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•7–1.
If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear stress on the web at A.
Indicate the shear-stress components on a volume element
located at this point.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a,
QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3
Applying the shear formula,
VQA
20(103)[0.64(10 - 3)]
=
tA =
It
0.2501(10 - 3)(0.02)
= 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in
Fig. b.
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7–2. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the maximum shear stress in the beam.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
From Fig. a.
Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3
The maximum shear stress occurs at the points along neutral axis since Q is
maximum and thicknest t is the smallest.
tmax =
VQmax
20(103) [0.865(10 - 3)]
=
It
0.2501(10 - 3) (0.02)
= 3.459(106) Pa = 3.46 MPa
Ans.
473
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7–3. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear force resisted by the web
of the beam.
200 mm
A
20 mm
20 mm
B
V
300 mm
200 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4
12
12
For 0 … y 6 0.15 m, Fig. a, Q as a function of y is
Q = ©y¿A¿ = 0.16 (0.02)(0.2) +
1
(y + 0.15)(0.15 - y)(0.02)
2
= 0.865(10 - 3) - 0.01y2
For 0 … y 6 0.15 m, t = 0.02 m. Thus.
t =
20(103) C 0.865(10 - 3) - 0.01y2 D
VQ
=
It
0.2501(10 - 3) (0.02)
=
E 3.459(106) - 39.99(106) y2 F Pa.
The sheer force resisted by the web is,
0.15 m
Vw = 2
L0
0.15 m
tdA = 2
L0
C 3.459(106) - 39.99(106) y2 D (0.02 dy)
= 18.95 (103) N = 19.0 kN
Ans.
474
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*7–4. If the T-beam is subjected to a vertical shear of
V = 12 kip, determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flangeweb junction AB. Sketch the variation of the shear-stress
intensity over the entire cross section.
4 in.
4 in.
3 in.
4 in.
B
6 in.
A
V ⫽ 12 kip
Section Properties:
y =
INA =
1.5(12)(3) + 6(4)(6)
©yA
=
= 3.30 in.
©A
12(3) + 4(6)
1
1
(12) A 33 B + 12(3)(3.30 - 1.5)2 +
(4) A 63 B + 4(6)(6 - 3.30)2
12
12
= 390.60 in4
Qmax = y1œ A¿ = 2.85(5.7)(4) = 64.98 in3
QAB = y2œ A¿ = 1.8(3)(12) = 64.8 in3
Shear Stress: Applying the shear formula t =
tmax =
VQ
It
VQmax
12(64.98)
=
= 0.499 ksi
It
390.60(4)
Ans.
(tAB)f =
VQAB
12(64.8)
=
= 0.166 ksi
Itf
390.60(12)
Ans.
(tAB)W =
VQAB
12(64.8)
=
= 0.498 ksi
I tW
390.60(4)
Ans.
475
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•7–5.
If the T-beam is subjected to a vertical shear of
V = 12 kip, determine the vertical shear force resisted by
the flange.
4 in.
4 in.
3 in.
4 in.
B
6 in.
A
V ⫽ 12 kip
Section Properties:
y =
©yA
1.5(12)(3) + 6(4)(6)
=
= 3.30 in.
©A
12(3) + 4(6)
INA =
1
1
(12) A 33 B + 12(3)(3.30 - 1.5)2 +
(4) A 63 B + 6(4)(6 - 3.30)2
12
12
= 390.60 in4
Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2
Shear Stress: Applying the shear formula
t =
VQ
12(65.34 - 6y2)
=
It
390.60(12)
= 0.16728 - 0.01536y2
Resultant Shear Force: For the flange
Vf =
tdA
LA
3.3 in
=
L0.3 in
A 0.16728 - 0.01536y2 B (12dy)
= 3.82 kip
Ans.
476
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7–6. If the beam is subjected to a shear of V = 15 kN,
determine the web’s shear stress at A and B. Indicate the
shear-stress components on a volume element located
at these points. Show that the neutral axis is located at
y = 0.1747 m from the bottom and INA = 0.2182110-32 m4.
