw
120
Mmax =
5.52
Referring to Appendix B, try W610 × 101:
Sx = 2520 × 103 mm3 .
Because the maximum moments for the applied load and
the dead weight of the beam (wd = (101)(9.81) = 990.8
N/m) do not occur at the same cross section, we must
redraw the diagrams using a distributed load with
intensity 2.99 kN/m.
14 kN/m
70 kN
70 kN. m
B
A
Beam AB
S=
5m
2m
35 kN
C
B
35 kN 35 kN
35 kN
Mmax = 70 kN·m occurs at A
70 × 103
Mmax
=
= 466.7×10−6 m3 = 466.7×103 mm3
σw
150 × 106
From Appendix B: W360×32.9 has S = 475 × 103 mm3 ,
w = 32.9(9.81) = 322.9 N/m.
The contribution of weight to the moment at A is
wL2
322.7(2.0)2
=
= 645.5 N · m
2
2
102
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so that the actual maximum stress is
70 × 103 + 645.5
σ max =
= 148.7 × 106 Pa < σ w
475 × 10−6
5m
30 kN
∴ Choose W360×32.9, resulting in σ max = 148.7 MPa ◭
Beam BC
20 kN/m
A
5m
D
12 kN/m
E
5m
260 kN 30 kN 30 kN
30 kN
Beam DE
Mmax = 30 (2.5) − 12 (2.5) (1.25) = 37.5 kN·m
Mmax
37.5 × 103
S=
=
109 = 312.5 × 103 mm3
6
σw
120 × 10
Mmax occurs at midspan
Mmax = 35(2.5) − 14(2.5)(1.25) = 43.75 kN · m
43.75 × 103
Mmax
=
= 291.7 × 10−6 m3
S =
σw
150 × 106
= 291.7 × 103 mm3
Referring to Appendix B, try W310 × 28.3:
Sx = 349 × 103 mm3 , and m/L = 28.3 kg/m. (Due to
weight of beam: Mmax = wL2 /8, also occurs at midspan.)
28.3 (9.81) (5)2
37.5 × 103 +
8
3
= 38.37 × 10 N · m
From Appendix B: W310×28.3 has S = 349 × 103 mm3 ,
w = 28.3(9.81) = 277. 6 N/m.
The contribution of weight to Mmax is
Mmax =
wL2
277.6(5)2
=
= 867.5 N · m
8
8
38.37 × 103
Mmax
= 109.9×106 Pa = 109.9 MPa
=
S
349 × 10−6
Because 109.9 MPa < 120 MPa, the lightest W-shape is:
W310 × 28.3; actual σ max is 109.9 MPa◭
σ max =
Hence the actual maximm stress is
43.75 × 103 + 867.5
Mmax
=
σ max =
S
349 × 10−6
6
= 127.8 × 10 Pa < σ w
Note that when the weight of DE is included, the force at
D becomes 30 + [0.0283 (9.81) (5) /2] = 30.7 kN.
∴ Choose W310×28.3, resulting in σ max = 127.8 MPa ◭
5.55
5.53
Rectangular cross section 60 mm wide and 140 mm high.
The vertical shear force is 32 kN.
Steel beams spaced 4 m apart. What is the lightest
S-shape if σ w = 90 MPa.
(a) Shear stress at neutral axis.
32 × 103
3
3V
=
= 5.71×106 Pa = 5.71 MPa ◭
τ=
2A
2 (0.060) (0.140)
4.5 m
R
1.5 m
(b) Shear stress 30 mm above the neutral axis.
w0
bh3
12
3
(0.060) (0.140)
I =
12
I = 13.72 × 10−6 m4
I =
w0 = (1000) (4.5) (4) = 18000 kg/m = 176580 N/m
1
1
R = w0 L = (176580) (4.5) = 397305 N
2
2
Mmax = 1.5R = 1.5 (397305) = 595957.5 N · m
S=
595957.5 × 103
Mmax
=
= 3311 × 103 mm3
σw
180
V A′ y′
Ib
32 × 103 (0.060 × 0.040) (0.050)
τ =
(13.72 × 10−6 ) (0.060)
τ =
From Appendix B, choose S610 × 158 (S = 3930 × 103
mm3 ) ◭
τ = 4.66 × 106 Pa = 4.66 MPa ◭
5.54
Select the lightest W-shape for each beam if σ w = 120
MPa. What is the actual maximum bending stress in each
beam?
103
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5.56
5.59
For a circular cross section show that τ = 4V / 3πr2 .
Find the maximum shear stress in the beam. V = 100 kN.
′ ′
VAy
Ib
V πr2 /2 (4r/3π)
τ =
(πr4 /4) (2r)
4 V
Q.E.D.
τ =
3 πr2
Maximum shear stress occurs at NA.
τ =
Ay = ΣAy ′ ; A1 = A2 = 9000 mm2
5.57
18000 y = (9000) (15) + (9000) (180) gives y = 97.5 mm
#
"
(300) (30)3
2
+ (9000) (97.5 − 15)
I=
12
"
#
3
(30) (300)
2
+
+ (9000) (180 − 97.5) = 190.7 × 106 mm4
12
330 − 97.5
= 810844 mm3
Q = A′ y ′ = (330 − 97.5) (30)
2
100 × 103 (810844)
τ max =
= 14.2 MPa ◭
(190.7 × 106 ) (30)
Find maximum value of length L if τ w = 0.8 MPa
Vmax = 900L N
240 (300)3
180 (240)3
I =
−
12
12
= 332.64 × 106 mm4
Q = (150 × 240 × 75) − (180 × 120 × 60)
= 1.404 × 106 mm3
τ = V Q/Ib becomes:
(900L) 1.404 × 106
0.8 =
(332.64 × 106 ) (60)
5.60
Solving gives L = 12.6 m ◭
Find smallest allowable value of b if τ w = 1.0 MPa.
5.58
Determine vertical shear stress at (a) neutral axis; (b) 4
in. above NA. V = 9000 N
From solution to Prob. 5.57, at neutral axis:
I = 332.64 × 106 mm4 and Q = 1.404 × 106 mm3
(a) At neutral axis: b = 60 mm
From above shear diagram, Vmax = 20 kN.
(9000) 1.404 × 106
VQ
=
τ =
Ib
(332.64 × 106 ) (60)
τ = 0.63 MPa ◭
3 Vmax
τw =
2 A
3 20 × 103
1.0 × 10 =
2
b2
6
Solving gives b = 0.1732 m = 1.732 mm ◭
(b) At 120 mm above NA:
Q = (240 × 30 × 135) = 972000 mm3
At 120 mm, b = 60 mm
τ=
VQ
(9000) (972000)
=
= 0.44 MPa ◭
Ib
(332.64 × 106 ) (60)
104
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(a) Maximum shear stress occurs at NA.
100 × 103 280 × 10−6
Vmax Q
τ max =
=
Ib
(45.87 × 10−6 ) (0.020)
5.61
= 30.52 × 106 Pa = 30.52 MPa ◭
(b) Percentage of shear force carried by the web.
Shear stress just below junction of flange and web
Beam (a)
I=2
Q = A1 y ′1 + A2 y ′2 = 120 × 20 × 90 = 216 × 103 mm3
100 × 103 216 × 10−6
Vmax Q
=
τ =
Ib
(45.87 × 10−6 ) (0.020)
t(4a)3
80 3
+ 4(at)(2a)2 =
a t
12
3
= 23.55 × 106 Pa = 23.55 MPa
Q is the first moment of the area drawn with thick line:
Shear force carried by the web
Q = (2at)a + (at)(2a) = 4a2 t
V (4a2 t)
3 V
VQ
=
= ◭
τ max =
80 3
It
20 at
a t t
3
2
(30.52 − 23.55) = 28.20 MPa
3
Vweb = τ avg Aweb = 28.20 × 106 (0.160 × 0.020)
τ avg = 23.55 +
= 90.2 × 103 N = 90.2 kN
Beam (b)
I = Ibeam (a) + 2
t(4a)3
80 3
32
112 3
=
a t + a3 t =
a t
12
3
3
3
%=
Q is the same as for beam (a) because AB contributes
nothing (QAB = 0).