200 mm
A
30 mm
25 mm
V
(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)
y =
= 0.1747 m
0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)
I =
1
(0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2
12
+
1
(0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2
12
+
1
(0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182 (10 - 3) m4
12
B
250 mm
30 mm
125 mm
œ
QA = yAA
= (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10 - 3) m3
QB = yABœ = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10 - 3) m3
tA =
15(103)(0.7219)(10 - 3)
VQA
= 1.99 MPa
=
It
0.218182(10 - 3)(0.025)
Ans.
tB =
VQB
15(103)(0.59883)(10 - 3)
= 1.65 MPa
=
It
0.218182(10 - 3)0.025)
Ans.
7–7. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the maximum shear stress in the beam.
200 mm
A
30 mm
25 mm
V
B
250 mm
30 mm
Section Properties:
I =
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10) - 3 m3
tmax =
VQ
30(10)3(1.0353)(10) - 3
= 4.62 MPa
=
It
268.652(10) - 6 (0.025)
Ans.
477
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*7–8. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the shear force resisted by the web
of the beam.
200 mm
A
30 mm
1
1
(0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4
12
12
I =
Q = a
25 mm
V
B
0.155 + y
b (0.155 - y)(0.2) = 0.1(0.024025 - y2)
2
250 mm
30(10)3(0.1)(0.024025 - y2)
tf =
268.652(10)
-6
30 mm
200 mm
(0.2)
0.155
Vf =
L
tf dA = 55.8343(10)6
L0.125
= 11.1669(10)6[ 0.024025y -
(0.024025 - y2)(0.2 dy)
1 3 0.155
y ]
2 0.125
Vf = 1.457 kN
Vw = 30 - 2(1.457) = 27.1 kN
Ans.
•7–9. Determine the largest shear force V that the member
can sustain if the allowable shear stress is tallow = 8 ksi.
3 in.
1 in.
V
3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
= 1.1667 in.
1 (5) + 2 (1)(2)
I =
1
(5)(13) + 5 (1)(1.1667 - 0.5)2
12
+ 2a
1
b (1)(23) + 2 (1)(2)(2 - 1.1667)2 = 6.75 in4
12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
tmax = tallow =
8 (103) = -
VQmax
It
V (3.3611)
6.75 (2)(1)
V = 32132 lb = 32.1 kip
Ans.
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7–10. If the applied shear force V = 18 kip, determine the
maximum shear stress in the member.
3 in.
1 in.
V
3 in. 1 in.
1 in.
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
= 1.1667 in.
1 (5) + 2 (1)(2)
I =
1
(5)(13) + 5 (1)(1.1667 - 0.5)2
12
+ 2a
1
b (1)(23) + 2 (1)(2)(2 - 1.1667) = 6.75 in4
12
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
tmax =
18(3.3611)
VQmax
=
= 4.48 ksi
It
6.75 (2)(1)
Ans.
7–11. The wood beam has an allowable shear stress of
tallow = 7 MPa. Determine the maximum shear force V that
can be applied to the cross section.
50 mm
50 mm
100 mm
50 mm
200 mm
V
50 mm
I =
1
1
(0.2)(0.2)3 (0.1)(0.1)3 = 125(10 - 6) m4
12
12
tallow =
7(106) =
VQmax
It
V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)]
125(10 - 6)(0.1)
V = 100 kN
Ans.
479
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*7–12. The beam has a rectangular cross section and is
made of wood having an allowable shear stress of tallow =
200 psi. Determine the maximum shear force V that can be
developed in the cross section of the beam. Also, plot the
shear-stress variation over the cross section.
V
12 in.
8 in.
Section Properties The moment of inertia of the cross-section about the neutral axis is
I =
1
(8) (123) = 1152 in4
12
Q as the function of y, Fig. a,
Q =
1
(y + 6)(6 - y)(8) = 4 (36 - y2)
2
Qmax occurs when y = 0. Thus,
Qmax = 4(36 - 02) = 144 in3
The maximum shear stress occurs of points along the neutral axis since Q is
maximum and the thickness t = 8 in. is constant.
tallow =
VQmax
;
It
200 =
V(144)
1152(8)
V = 12800 16 = 12.8 kip
Ans.