Vweb
90.2
(100) =
(100) = 90.2% ◭
V
100
5.63
V (4a2 t)
3 V
VQ
=
=
◭
∴ τ max =
112 3
It
28 at
a t t
3
Solve Prob. 5.62 if height of web is 200 mm instead of
160 mm.
5.62
The vertical shear force is V = 100 kN. Compute (a) τ max ,
and (b) percentage of V carried by the web.
3
3
(120) (240)
(100) (200)
−
= 71.57 × 106 mm4
12
12
Q = A1 y′1 + A2 y′2 = (120 × 20 × 110) + (20 × 100 × 50)
= 364 × 103 mm3
I =
3
(a) Maximum shear stress occurs at NA.
100 × 103 364 × 10−6
Vmax Q
=
τ max =
Ib
(71.57 × 10−6 ) (0.020)
3
(120) (200)
(100) (160)
−
= 45.87 × 106 mm4
12
12
Q = A1 y ′1 + A2 y ′2 = (120 × 20 × 90) + (80 × 20 × 40)
= 280 × 103 mm3
I =
= 25.43 × 106 Pa = 25.43 MPa ◭
105
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(b) Percentage of shear force carried by the web.
5.66
Shear stress just below junction of flange and web
Find the shear stress at section C, 30 mm above bottom of
beam.
Q = A1 y ′1 + A2 y ′2 = 120 × 20 × 110 = 264 × 103 mm3
100 × 103 264 × 10−6
Vmax Q
τ =
=
Ib
(71.57 × 10−6 ) (0.020)
= 18.44 × 106 Pa = 18.44 MPa
Shear force carried by the web
2
(25.43 − 18.44) = 23.10 MPa
3
Vweb = τ avg Aweb = 23.10 × 106 (0.200 × 0.020)
τ avg = 18.44 +
= 92.4 × 103 N = 92.4 kN
%=
Q = A′ y ′ = (100 × 30) (62.5 − 15) = 142.5 × 103 mm3
At section C, V = 4.667 kN.
Vweb
92.4
(100) =
(100) = 92.4% ◭
V
100
τ =
4.667 × 103 142.5 × 10−6
VQ
=
Ib
(19.3 × 10−6 ) (0.100)
τ = 345 × 103 Pa = 345 kPa ◭
5.64
Determine the shear force that causes τ max = 1 MPa.
5.67
Q = (142.5 × 180 × 90) − (120 × 150 × 75)
Q = 958500 mm3
3
3
(142.5) (360)
(120) (300)
I =
−
12
12
6
4
I = 284 × 10 mm
Plot the shear stress at 30 mm intervals on the section
that carries the maximum shear stress.
τ max Ib
Q
(1) 284 × 106 (22.5)
= 6666.7 N ◭
V =
958500
V =
5.65
τ=
Determine largest safe
value ofw0 if τ max = 2 MPa
22.5
Q = (150 × 22.5) 60 +
2
+ (60 × 22.5) (30)
Q = 281 × 103 mm3
(2250) Q
Q
VQ
=
= 28.125 × 10−6
Ib
(80 × 106 ) b
b
(Q = A′ y′ )
Dist. from top
Q τ = 28.125 × 10−6
∗
3
(mm)
(Q/b) (kPa)
mm
30
(30 × 30) (150)
127
60
(30 × 60) (135)
211
90
(30 × 90) (120)
309
120
(30 × 120) (105)
354
150
(30 × 150) (90)
380
165 (NA)
(165 × 30) (82.5)
383 (τ max ) ◭
180
(30 × 180) (75)
380
210
(30 × 210) (60)
354
240−
(30 × 240) (45)
304
240+
(30 × 240) (45)
76
τ max Ib
Q
(2) 38 × 106 (22.5)
2 w0 =
281 × 103
Vmax = 2 w0 N =
Solving gives w0 = 3043 N/m ◭
∗− In flange, b = 30 mm, at 8+ , b = 120 mm
106
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5.70
Calculate the maximum shear stress in the beam.
15 kN/m
6m
225 kN
18 m
135
V
(kN)
5.68
204 mm
6m
225 kN
20.3
mm
90
–90
–135
622 mm
27.7 mm
For one S610 × 180:
I = 1320 × 106 mm4
Q = Qflange + Qweb
27.7
Q = (204) (27.7) 311 −
2
1
(311 − 27.7)
+ (20.3) (311 − 27.7)
2
What is the maximum safe value of P if τ w = 8 MPa?
Q = 2.49 × 106 mm3
For both S18 × 70:
135 × 103 2.49 × 106
Vmax (2Q)
Vmax Q
=
=
(2I) (2b)
2Ib
2 (1320 × 106 ) (20.3)
= 6.27 MPa ◭
τ max =
τ max Ib
Vmax =
Q
8 × 106 15.52 × 10−6 (0.020)
1.5P =
0.020 × 0.110 × 0.055
Solving gives P = 13 680 N = 13.68 kN ◭
5.71
Find the maximum shear stress for a beam with the cross
section shown if Vmax = 220 MPa.
#
"
3
3
(120) (60)
(90) (120)
2
I =
+2
+ (90 × 120) (90)
12
12
5.69
Determine the largest permissible value of w0
if τ w = 1.2 MPa.
= 203 × 106 mm4
At neutral axis (b = 120 mm)
Q = (30 × 120) (15) + (90 × 120) (90)
Q = 1.026 × 106 mm3
Vmax Q
Ib
180 × 103 1.026 × 106
τ max =
(203 × 106 ) (120)
τ max = 7.58 MPa
τ max =
3
3
(120) (160)
(80) (120)
−
= 29.44 × 106 mm4
12
12
Q = (120 × 80 × 40) − (80 × 60 × 30)
Q = 240 × 103 mm3
1.2 × 106 29.44 × 10−6 (0.040)
τ max Ib
1.5w0 =
Vmax =
Q
240 × 10−6
I =
At 30 mm above or below neutral axis (b = 90 mm)
Q = (90 × 120) (90) = 972000 mm3
180 × 103 (972000)
Vmax Q
τ max =
=
= 9.58 MPa
Ib
(203 × 106 ) (90)
Comparing above values, τ max = 9.58 MPa ◭
Solving gives w0 = 3925 N/m = 3.93 kN/m ◭
107
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5.72
5.74
What is the maximum shear stress in a beam with the
cross section shown (C250 × 45 section with fin attached)
caused by the shear force V = 200 kN?
(a)
2h/3− y
12
76
A'
8.96
NA
I = 4.41 × 106 mm4 11.1
2h/3
b
y
77
h/3
a
h
a
2
h−y
3
2
1
a 2
2
A′ = b
h−y =
h−y
2
3
2h 3
1
2
y′ = y +
h−y
(Centroidal coordinate of A′ )
3 3
2 a 2
1 2
′ ′
Q = Ay =
h−y
y+
h−y
2h 3
3 3
1 2
1 2
Q
=
h−y y+
h−y
b
2 3
3 3
2
1
2 2 1
1
1
=
h−y
h+ y =
h + hy − y 2
3
9
3
27
9
3
Shear stress at base of fin (b = 12 mm)
76
= 42.83 × 103 mm3
Q = (12 × 76) 8.96 +
2
200 × 103 42.83 × 10−6
VQ
τ =
=
Ib
(4.41 × 10−6 ) (0.012)
b =
τ = 161.9 × 106 Pa = 161.9 MPa
Shear stress at neutral axis [b = 2(11.1) = 22.2 mm]
77 − 8.96
= 51.39 × 103 mm3
Q = 22.2 (77 − 8.96)
2
200 × 103 51.39 × 10−6
VQ
=
τ =
Ib
(4.41 × 10−6 ) (0.0222)
Q/b is maximized when d(Q/b)/dy = 0:
τ = 105.0 × 106 Pa = 106.0 MPa
Comparing above values, τ max = 161.9 MPa ◭
2
1
h− y =0
9
3
y=
1
h
6
1
1
h+ h
9
18
Q.E.D.
(b)
∴
5.73
2h/3−y
y
A'
Q
b
=
max
τ max =
2h/3
1
2
h− h
3
6
=
1 2
h
12
VQ
h2 /12
V
=V 3
=3
◭
Ib
ah /36
ah
b
5.75
h/3
a
Simply supported beam (length L, width b, and height h)
carries a concentrated load P at midspan. If L = 10h,
determine the ratio σ max /τ max .