Thus, the shear stress distribution as a function of y is
t =
12.8(103) C 4(36 - y2) D
VQ
=
It
1152 (8)
=
E 5.56 (36 - y2) F psi
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7–13. Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 20 kN.
12 mm
Section Properties:
INA
60 mm
1
1
=
(0.12) A 0.0843 B (0.04) A 0.063 B
12
12
V
= 5.20704 A 10 - 6 B m4
12 mm
80 mm
Qmax = ©y¿A¿
20 mm
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)
= 87.84 A 10 - 6 B m3
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Applying the shear formula
tmax =
VQmax
It
20(103)(87.84)(10 - 6)
=
5.20704(10 - 6)(0.08)
= 4 22 MPa
Ans.
7–14. Determine the maximum shear force V that the
strut can support if the allowable shear stress for the
material is tallow = 40 MPa.
12 mm
60 mm
Section Properties:
INA =
V
1
1
(0.12) A 0.0843 B (0.04) A 0.063 B
12
12
12 mm
= 5.20704 A 10 - 6 B m4
80 mm
Qmax = ©y¿A¿
20 mm
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)
= 87.84 A 10 - 6 B m3
Allowable shear stress: Maximum shear stress occurs at the point where the neutral
axis passes through the section.
Applying the shear formula
tmax = tallow =
40 A 106 B =
VQmax
It
V(87.84)(10 - 6)
5.20704(10 - 6)(0.08)
V = 189 692 N = 190 kN
Ans.
481
20 mm
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7–15. Plot the shear-stress distribution over the cross
section of a rod that has a radius c. By what factor is the
maximum shear stress greater than the average shear stress
acting over the cross section?
c
y
V
x = 2c2 - y2 ;
p 4
c
4
I =
t = 2 x = 2 2c2 - y2
dA = 2 x dy = 22c2 - y2 dy
dQ = ydA = 2y 2c2 - y2 dy
x
Q =
Ly
2y2c2 - y2 dy = -
3 x
2
2 2
2
(c - y2)2 | y = (c2 - y2)3
3
3
3
V[23 (c2 - y2)2]
VQ
4V 2
t =
=
=
[c - y2)
p 4
2
2
It
3pc4
( 4 c )(2 2c - y )
The maximum shear stress occur when y = 0
tmax =
4V
3 p c2
tavg =
V
V
=
A
p c2
The faector =
tmax
=
tavg
4V
3 pc2
V
pc2
=
4
3
Ans.
482
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*7–16. A member has a cross section in the form of an
equilateral triangle. If it is subjected to a shear force V,
determine the maximum average shear stress in the member
using the shear formula. Should the shear formula actually be
used to predict this value? Explain.
I =
V
1
(a)(h)3
36
y
h
;
=
x
a>2
Q =
a
LA¿
Q = a
y =
y dA = 2c a
2h
x
a
1
2
2
b (x)(y) a h - yb d
2
3
3
4h2
2x
b (x2)a 1 b
a
3a
t = 2x
t =
t =
V(4h2>3a)(x2)(1 - 2x
VQ
a)
=
It
((1>36)(a)(h3))(2x)
24V(x - a2 x2)
a2h
24V
4
dt
= 2 2 a 1 - xb = 0
a
dx
ah
At x =
y =
a
4
h
2h a
a b =
a 4
2
tmax =
24V a
2 a
a b a1 - a b b
a 4
a2h 4
tmax =
3V
ah
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.
483
h
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•7–17.
Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 600 kN.