2
b = a
3 1 2
2
2
2
1
′
h =
a
h = ah
A = b
2
3
2 3
3
9
1
2
2
y′ =
h = h
3 3
9
4 2
2
2
ah
h =
ah
Q = A′ y ′ =
9
9
81
4 2
ah
V
VQ
8V
81
τ =
= 3 =
◭
Ib
3 ah
2
ah
a
36
3
From V -M diagrams: Vmax =
P
PL
Vmax L
; Mmax =
=
2
4
2
Using τ max = (3/2) Vmax /A,
Vmax =
2
τ max bh (1)
3
Using σ max = Mmax /S = 6Mmax/bh2 , we find
Mmax = σ max bh2 /6, which gives
Vmax L
σ max bh2
=
2
6
or Vmax =
σ max bh2
3L
108
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Substituting L = 10h,
Vmax =
σ max bh2
σ max bh
=
3 (10h)
30
(2)
Equating Eqs. (1) and (2) ,
σ max bh
2
τ max bh =
3
30
σ max
= 20 ◭
τ max
w0 L2
w0 L L
w0 L
=
; Mmax =
2
2
3
6
4
2
πd
d3
4d
πd
I =
=
Q = A′ y ′ =
64
8
6π
12
16w0 L2
Mmax c
d
w0 L2
σ max =
/ πd4 /64 =
=
I
6
2
3πd3
3 4 w0 L
d
πd
8w0 L
Vmax Q
=
/
(d) =
τ max =
Ib
2
12
64
3πd2
2L
3πd2
16w0 L2
σ max
=
=
◭
3
τ max
3πd
8w0 L
d
Vmax =
5.76
Determine the maximum allowable value of w0 given:
glue: τ w = 0.5 MPa, wood: τ w = 0.8 MPa and σ w = 10
MPa.
5.78
Vmax = 1.25 w0 N;
1
Mmax = (1.25) (1.25 w0 ) = 0.78125 w0 N · m
2
3
bh3
(150) (250)
I =
=
= 195.3 × 106 mm4
12
12
Determine the maximum value of W if σ max = 10 MPa
and τ max = 0.5 MPa
Shear stress in glue (max. at 100 mm from top or bottom)
τ Ib
Q
(0.5) 195 × 106 (150)
1.25 w0 =
(150 × 100 × 75)
Vmax =
or w0 = 10416 N/m
Shear stress in wood (max. at NA: τ max = 3Vmax / (2A) )
2Aτ max
Vmax =
3
2 (150 × 250) (0.8)
1.25 w0 =
3
From shear diagram:
Vmax = 2W N
W
45W
1
Mmax =
(9) W +
=
N·m
2
4
8
or w0 = 16000 N/m
Bending stress in wood (max. at top or bottom:
Bending stress Using σ max = Mmax /S = 6Mmax /bh2 gives
σ max = Mmax c/I )
σ max bh2
Mmax =
6
45W
(10)
(270) (450)2
× 103 =
8
6
6
(10) 195.3 × 10
σ max I
=
c
125
6
(10)
195.3
×
10
2.5 w0 × 103 =
or w0 = 6249.6 N/m
125
Mmax =
Comparing above values, (w0 )max = 6249.6 N/m ◭
or W = 16200 N
Shear stress Using τ max = 3Vmax / (2A) gives
2Aτ max
3
2 (270 × 450) (0.5)
or W = 20250 N
2W =
3
Comparing above values, Wmax = 16200 N ◭
Vmax =
5.77
Find ratio σ max /τ max in terms of L and d. Assume τ max
occurs at NA and it is uniformly distributed.
109
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5.79
Vmax = P N when x = 0
PL
P (8)
L
Mmax =
=
= 2P N · m when x = = 4 m
4
4
2
A W250 × 49.1 cantilever beam carries a uniformly
distributed load of intensity w0 over its entire length 4-m
length. Determine the ratio σ max /τ max .
Mmax c
(2P )(0.2)
20 × 106 =
I
573.3 × 10−6
P = 28.7 × 103 N
From Appendix B: S = 574×10−6 m3 , I = 71.2×10−6 m4
σw =
Vmax = w0 L = 4w0 N, Mmax = w0 L2 /2 = 8w0 N · m
τw =
11.0
247
Vmax Q
Ib
5 × 106 =
P = 30.2 × 103 N
7.37
(P )(1.90 × 10−3 )
(573.3 × 10−6 )(0.02)
∴ The largest safe value of the force is P = 28.7 kN ◭
202
Bending stress
σ max =
5.81
8w0
Mmax
= 13 937w0
=
S
574 × 10−6
t
300 mm
Shear stress
t
247 11.0
Q = (11.0) (202)
−
2
2
247
1
247
− 11.0 (7.37)
− 11.0
+
2
2
2
3
240 mm
d
180
mm
ΣAi di
(300t)(240) + (120t)(240) + (480t)(120)
=
ΣAi
300t + 120t + 480t
= 176 mm
I = Σ I i + Ai (di − d)2
= 0 + (300t)(240 − 176)2 + 0 + (120t)(240 − 176)2
(2t)(240)3
2
+
+ (480t)(120 − 176)
12
d =
3
= 308.8 × 10 mm
(4w0 ) 308.8 × 10−6
VQ
=
= 2354w0
τ max =
Ib
(71.2 × 10−6 ) (0.00737)
Ratio of stresses
σ max
13 937w0
=
= 5.93 ◭
τ max
2354w0
= (5.5 × 106 )t mm4
Q = 300t(240 − 176) = 19200t mm3
(1st moment of shaded area)
5.80
(a)
50
Mmax = P L = 12000(5) = 60000 N · m
(60000 × 103 ) (176)
Mmax c
50 =
σ max =
I
(5.5 × 106 )t
t = 38.4 mm ◭
100
100
NA 20
I =
20(400)3
+4
12
400
50(100)3
+ (50 × 100)(150)2
12
= 573.3 × 106 mm4 = 573.3 × 10−6 m4
(b)
Vmax = P = 12 kN
Vmax Q
(12 × 103 )(19200t)
τ glue =
=
= 0.35 MPa ◭
Ib
(5.5 × 106 )t(120)
Q = (20 × 200)(100) + 2(50 × 100)(150)
= 1.90 × 106 mm3 = 1.90 × 10−3 m3
Note that this result is independent of t.
110
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5.82
τ Ib
|Vmax | =
Q
4w0 =
1.4 × 106
Solving gives w0 = 2201 N/m.
222.1 × 10−6 (0.025)
882.8 × 10−6
Comparing above values, w0 max = 1851 N/m ◭
5.84
(σ T )max = 100 MPa, (σ C )max = 160 MPa and τ w = 40
MPa (a) Show d = 65 mm and I = 50 × 106 mm4 .(b)
Determine the largest allowable value of W.
(a)
[(43200) − (27000)] y = 43200 (90) − 27000 (105)
mm. Q.E.D.
270(300)3 240(180)3
−
= 490.86 × 106 mm4
12
12
Q = (270 × 150)(75) − (240 × 90)(45) = 2.065 × 106 mm3
I =
Taking axis at the top of cross section:
3
3
(180) (150)
(240) (180)
2
−
− (16200) (180 − 65)
3
3
= 49.8 × 106 mm4 Q.E.D.
I =
(P × 103 )(150)
P = 32724 N
(490.86 × 106 )
(P/2)(2.065 × 106 )
4=
P = 57049 N
(490.86 × 106 ) (30)
Mmax c
I
Vmax Q
τ max =
Ib
10 =
σ max =
y = 65
∴ The largest safe value is P = 32724 N ◭
5.83
Determine the largest allowable value of w0 if σ w = 10
MPa and τ w = 1.4 MPa.
3
(b) ctop = 180 − 65 = 115 mm, cbot = 65 mm
3
(200) (300)
(175) (250)
−
= 222.1 × 106 mm4
12
12
Q = (200 × 150 × 75) − (175 × 125 × 62.5)
Q = 882.8 × 103 mm3
I =
Use σ top = −M ctop /I and σ bot = M cbot /I.
At x = 2 m
M = −2W N·m (T in top, C in bottom)
−2W × 103 (115)
W = 21652 N
100 = −
49.8 × 106
−2W × 103 (65)
−160 =
W = 61292 N
49.8 × 106
At x = 5 m:
M = W N·m (C in top, T in bottom)
W × 103 (115)
W = 69287 N
−160 = −
49.8 × 106
W × 103 (65)
100 =
W = 76615 N
49.8 × 106
1
(4) (4 w0 ) = 8w0 N · m
2
10 × 106 222.1 × 10−6
σw I
8w0 =
|Mmax | =
c
0.150
Solving gives w0 = 1851 N/m.
Shear stress at x = 2 m: Vmax = 2W N. Use τ = V Q/Ib.
|Vmax | = 4 w0 N, |Mmax | =
40 =
(2W ) (2) [30 × 115 × (115/2)]
(49.8 × 106 ) (60)
W = 150623 N
Comparing above results, Wmax = 21652 N ◭
111
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5.85
For the thin-walled tube, (a) find largest allowable w0
if σ w = 100 MPa and (b) compute the corresponding
maximum shear stress.
(b) ctop = 165 − 48.75 = 116.25 mm, cbot = 48.75 mm
Use σ top = −Mmax ctop /I and σ bot = Mmax cbot /I.
Mmax = P N·m (C in top, T in bottom,)
(P ) 116.25 × 10−3
6
−140 × 10 = −
11.918 × 10−6
(P ) 48.75 × 10−3
6
48 × 10 =
11.918 × 10−6
Comparing above results, Pmax = 11.73 kN◭
(a) Maximum allowable w0
σI
c
100 × 106
7.564 × 10−6
0.100
w0 = 10 506 N/m = 10.51 kN/m ◭
0.720 w0 =
P = 11 730 N
Vmax = P N. Use τ = V Q/Ib.
165 − 48.75
= 101.36 × 103 mm3
Q = (15) (165 − 48.75)
2
(P ) 101.36 × 10−6
6
30 × 10 =
P = 52 910 N
(11.918 × 10−6 ) (0.015)
i
π h
4
4
I =
(0.200) − (0.195) = 7.564 × 10−6 m4
64
πr2 4r
2r3
d3
′ ′
=
=
Q = Ay =
2
3π
3
12
i
1 h
3
3
Q =
(0.200) − (0.195) = 48.76 × 10−6 m3
12
M=
P = 14 350 N
5.87
(b) Maximum shear stress
Determine largest safe value of w0 if σ w = 24 MPa and
τ w = 1 MPa
Vmax = 1.5w0 = 1.5 (10.51) = 15.77 kN
15.77 × 103 48.76 × 10−6
Vmax Q
τ max =
=
Ib
(7.564 × 10−6 ) (5 × 10−3 )
τ max = 20.3 × 106 Pa = 20.3 MPa ◭
5.86
(σ T )max = 48 MPa, (σ C )max = 140 MPa and τ w = 30
MPa (a) Show d = 48.75 mm and I = 11.918 mm4 .(b)
Determine the maximum allowable value of P.
3
3
(210) (180)
(240) (240)
−
= 174.4 × 106 mm4
12
12
Q = (240 × 30 × 105) + (30 × 90 × 45) = 877.5 × 103 mm3
8w0 × 103 (120)
Mmax c
σ max =
24 =
I
174.4 × 106
(a) [2 (150 × 15)] y = (150 × 15) (7.5 + 90)
I=
y = 48.75 mm Q.E.D.
(150) (15)3
2
+ (150 × 15) (48.75 − 7.5)
12
3
(15) (150)
2
+
+ (15 × 150) (90 − 48.75)
12
I = 11.918 × 106 mm2 Q.E.D.
I =
Solving gives w0 = 4360 N/m
112
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Vmax Q
τ max =
Ib
(4w0 ) 877.5 × 103
1=
(174.4 × 106 ) (30)
2P
kN
12 +
3
2P
1
2P
Mmax = 12 +
(1) − (1) (8) = 8 +
kN · m
3
2
3
Bending stress: σ w = 6Mmax / bh2 , or Mmax = σ w bh2 /6
10 × 106 (0.150) (0.250)2
2P
3
× 10 =
8+
3
6
Vmax =
Solving gives w0 = 1490.6 N/m
Comparing above values, (w0 )max = 1490.6 N/m ◭
5.88
Solving yields P = 11.44 kN
Find the smallest allowable cross-sectional dimensions of
the square beam if σ w = 8 MPa and τ w = 1.0 MPa.
Shear stress: τ w = 3Vmax / (2A) , or Vmax = 2τ w A/3
2P
2
1.2 × 106 (0.150 × 0.250)
× 103 =
12 +
3
3
Solving yields P = 27.0 kN
Comparing above values, Pmax = 11.44 kN ◭
5.90
(σ T )max = 40 MPa, (σ C )max = 80 MPa and τ w = 24 MPa
(a) Show d = 50 mm and I = 15.96 mm4 .
(b) Determine the largest allowable value of W.
(a) A = 2 (20 × 140) + (140 × 20)
A = 5600 + 2800 = 8400 mm2
Vmax = 4 kN, Mmax = 4 kN · m
Bending stress: σ w = 6Mmax / bh2 = 6Mmax/b3 which
gives
r
r
3
3 6 (4 × 10 )
3 6Mmax
=
= 0.1442 m = 144.2 mm
b=
6
σw
8 × 10
Shear stress: τ w = 3Vmax / (2A) = 3Vmax / 2b2 which gives
s
r
3Vmax
3 (4 × 103 )
=
b=
= 0.0775 m = 77.5 mm
2τ w
2 (1.0 × 106 )
(8400) y = 5600 (70) + 2800 (10)
mm. Q.E.D.
y = 50.0
3
I = 2
(20) (140)
2
+ (20 × 140) (20)
12
!
(140) (20)3
+ (140 × 20) (40)2
12
I = 15.96 × 106 mm2 Q.E.D.
+
Comparing above values, the smallest dimensions are
(144.2 mm×144.2 mm) ◭
5.89
Find maximum allowable value of P if σ w = 10 MPa and
τ w = 1.2 MPa
(b) ctop = 90 mm., cbot = 50 mm
Use σ top = −M ctop /I and σ bot = M cbot /I.
At x = 1 m:
Mmax = −W N·m (T in top, C in bottom)
(−W ) (0.090)
15.96 × 10−6
(−W
) (.050)
−80 × 106 =
15.96 × 10−6
40 × 106 = −
W = 7093 N
W = 25 540 N
113
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At x = 3 m:
M = 2W N·m (C in top, T in bottom)
(2W ) (0.090)
15.96 × 10−6
(2W ) (0.050)
40 × 106 =
15.96 × 10−6
−80 × 106 = −
Beam (b)
I = 959 × 106 mm4 (Same as for beam (a))
Q = (300 × 60)(150) = 2.7 × 106 mm3
Fw I
(2 × 1250)(959 × 106 )
e =
=
= 22.2 mm ◭
VQ
40000(2.7 × 106 )
W = 7093 N
W = 6384 N
Shear stress at x = 1+ m: Vmax = 3W N.
[τ max = Vmax Q/Ib]
24×106 =
(3W ) (2) (0.020 × 0.090) (0.045)
(15.96 × 10−6 ) (0.040)
W = 31 530 N
5.93
Comparing above results, Wmax = 6.38 kN ◭
25
50
25
90
5.91
90
50
NA
Cross section is three 80-mm by 200-mm planks bolted
together. Loading causes τ = 1.2 MPa. Determine largest
permissible spacing if Fw = 6.5 kN.
QN A = (80 × 100 × 50)
+ (200 × 80 × 140)
I = 2
QN A = 2640 × 103 mm3
Qf lange = (200 × 80 × 140)
3
25(180)3 50(90)3
+
+ (50 × 90)(95)2
12
12
= 111.6 × 106 mm4 = 111.6 × 10−6 m4
Q = (50 × 90)(95) = 472.5 × 103 mm3 = 472.5 × 10−6 m3
3
Qf lange = 2240 × 10 mm
The shear force in the bolt is (note that the bolt is in
double shear)
Fw I
Fw I
=
V Qf lange
(τ Ib/QN A) Qf lange
6.5 × 103 2640 × 10−6
Fw QN A
=
e =
τ bQf lange
(1.2 × 106 ) (0.080) (2240 × 10−6 )
= 0.0798 m = 79.8 mm ◭
e =
F =
1 V Qe
1 (3 × 103 )(472.5 × 10−6 )(0.10)
= 635.1 N
=
2 I
2
111.6 × 10−6
πd2
πd2
635.1 = (5 × 106 )
4
4
d = 0.01272 m = 12.72 mm ◭
F = τw
5.92
300 mm × 60 mm
NA
(a)
5.94
240 mm × 60 mm
360 mm × 60 mm
180 mm × 60 mm
NA
A simply-supported 6 m plank has the cross section
shown. The working stress for bending is σ w = 6 MPa and
the allowable shear force in each screw is Fw = 1250 N
(e = 150 mm). What limit should be placed on the weight
W of a person who walks across the plank?