30 mm
150 mm
V
100 mm
100 mm
100 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275(10 - 3) m4
12
12
From Fig. a,
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1)
= 1.09125(10 - 3) m3
The maximum shear stress occurs at the points along the neutral axis since Q is
maximum and thickness t = 0.1 m is the smallest.
tmax =
VQmax
600(103)[1.09125(10 - 3)]
=
It
0.175275(10 - 3) (0.1)
= 37.36(106) Pa = 37.4 MPa
Ans.
484
30 mm
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7–18. Determine the maximum shear force V that the strut
can support if the allowable shear stress for the material is
tallow = 45 MPa.
30 mm
150 mm
V
100 mm
100 mm
100 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4
12
12
From Fig. a
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1)
= 1.09125 (10 - 3) m3
The maximum shear stress occeurs at the points along the neutral axis since Q is
maximum and thickness t = 0.1 m is the smallest.
tallow =
VQmax
;
It
45(106) =
V C 1.09125(10 - 3) D
0.175275(10 - 3)(0.1)
V = 722.78(103) N = 723 kN
Ans.
485
30 mm
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7–19. Plot the intensity of the shear stress distributed over
the cross section of the strut if it is subjected to a shear force
of V = 600 kN.
30 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.3)(0.213) (0.2)(0.153) = 0.175275 (10 - 3) m4
12
12
For 0.075 m 6 y … 0.105 m, Fig. a, Q as a function of y is
Q = y¿A¿ =
1
(0.105 + y) (0.105 - y)(0.3) = 1.65375(10 - 3) - 0.15y2
2
For 0 … y 6 0.075 m, Fig. b, Q as a function of y is
Q = ©y¿A¿ = 0.09 (0.03)(0.3) +
1
(0.075 + y)(0.075 - y)(0.1) = 1.09125(10 - 3) - 0.05 y2
2
For 0.075 m 6 y … 0.105 m, t = 0.3 m. Thus,
t =
600 (103) C 1.65375(10 - 3) - 0.15y2 D
VQ
= (18.8703 - 1711.60y2) MPa
=
It
0.175275(10 - 3) (0.3)
At y = 0.075 m and y = 0.105 m,
t|y = 0.015 m = 9.24 MPa
ty = 0.105 m = 0
For 0 … y 6 0.075 m, t = 0.1 m. Thus,
t =
VQ
600 (103) [1.09125(10 - 3) - 0.05 y2]
= (37.3556 - 1711.60 y2) MPa
=
It
0.175275(10 - 3) (0.1)
At y = 0 and y = 0.075 m,
t|y = 0 = 37.4 MPa
ty = 0.075 m = 27.7 MPa
The plot shear stress distribution over the cross-section is shown in Fig. c.
486
150 mm
V
100 mm
100 mm
100 mm
30 mm
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*7–20. The steel rod is subjected to a shear of 30 kip.
Determine the maximum shear stress in the rod.
The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is
p
p
I = r4 = (24) = 4 p in4
4
4
30 kip
dQ = ydA = y (2xdy) = 2xy dy
1
However, from the equation of the circle, x = (4 - y2)2 , Then
1
dQ = 2y(4 - y2)2 dy
Thus, Q for the area above y is
2 in
1
2y (4 - y2)2 dy
Ly
3 2 in
2
= - (4 - y2)2 y
3
=
3
2
(4 - y2)2
3
1
Here, t = 2x = 2 (4 - y2)2 . Thus
30 C 23 (4 - y2)2 D
VQ
=
t =
1
It
4p C 2(4 - y2)2 D
3
t =
5
(4 - y2) ksi
2p
By inspecting this equation, t = tmax at y = 0. Thus
¿=
tmax
A
2 in.
Q for the differential area shown shaded in Fig. a is
Q =
1 in.
20
10
= 3.18 ksi
=
p
2p
Ans.
487
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•7–21.
The steel rod is subjected to a shear of 30 kip.
Determine the shear stress at point A. Show the result on a
volume element at this point.
1 in.
A
The moment of inertia of the circular cross-section about the neutral axis (x axis) is
I =
2 in.
p 4
p
r = (24) = 4p in4
4
4
30 kip
Q for the differential area shown in Fig. a is
dQ = ydA = y (2xdy) = 2xy dy
1
However, from the equation of the circle, x = (4 - y2)2 , Then
1
dQ = 2y (4 - y2)2 dy
Thus, Q for the area above y is
2 in.
1
Q =
Ly
= -
2y (4 - y2)2 dy
2 in.
3
3
2
2
(4 - y2)2 `
= (4 - y2)2
3
3
y
1
Here t = 2x = 2 (4 - y2)2 . Thus,
30 C 23 (4 - y2)2 D
VQ
=
t =
1
It
4p C 2(4 - y2)2 D
3
t =
5
(4 - y2) ksi
2p
For point A, y = 1 in. Thus
tA =
5
(4 - 12) = 2.39 ksi
2p
Ans.
The state of shear stress at point A can be represented by the volume element
shown in Fig. b.
488
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7–22. Determine the shear stress at point B on the web of
the cantilevered strut at section a–a.
2 kN
250 mm
a
250 mm
4 kN
300 mm
a
20 mm
70 mm
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
y =
= 0.03625 m
(0.05)(0.02) + (0.07)(0.02)
I =
+
B
20 mm
50 mm
1
(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2
12
1
(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4
12
yBœ = 0.03625 - 0.01 = 0.02625 m
QB = (0.02)(0.05)(0.02625) = 26.25(10 - 6) m3
tB =
6(103)(26.25)(10 - 6)
VQB
=
It
1.78622(10 - 6)(0.02)
= 4.41 MPa
Ans.
7–23. Determine the maximum shear stress acting at
section a–a of the cantilevered strut.
2 kN
250 mm
a
250 mm
4 kN
300 mm
a
20 mm
70 mm
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
= 0.03625 m
(0.05)(0.02) + (0.07)(0.02)
I =
1
(0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2
12
+
20 mm
50 mm
1
(0.02)(0.073) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625(10 - 6) m4
12
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10 - 6) m3
tmax =
B
VQmax
6(103)(28.8906)(10 - 6)
=
It
1.78625(10 - 6)(0.02)
= 4.85 MPa
Ans.
489
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*7–24. Determine the maximum shear stress in the T-beam
at the critical section where the internal shear force is
maximum.
10 kN/m
A
1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
150 mm
The neutral axis passes through centroid c of the cross-section, Fig. c.
'
0.075(0.15)(0.03) + 0.165(0.03)(0.15)
© y A
=
y =
©A
0.15(0.03) + 0.03(0.15)
150 mm
1
(0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2
12
+
1
(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2
12
= 27.0 (10 - 6) m4
From Fig. d,
Qmax = y¿A¿ = 0.06(0.12)(0.03)
= 0.216 (10 - 3) m3
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness t = 0.03 m is the smallest.
tmax =
27.5(103) C 0.216(10 - 3) D
Vmax Qmax
=
It
27.0(10 - 6)(0.03)
= 7.333(106) Pa
= 7.33 MPa
Ans.
490
30 mm
30 mm
= 0.12 m
I =
B
C
The FBD of the beam is shown in Fig. a,
1.5 m
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•7–25.
Determine the maximum shear stress in the
T-beam at point C. Show the result on a volume element
at this point.
10 kN/m
A
B
C
1.5 m
3m
150 mm
150 mm
30 mm
using the method of sections,
+ c ©Fy = 0;
VC + 17.5 -
1
(5)(1.5) = 0
2
VC = -13.75 kN
The neutral axis passes through centroid C of the cross-section,
0.075 (0.15)(0.03) + 0.165(0.03)(0.15)
©yA
=
©A
0.15(0.03) + 0.03(0.15)
y =
= 0.12 m
I =
1
(0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2
12
+
1
(0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2
12
= 27.0 (10 - 6) m4
Qmax = y¿A¿ = 0.06 (0.12)(0.03)
= 0.216 (10 - 3) m3 490
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness t = 0.03 m is the smallest.
tmax =
30 mm
13.75(103) C 0.216(10 - 3) D
VC Qmax
=
It
27.0(10 - 6) (0.03)
= 3.667(106) Pa = 3.67 MPa
Ans.