(b)
Beam (a)
300(360)3 180(240)3
−
= 959 × 106 mm4
12
12
Q = (180 × 60)(150)
= 1.62 × 106 mm3 (1st moment of shaded area)
(2 × 1250)(959 × 106 )
Fw I
=
= 37 mm ◭
e =
VQ
40000(1.62 × 106 )
I =
"
#
"
#
(105) (45)3
(360) (15)3
2
I = 2
+2
+ (5400) (30)
12
12
= 11.51 × 106 mm4
114
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5.96
WL
W (6)
=
4
4
Mmax = 1.5W N · m. ( when person is at midspan)
Vmax = W N (when person is over a support)
Mmax =
The diameter of rivets is 22 mm. The total moment of
inertia about the NA is I = 4770 × 106 mm4 . The working
stresses are τ w = 100 MPa for rivet shear and (σ b )w = 280
MPa for bearing of the web plate. The shear force is
V = 450 kN. Determine the largest allowable spacing of
the rivets.
If bending governs:
(6) 11.51 × 106
(1.5 × 10 )W =
(22.5 + 15)
W = 1227.7 N (too large)
σw I
Mmax =
c
3
If force in screws governs:
(2Fw ) I
V =
Qe
(2 × 1250) 11.51 × 106
W =
(360 × 15 × 30) (150)
W = 1184.2 N ◭
Q = Qangles + Qplate
Q = 2 (2430) (491.1) + (300 × 10) (515)
= 3932 × 103 mm3
If rivet shear governs, allowable force carried by a rivet:
5.95
2
π (0.022)
100 × 106
4
Fw = 38.01 × 103 N = 38.01 kN (too large)
Fw = Aτ w =
The load P causes σ max = 10 MPa. The bolts (e = 360
mm) are tightened to a tensile stress σ T = 108 MPa.
Determine the bolt diameters d so that the shear between
the planks can be transmitted by friction only (µ = 0.3).
If bearing stress in the plate governs, allowable force
carried by a rivet (double shear):
3
I=
1
1
drivet tweb (σ b )w = (0.022) (0.010) 280 × 106
2
2
Fw = 30.80 × 103 N = 30.80 kN (use this)
(180) (360)
= 699.8 × 106 mm4
12
Fw =
2 × 30.80 × 103 4770 × 10−6
(2Fw ) I
=
e =
VQ
(450 × 103 ) (3932 × 10−6 )
e = 0.1661 m = 166.1 mm ◭
5.97
P
P
N, Mmax = (2.5) = 1.25P N · m
Vmax =
2
2
σ max I
Mmax =
c
Two C380 × 60 channels, drivet = 19 mm, e = 200 mm.
What is largest allowable shear force V in the beam if
τ w = 100 MPa for rivet shear, and (σ b )w = 220 MPa for
bearing in the channels.
From Appendix B for one C380 × 60:
(10) 699.8 × 106
(1.25 × 10 )P =
180
P = 31102 N (V = 15551 N)
3
A = 7610 mm2
mm
The required shear force is
F =
V Qe
(15551) (180 × 120 × 120) (360)
= 20736 N
=
I
699.8 × 106
2
πd
F = µN = µσ T A 20736 = (0.3) (108)
4
I x = 3.82 × 106 mm4
tweb = 13.2
x
19.8
For the composite section:
h
i
2
Ix = 2 I x + Ad2 = 2 3.82 × 106 + (7610) (19.8)
Solving gives d = 28.5 mm ◭
Ix = 13.607 × 106 mm4
Q = (7610) (19.8) = 150.68 × 103 mm3
115
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If rivet shear governs, allowable force carried by a rivet:
5.99
2
π (0.019)
100 × 106
4
Fw = 28.35 × 103 N = 28.35 kN (use this)
Fw = Arivet τ w =
The beam is fabricated with 15-mm rivets. The maximum
shear force is V = 65 kN. (a) Determine largest spacing e
if τ w = 100 MPa. (b) Compute the corresponding
maximum bearing stress exerted by the rivets.
For C180 × 14.6:
If bearing stress in the plate governs, allowable force
carried by a rivet (single shear):
Fw = drivet tweb (σ b )w = (0.019) (0.0132) 220 × 106
I 1 = 0.398 × 106 mm4
Fw = 55.18 × 103 N = 55.18 kN (too large)
2 × 28.35 × 103 13.607 × 10−6
(2Fw ) I
V =
=
eQ
(0.200) (150.68 × 10−6 )
A1 = 1850 mm2
y 1 = (251/2) + 5.33 − 13.7 = 117.13 mm
For W250 × 17.9:
I 2 = 22.4 × 106 mm4
V = 25.6 × 103 N = 25.6 kN ◭
A2 = 2280 mm2
5.33
5.98
NA _y
C2
Two C380 × 60 channels riveted back to back and joined
with 255 by 15 mm plates. drivet = 22.5 mm Vmax = 120
MPa (a) Determine largest spacing e if τ w = 30 MPa. (b)
Compute the corresponding maximum bearing stress in
the channels.
From Appendix B for one C380 × 60: I = 145 × 106 mm4
255 mm
15 mm
_
y1
251
5.33
(1850) (117.13)
ΣAi y i
=
= 52.47 mm
ΣAi
1850 + 2280
2
IN A = 0.398 × 106 + (1850) (117.13 − 52.47)
2
+ 22.4 × 106 + (2280) (52.47)
y =
16.5 mm
(a)
381 mm
13.7
C1
C
y2 = 0
IN A = 36.81 × 106 mm4
Q = 1850 (117.13 − 52.47) = 119.62 × 103 mm3
!
π (0.015)2
6
= 17 671 N
Fw = τ w Arivet = 100 × 10
4
(2 × 17 671) 36.81 × 10−6
(2Fw ) IN A
=
e =
VQ
(65 × 103 ) (119.62 × 10−6 )
e = 0.1676 m = 167.3 mm ◭
13.2 mm
89.4 mm
For the composite section: IN A = 2Ichannel +2Iplate :
3
bh
2
IN A = 2I + 2
+ Aplate d
12
#
"
3
(255)
(15)
2
+ (255 × 15) (187.5)
IN A = 2 145 × 106 + 2
12
(b)
Note that t = 5.33 mm for both shapes
σb =
IN A = 559 × 106 mm4
Fw
17 671
=
drivet t
(0.015) (0.00533)
= 221 × 106 Pa = 221 · MPa ◭
Q = (255 × 15) (187.5) = 0.717 × 106 mm3
(a) The allowable shear force in one rivet:
5.100
2
π (22.5)
= 11928 N
4
(2 × 11928) 559 × 106
(2Fw ) IN A
=
= 186 mm ◭
e=
VQ
(100 × 103 ) (0.717 × 106 )
11928
Fw
=
= 32 MPa ◭
(b) σ b =
drivet tflange
(22.5) (16.5)
Fw = τ A = (30)
Find maximum tensile and compressive bending stresses.
Neglect the weight of the beam.
y=
ΣAi y i
(200 × 40) (100 + 220)
=
= 160.0 mm
ΣAi
2 (200 × 40)
116
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#
(40) (200)3
2
+ (8000) (60)
I =
12
"
#
3
(200) (40)
2
+
+ (8000) (60)
12
"
5.102
22.5 mm
180 mm
180 mm
244.6 mm
I = 85.33 × 106 mm4
22.5 mm
64.6 mm
NA
360 mm
d = 355.4 mm
60 mm
ΣAi di
(600 × 225)(300) − (360 × 180)(240)
=
ΣAi
600 × 225 − 360 × 180
= 355.4 mm
I = Σ I i + Ai (di − d)2
d =
Mmax = 1.8 kN·m, compression in top, tension in bottom
1.8 × 103 80 × 10−3
M ctop
σ top = −
=−
I
85.33 × 10−6
6
σ top = −1.688 × 10 Pa
225(600)3
+ (600 × 225)(300 − 355.4)2
12
180(360)3
− (360 × 180)(240 − 355.4)2
−
12
= 2901.5 × 106 mm4
244.6
64.6
Q = (244.6 × 225)
− (64.6 × 180)
2
2
6
3
= 6.36 × 10 mm
=
1.8 × 103 160 × 10−3
M cbot
=
σ bot =
I
85.33 × 10−6
6
σ bot = 3.38 × 10 Pa
Therefore, the maximum tensile and compressive bending
stresses are
VQ
V (6.36 × 106 )
5=
Ib
(2901.5 × 106 )(45)
V = 102647 N = 102.65 kN ◭
τ max =
(σ T )max = 3.38 MPa and (σ C )max = 1.688 MPa ◭
5.103
5.101
Analysis of ABC
Steel beam (E = 200 GPa), base b = 30 mm, height
h = 7.5 mm. When the couple M0 is applied, the radius of
curvature of the NA is 500 mm. Determine the value of
M0. Neglect the weight of the beam.
15 kN
A
7.5 kN . m B
0.5 m
1.0 m
10 kN
5 kN
5
V
(kN)
3
(30) (7.5)
I=
= 1054.68 mm4
12
C
–10
200 × 103 (1054.68)
EI
M0 =
=
ρ
5000
M0 = 42187 N · mm = 42.187 N · m ◭
2.5
M
(kN . m)
60(80)3 50(70)3
−
= 1.1308 × 106 mm4
12
12
= 1.1308 × 10−6 m4
I =
Mmax = 10 kN · m (just to the right of B)
(10 × 103 )(0.04)
Mmax ymax
=
σ max =
I
1.1308 × 10−6
6
= 353.7 × 10 Pa ◭
117
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Analysis of BDE
5.106
Mmax = 7.5 kN · m (in segment BD)
π(0.075)3
πd3
=
= 41.42 × 10−6 m2
S =
32
32
7.5 × 103
Mmax
= 181.1 × 106 Pa
=
σ max =
S
41.42 × 10−6
5.104
Determine the magnitude and location of the maximum
bending stress. Neglect the weight of the beam.
150
150
75
NA
75
σ max =
150
20
6 10 500 × 103
6Mmax
= 16.2 MPa ◭
=
2
bh2
(120) (180)
(Occurs at top or bottom of beam, just to the left of D)
3
3
40(150)
150(300)
+
= 348.8 × 106 mm4
12
12
= 348.8 × 10−6 m4
Q = (150 × 150)(75) + (40 × 75)(37.5) = 1.80 × 106 mm3
I=
−3
5.107
Find (σ T )max and (σ C )max , ctop = 210 mm, cbot = 60 mm.
3
= 1.80 × 10 m (1st moment of shaded area)
τ w Ib
(2 × 106 )(348.8 × 10−6 )(0.04)
V =
= 15 500 N ◭
=
Q
1.80 × 10−3
5.105
The width of the stepped beam is 60 mm Determine the
maximum bending stress due to the 3600-N·m couple.
Neglect the weight of the beam.
At section B
M = −12 000 N·m (tension in top, compression in
bottom)
M ctop
I
−12 000 × 103 (210)
= 37 MPa
σ top = −
68 × 106
−12 000 × 103 (60)
M cbot
σ bot =
= −10.6 MPa
=
I
68 × 106
σ top = −
Just to the left of B
M = −2880 N·m
6 2880 × 103
6 |M |
= 35.6 MPa
=
σ max =
2
bh2
(60) (90)
At section D
M = 6000 N·m (compression in top, tension in bottom)
M ctop
I
6000 × 103 (210)
σ top = −
= −18.5 MPa
68 × 106
6000 × 103 (60)
M cbot
= 53 MPa
=
σ bot =
I
68 × 106
σ top = −
Just to the right of B
M = 720 N·m
6 720 × 103
6M
σ= 2 =
= 20 MPa
2
bh
(60) (60)
Comparing above two values, σ max = 35.6 MPa ◭
118
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Above results give
(σ T )max = 37 MPa ◭
Including the weight of the beam:
(σ C )max = 18.5 MPa ◭
5000 N
5000 N
2224.8 N
370.8 N/m
3708 N
2224.8 N
6m
10 m
6m
5.108
9078.8 N
Find largest allowable value of P if (σ T )w = 90 MPa and
(σ C )w = 108 MPa, ctop = 102 mm, cbot = 138 mm,
I = 186 × 106 mm4 .
9078.8 N
7224.8
V (N)
5000
1854
–1854
Location of Mmax
–5000
–7224.8
5000 + 7224.8
(6) = 36674.4 N · m
2
Mmax
36674.4 × 103
σ max =
= 91 MPa
=
S
403 × 103
Mmax =
At section B
M = − 3P N·m (tension in top, compression in bottom)
−3P × 103 102
M ctop
=−
P = 54706 N
90 = −
I
186 × 106
−3P × 103 (138)
M cbot
P = 48522 N
=
−108 =
I
186 × 106
S250 × 37.8; actual maximum bending stress is 91 MPa ◭
5.110
At section D
Determine the largest value of W if σ w = 120 MPa for the
S380 × 74 section.
M = + 1.5P N·m (compression in top, tension in
bottom)
1.5P × 103 102
M ctop
P = 131294 N
−108 = −
=−
I
186 × 106
1.5P × 103 (138)
M cbot
P = 80870 N
=
90 =
I
186 × 106
Comparing above results, Pmax = 48522 N = 48.52 kN ◭
Mmax =
5.109
From Appendix B
Neglecting the weight of the beam:
Mmax = σ w Sx
1
(2.3) (2.3 W ) = 2.645 W N · m
2
For S380 × 74, Sx = 1060 × 103 mm3
2.645 W = 120 × 106 1060 × 10−6
W = 48 100 N = 48.1 kN ◭
Mmax = 5000(6) = 30000 N · m at the supports
Mmax
30000 × 103
S =
=
= 277.8 × 103 mm3
σw
108
Choose S250 × 37.8 which has S = 403 × 103 mm3 and
weighs 37.8 kg/m
weight = 37.8 × 9.81 = 370.8 N/m
119
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At section B and D
5.111
M = −4.5 w0 N/m (T in top, C in bottom)
Find largest allowable value of w0 if (σ T )w = 20 MPa and
(σ C )w = 80 MPa, ctop = 180 mm, cbot = 50 mm,
I = 36 × 106 mm4 .
M ctop
(−4.5 w0 ) (0.180)
=−
I
36 × 10−6
w0 = 889 N/m
(−4.5 w0 ) (0.050)
M cbot
=
−80 × 106 =
I
36 × 10−6
w0 = 12 800 N/m
20 × 106 = −
Comparing above results, (w0 )max = 889 N/m ◭
5.113
Simply supported beam with uniformly distributed load
12000 N/m over entire 6 m length. Find lightest S-shape if
σ w = 108 MPa. What is the actual stress in the beam
selected?
w0 L2 (12000 × 10−3 )(6000)2
=
= 54 × 106 N · mm
Mmax =
8
8
Mmax
54 × 106
Smin =
=
= 0.5 × 106 mm3
σw
108
At section B and D
M = −0.5 w0 N/m (T in top, C in bottom)
(−0.5 w0 ) (0.180)
M ctop
=−
I
36 × 10−6
w0 = 8000 N/m
M cbot
(−0.5 w0 ) (0.050)
−80 × 106 =
=
I
36 × 10−6
w0 = 115 200 N/m
20 × 106 = −
At midspan
Referring to Appendix B, try S310 × 47.3: Sx = 593 × 103
mm3 . Because the maximum moments for both the
applied load and the dead weight occur at midspan, they
can be added.