491
1.5 m
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7–26. Determine the maximum shear stress acting in the
fiberglass beam at the section where the internal shear
force is maximum.
200 lb/ft
150 lb/ft
D
A
6 ft
6 ft
2 ft
4 in.
6 in.
0.5 in.
4 in.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, Vmax = 878.57 lb.
Section Properties:
INA =
1
1
(4) A 7.53 B (3.5) A 63 B = 77.625 in4
12
12
Qmax = ©y¿A¿
= 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Applying the shear formula
tmax =
=
VQmax
It
878.57(12.375)
= 280 psi
77.625(0.5)
Ans.
492
0.75 in.
0.75 in.
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7–27. Determine the shear stress at points C and D
located on the web of the beam.
3 kip/ft
D
A
C
B
6 ft
6 ft
6 in.
0.75 in.
The FBD is shown in Fig. a.
Using the method of sections, Fig. b,
+ c ©Fy = 0;
18 -
1
(3)(6) - V = 0
2
V = 9.00 kip.
The moment of inertia of the beam’s cross section about the neutral axis is
I =
1
1
(6)(103) (5.25)(83) = 276 in4
12
12
QC and QD can be computed by refering to Fig. c.
QC = ©y¿A¿ = 4.5 (1)(6) + 2 (4)(0.75)
= 33 in3
QD = y3œ A¿ = 4.5 (1)(6) = 27 in3
Shear Stress. since points C and D are on the web, t = 0.75 in.
tC =
VQC
9.00 (33)
=
= 1.43 ksi
It
276 (0.75)
Ans.
tD =
VQD
9.00 (27)
=
= 1.17 ksi
It
276 (0.75)
Ans.
493
6 ft
1 in.
C
D
4 in.
4 in.
6 in.
1 in.
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*7–28. Determine the maximum shear stress acting in the
beam at the critical section where the internal shear force
is maximum.
3 kip/ft
D
A
C
B
6 ft
6 ft
6 in.
The FBD is shown in Fig. a.
The shear diagram is shown in Fig. b, Vmax = 18.0 kip.
0.75 in.
6 ft
1 in.
C
D
4 in.
4 in.
6 in.
1 in.
The moment of inertia of the beam’s cross-section about the neutral axis is
I =
1
1
(6)(103) (5.25)(83)
12
12
= 276 in4
From Fig. c
Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75)
= 33 in3
The maximum shear stress occurs at points on the neutral axis since Q is the
maximum and thickness t = 0.75 in is the smallest
tmax =
Vmax Qmax
18.0 (33)
=
= 2.87 ksi
It
276 (0.75)
Ans.
494
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7–30. The beam has a rectangular cross section and is
subjected to a load P that is just large enough to develop a
fully plastic moment Mp = PL at the fixed support. If the
material is elastic-plastic, then at a distance x 6 L the
moment M = Px creates a region of plastic yielding with
an associated elastic core having a height 2y¿. This situation
has been described by Eq. 6–30 and the moment M is
distributed over the cross section as shown in Fig. 6–48e.
Prove that the maximum shear stress developed in the beam
is given by tmax = 321P>A¿2, where A¿ = 2y¿b, the crosssectional area of the elastic core.
P
x
Plastic region
2y¿
h
b
Elastic region
Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element in
the plastic zone and write the equation of equilibrium.
; ©Fx = 0;
tlong A2 + sg A1 - sg A1 = 0
tlong = 0
This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the
corresponding transverse stress, tmax, is also equal to zero in the plastic zone.
Therefore, the shear force V = P is carried by the malerial only in the elastic zone.