For S310 × 47.3, mass = 47.3 kg/m
wd L2
w0 L2
+
= 54 × 106
Mmax =
8
8
(47.3) (9.81) 10−3 (6000)2
+
8
= 54 × 106 + 2.088 × 106 = 56.088 × 106 N · mm
M = +7.5 w0 N · m (C in top, T in bottom)
7.5 w0 (0.180)
M ctop
w0 = 2133 N/m
=−
I
36 × 10−6
M cbot
7.5 w0 (0.050)
20 × 106 =
w0 = 1920 N/m
=
I
36 × 10−6
−80 × 106 = −
Comparing above results, (w0 )max = 1920 N/m ◭
56.088 × 106
Mmax
=
= 94.6 MPa
S
593 × 103
Because 94.6 MPa is less than the working stress (108
MPa) the lightest S-shape is
σ max =
5.112
Solve Prob. 5.107 using b = 3 m as the overhang.
(Find largest allowable value of w0 if (σ T )w = 20 MPa and
(σ C )w = 80 MPa, ctop = 180 mm, cbot = 50 mm,
I = 36 × 106 mm4 ).
S310×47.3; actual maximum bending stress is 94.6 MPa ◭
5.114
Find the lightest W-shape if σ w = 120 MPa and calculate
the actual maximum bending stress in the beam selected.
40 kN
20 kN/m
5m
5m
120 kN
120 kN
120
20
V
(kN)
Because the bending moment is negative throughout the
beam, only the bending stresses at sections B (and D)
need be considered.
Mmax
−20
−120
120
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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1
(5) (120 + 20) = 350.0 kN · m
2
350.0 × 103
Mmax
=
Smin =
σw
120 × 106
−3
= 2.917 × 10 m3 = 2917 × 103 mm3
5.116
Mmax =
Vertical shear force is V = 60 kN. Determine the shear
stress (a) at the neutral axis, and (b) on the glued joint.
Referring to Appendix B, try W610 × 125:
3
Sx = 3210 × 10 mm
3
(200 × 40 × 20) + (20 × 100 × 90)
ΣAi y i
=
ΣAi
(200 × 40) + (20 × 100)
y = 34.0 mm
y =
m/L = 125 kg/m
wd = (m/L)g = 125(9.81) = 1226 N/m
Because the maximum moments for both the applied load
and the dead weight occur at midspan, they can be added.
wd L2
350.0 × 103 +
8
(1226) (10)2
= 365.3 × 103 N · m
= 350.0 × 103 +
8
365.3 × 103
Mmax
=
σ max =
S
3210 × 10−6
= 113.8 × 106 Pa = 113.8 MPa
Mmax =
"
#
3
(200) (40)
2
I =
+ (200 × 40) (34.0 − 20)
12
#
"
3
(20) (100)
2
+ (20 × 100) (90 − 34.0)
+
12
Because 113.8 MPa is less than the working stress (120
MPa), the lightest W-shape is
I = 10.57 × 106 mm4
(a) Shear stress at neutral axis (b = 200 mm)
W610 × 125; actual maximum stress is 113.8 MPa ◭
34.0
= 115.6 × 103 mm3
2 60 × 103 115.6 × 10−6
VQ
τ =
=
Ib
(10.57 × 10−6 ) (0.200)
Q = 200 × 34.0 ×
5.115
Find maximum shear stress
on the section if V = 60 kN.
τ = 3.28 × 106 Pa = 3.28 MPa ◭
(b) Shear stress on glue joint (b = 20 mm)
ΣAi y i
y =
ΣAi
(180 × 200 × 100) − (120 × 160 × 80)
y =
(180 × 200) − (120 × 160)
y = 122.86 mm
#
"
(180) (200)3
2
+ (180 × 200) (100 − 122.86)
I =
12
"
#
3
(120) (160)
2
−
+ (120 × 160) (80 − 122.86)
12
Q = 20 × 100 × (90 − 34) = 112.0 × 103 mm3
60 × 103 112.0 × 10−6
VQ
=
τ =
Ib
(10.57 × 10−6 ) (0.020)
τ = 31.8 × 106 Pa = 31.8 MPa ◭
5.117
The W360 × 262 section carries V = 650 kN. Calculate (a)
minimum shear stress in the web, (b) maximum shear
stress in the web, and (c) the percentage of V carried by
the web. (From Appendix B: I = 891 × 106 mm4 .)
I = 62.58 × 106 mm4
Maximum shear stress occurs at neutral axis (b = 60 mm)
122.86
= 452.8 × 103 mm3
Q = 2 30 × 122.86 ×
2
60 × 103 452.8 × 10−6
Vmax Q
=
τ max =
Ib
(62.58 × 10−6 ) (0.060)
33.3
81.01 MPa
90.31 MPa
386
319.4
21.1
τ max = 7.24 × 106 Pa = 7.24 MPa ◭
399
121
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(a) Minimum shear in web (just below junction of flange
and web)
5.119
386 − 33.3
= 2.343 × 106 mm3
2
650 × 103 2.343 × 10−3
VQ
τ =
=
Ib
(891 × 10−6 ) (0.0211)
Determine the ratio σ max /τ max .
Q = 399 × 33.3 ×
= 81.01 × 106 Pa = 81.01 MPa ◭
(b) Maximum shear in web (at neutral axis)
319.4 1 319.4
6
Q = 2.343 × 10 + (21.1)
2
2
2
= 2.612 × 106 mm3
650 × 103 2.612 × 10−3
VQ
τ =
=
Ib
(891 × 10−6 ) (0.0211)
Q = (240 × 30 × 165) + (150 × 30) (75) = 1.52 × 106 mm3
= 90.31 × 106 Pa = 90.31 MPa ◭
(c) Percentage of V carried by the web
Vmax =
2
(90.31 − 81.01) = 87.21 MPa
3
Vweb = τ avg Aweb = 87.21 × 106 (0.0211 × 0.3194)
τ avg = 81.01 +
Mmax =
w0 L2
8
w0 L2 c
Vmax Q
w0 LQ
Mmax c
=
τ max =
=
I
8I
Ib
2Ib
σ max
Lcb
(15000)(180)(30)
=
=
= 13.3 ◭
τ max
4Q
4(1.52 × 106 )
σ max =
= 587.7 × 103 N = 587.7 kN
%=
w0 L
2
Vweb
587.7
(100%) =
(100%) = 90.4 % ◭
V
650
5.118
5.120
Find the smallest allowable value of b if σ w = 8 MPa and
τ w = 1.0 MPa.
Find the largest value of W if σ w = 12 MPa
and τ w = 0.5 MPa.
(300) (360)3
(240) (300)3
−
= 626.4 × 106 mm4
12
12
Q = (300 × 180 × 90) − (240 × 150 × 75) = 2.16 × 106 mm3
I=
Vmax = 5000 N, Mmax = 5000 N · m
Bending stress: σ w = 6Mmax / bh2 = 6Mmax/b3 which
gives
r
r
6Mmax
6 (5000)
=
= 0.0612 m = 61.2 mm
b=
σw
8 × 106
Shear stress: τ w = 3Vmax / (2A) = 3Vmax / 2b2 which gives
s
r
3Vmax
3 (5000)
b=
=
= 0.0866 m = 86.6 mm
2τ w
2 (1.0 × 106 )
Comparing above values, bmin = 86.6 mm ◭
122
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Vmax = 0.79 W N Mmax = 2.925 W N · m
(12) 626.4 × 10
σw I
2.925 W × 1000 =
c
180
W = 14277 N
(0.5) 626.4 × 106 (60)
τ Ib
0.79 W =
Vmax =
Q
2.16 × 106
W = 11013 N
Mmax =
(a)
6
Mmax c
I
3P × 103 (120)
gives P = 10080 N
14 =
259.2 × 106
QN A = (240 × 120 × 60) − (120 × 60 × 30)
σ max =
= 1.512 × 106 mm3
V QN A
(10080/2) (1.512 × 106 )
τ max =
=
Ib
(259.2 × 106 ) (120)
= 0.245 MPa ◭
Comparing above values, Wmax = 11013 N ◭
(b) Q = 180 × 60 × 90 = 0.972 × 106 mm3
2 (1000) 259.2 × 106
2Fw I
=
= 105.8 mm ◭
e=
VQ
(10080/2) (0.972 × 106 )
5.121
The load W travels across the beam. Determine the
maximum value of W if σ w = 14 MPa and τ w = 1 MPa
5.123
(a) Find the maximum bending stress, and (b) the largest
permissible spacing of the bolts if Fw = 30 kN (each bolt).
From Appendix B, W200 × 100: A = 12 700 mm2 ,
Ix = 113 × 106 mm4 , depth = 229 mm.
3
(120) (540)
= 1574.6 × 106 mm4
12
W (12)
WL
=
= 3 W N · m (W at mispan)
Mmax =
4
4
Vmax = W N (W over a support)
3W × 103 (270)
Mmax c
W = 27215 N
14 =
σw =
I
1574.6 × 106
3V
3 (W )
τ max =
1=
W = 43200 N
2A
2 (120 × 540)
Comparing above results, Wmax = 27215 N ◭
I =
I = 2(Ix + Ad2 )
"
229
I = 2 113 × 106 + (12 700)
2
2 #
= 559.0 × 106 mm4
5.122
The maximum bending stress is σ max = 14 MPa. (a)
Determine τ max . (b) If Fw = 1000 N (each screw),
calculate the largest allowable spacing of the screws.
3
I=
Vmax = 150 N Mmax = 375 kN · m
375 × 103 (0.229)
Mmax c
=
(a) σ max =
I
559.0 × 10−6
6
σ max = 153.6 × 10 Pa = 153.6 MPa ◭
229
= 1.4542 × 106 mm3
(b)
Q = Ay = (12 700)
2
2 30 × 103 559.0 × 10−6
2Fw I
=
e =
VQ
(150 × 103 ) (1.4542 × 10−3 )
e = 0.1538 m = 153.8 mm ◭
3
(240) (240)
(120) (120)
−
= 259.2 × 106 mm4
12
12
Vmax =
P
N
2
Mmax =
P
(6) = 3P N · m
2
123
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5.124
1 X
(bi hi ) · yi
·
A i
X bi · (hi )3
2
I :=
+ bi · hi · (d − yi )
12
i
X
hi
H :=
d :=
A beam with the cross section shown carries a load w0
(N/m) over a simply supported span L = 4 m. (a) Find
largest value of w0 if σ max = 120 MPa. (b) Determine
largest spacing of the rivets (d = 10 mm) if τ w = 80 MPa.
For C100 × 10.8 (see Appendix B): A = 1370 mm2 ,
Ix = 0.177 × 106 mm4 .
7.5
102
i
7.5
11.7
c :=
C of channel
120
d
(H − d)
NA
σ max :=
"
2
I = 2 0.177 × 106 + 1370 (60 − 11.7) +
3
7.5 (120)
12
#
Mmax c
(2 w0 ) (0.060)
120 × 106 =
I
8.906 × 10−6
w0 = 8906 N/m ◭
Q = Ay = (1370) (60 − 11.7) = 66.17 × 103 mm3
π (0.010)2
= 6283 N
Fw = τ max Arivet = 80 × 106
4 2 (6283) 8.906 × 10−6
2Fw I
e =
=
VQ
(17 812) (66.17 × 10−6 )
e = 0.0950 m = 95.0 mm ◭
50
b :=  7.5  · mm
90

12
h :=  70  · mm
10


25
h :=  25  · mm
25
i := 1, 2..3
h2
h1
y2 := h1 +
y1 :=
2
2
h3
y3 := h1 + h2 +
2
X
b i · hi
A :=
C5.1 MathCad worksheet for Part
(a)

Distance from
neutral axis to
outermost fiber
M := 45 · N · m
Computations:
All coordinates are measures from the bottom of X section
(b) Vmax = 2 w0 = 2 (8906) = 17 812 N

Height of
X section
σ max = 4.028 × 107 Pa
Given:


100
b :=  75  · mm
50
σ max =
Given:

M ·c
I
H
2
otherwise
Moment of
inertia
C5.1 MathCad worksheet for Part
(b)
= 8.906 × 106 mm4
w0 (4)
w0 L
=
= 2 w0 N
Vmax =
2
2
2
w0 L2
w0 (4)
Mmax =
=
= 2 w0 N · m
8
8
(a)
if d ≥
Coordinate of
centroid
i
1 X
(bi hi ) · yi
·
A i
X bi · (hi )3
2
I :=
+ bi · hi · (d − yi )
12
i
X
hi
H :=
d :=
M := 2000·N·m
i
Computations:
All coordinates are measures from the bottom of X section
i := 1, 2..3
h2
h1
y2 := h1 +
y1 :=
Centroidal
2
2
coordinates
h3
y3 := h1 + h2 +
of rectangles
2
X
X-sectional
b i · hi
A :=
area
i
c :=
d
if d ≥
(H − d)
σ max :=
M ·c
I
H
2
otherwise
Centroidal
coordinates
of rectangles
X-sectional
area
Coordinate of
centroid
Moment of
inertia
Height of
X section
Distance from
neutral axis to
outermost fiber
σ max = 7.866 × 105 Pa
124
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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C5.2 MathCad worksheet for Part
(a)
C5.2 MathCad worksheet for Part
(b)
Given:
Given:
h1 := 30 · mm
L := 2 · m
h2 := 120 · mm
b := 25 · mm
w0 := 2000 · N · m−1
h(x) := h1 + (h2 − h1 ) ·
h1 := 38 · mm h2 := 112 · mm
L := 2 · m
b := 25 · mm
w0 := 2000 · N · m−1
x2
L2
h(x) := h1 + (h2 − h1 ) ·
Computations:
S(x) :=
Computations:
b · h(x)2
6
σ(x) :=
M (x) :=
w0 · x2
2
S(x) :=
σ(x) :=
Plotting range and increment
x := 0, 0.01 · L..L
1·108
1·108
8·107
8·107
6·107
4·107
2·107
x := 1.2 · m
Given
b · h(x)2
6
M (x)
S(x)
Stress (Pa)
Stress (Pa)
x := 0, 0.01 · L..L
0
x2
L2
M (x) :=
w0 · x2
2
M (x)
S(x)
Plotting range and increment
6·107
4·107
2·107
0
0.5
1
x (m)
1.5
0
2
Trial value for location of maximum stress
x := 1.4 · m
d
σ(x) = 0 x1 := Find(x) x1 = 1.1547 m
dx
σ max := σ(x1 ) σ max = 8.889 × 107 Pa
Given
0
0.5
1
x (m)
1.5
2
Trial value for location of maximum stress
d
σ(x) = 0 x1 := Find(x) x1 = 1.4332 m
dx
σ max := σ(x1 ) σ max = 8.535 × 107 Pa
125
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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C5.3 MathCad worksheet for Part
(a)
C5.3 MathCad worksheet for Part
(b)
Given:
Given:
h1 := 18 · mm
b := 4 · mm
P := 100000 · N
h2 := 30 · mm
L := 36 · m
h(x) := h1 + (h2 − h1 ) · sin
h1 := 26.55 · mm
b := 4 · mm
P := 100000 · N
π·x
L
h(x) := h1 + (h2 − h1 ) · sin
Computations:
b · h(x)2
S(x) :=
6
x := 0, 0.01 · L..L
π·x
L
Computations:
(L − x) · x
M (x) := P ·
L
σ(x) :=
h2 := 26.55 · mm
L := 36 · m
S(x) :=
M (x)
S(x)
b · h(x)2
6
M (x) := P ·
σ(x) :=
(L − x) · x
L
M (x)
S(x)
Plotting range and increment
x := 0, 0.01 · L..L
2·104
Plotting range and increment
3·104
1·104
Stress (Pa)
Stress (Pa)
1.5·104
5000
0
0
10
20
x (m)
2·104
1·104
30
0
0
10
20
x (m)
30
126
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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C5.4 MathCad worksheet
Given:
L := 8 · m
t := 0.5 · mm b := 4 · mm
h1 := 2 · mm h2 := 18 · mm
P := 4800 · N
h(x) := h1 + (h2 − h1 ) ·
x
L
Computations:
"
2 #
t · h(x)3
b · t3
h(x) + t
I(x) :=
+2·
+b·t·
12
12
2
h(x)
+t
2
h(x) + t t · h(x)2
+
Q(x) := b · t ·
2
8
c(x) :=
x := 0, 0.01 · L..L
(P · x) · c(x)
I(x)
P · Q(x)
τ (x) :=
I(x) · t
σ(x) :=
Plotting range and increment
1·104
Stress (Pa)
8000
6000
4000
2000
0
0
2
4
x (m)
6
8
127
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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