Section Properties:
INA =
1
2
(b)(2y¿)3 = b y¿ 3
12
3
Qmax = y¿ A¿ =
y¿
y¿ 2b
(y¿)(b) =
2
2
Maximum Shear Stress: Applying the shear formula
V A y¿2 b B
3
tmax
However,
VQmax
=
=
It
A¿ = 2by¿
tmax =
3P
‚
2A¿
A by¿ B (b)
2
3
3
=
3P
4by¿
hence
(Q.E.D.)
495
L
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7–31. The beam in Fig. 6–48f is subjected to a fully plastic
moment Mp . Prove that the longitudinal and transverse
shear stresses in the beam are zero. Hint: Consider an element
of the beam as shown in Fig. 7–4c.
P
x
Plastic region
2y¿
h
b
Elastic region
L
Force Equilibrium: If a fully plastic moment acts on the cross section, then an
element of the material taken from the top or bottom of the cross section is
subjected to the loading shown. For equilibrium
; ©Fx = 0;
sg A1 + tlong A2 - sg A1 = 0
tlong = 0
Thus no shear stress is developed on the longitudinal or transverse plane of the
element. (Q. E. D.)
*7–32. The beam is constructed from two boards fastened
together at the top and bottom with two rows of nails
spaced every 6 in. If each nail can support a 500-lb shear
force, determine the maximum shear force V that can be
applied to the beam.
6 in.
6 in.
2 in.
2 in.
V
6 in.
Section Properties:
I =
1
(6) A 43 B = 32.0 in4
12
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow: There are two rows of nails. Hence, the allowable shear flow
2(500)
= 166.67 lb>in.
q =
6
q =
166.67 =
VQ
I
V(12.0)
32.0
V = 444 lb
Ans.
496
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•7–33.
The beam is constructed from two boards
fastened together at the top and bottom with two rows of
nails spaced every 6 in. If an internal shear force of
V = 600 lb is applied to the boards, determine the shear
force resisted by each nail.
6 in.
6 in.
2 in.
2 in.
Section Properties:
I =
1
(6) A 43 B = 32.0 in4
12
V
6 in.
Q = y¿A¿ = 1(6)(2) = 12.0 in4
Shear Flow:
q =
VQ
600(12.0)
=
= 225 lb>in.
I
32.0
There are two rows of nails. Hence, the shear force resisted by each nail is
q
225 lb>in.
F = a bs = a
b(6 in.) = 675 lb
2
2
Ans.
7–34. The beam is constructed from two boards fastened
together with three rows of nails spaced s = 2 in. apart. If
each nail can support a 450-lb shear force, determine the
maximum shear force V that can be applied to the beam. The
allowable shear stress for the wood is tallow = 300 psi.
s
s
1.5 in.
The moment of inertia of the cross-section about the neutral axis is
I =
V
1
(6)(33) = 13.5 in4
12
6 in.
Refering to Fig. a,
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 6 in.
tallow =
VQmax
;
It
300 =
V(6.75)
13.5(6)
V = 3600 lb = 3.60 kips
Shear Flow: Since there are three rows of nails,
F
450
b = 675 lb>in.
qallow = 3 a b = 3 a
s
2
VQA
V(6.75)
;
675 =
qallow =
I
13.5
V = 1350 lb = 1.35 kip
497
Ans.
1.5 in.
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7–35. The beam is constructed from two boards fastened
together with three rows of nails. If the allowable shear
stress for the wood is tallow = 150 psi, determine the
maximum shear force V that can be applied to the beam.
Also, find the maximum spacing s of the nails if each nail
can resist 650 lb in shear.
s
s
1.5 in.
V
6 in.
The moment of inertia of the cross-section about the neutral axis is
I =
1
(6)(33) = 13.5 in4
12
Refering to Fig. a,
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 6 in.
tallow =
VQmax
;
It
150 =
V(6.75)
13.5(6)
V = 1800 lb = 1.80 kip
Since there are three rows of nails, qallow = 3 a
qallow =
VQA
;
I
Ans.
F
650
1950 lb
b = 3¢
b
≤ = a
s
s
s
in.
1800(6.75)
1950
=
s
13.5
s = 2.167 in = 2
1
in
8
Ans.
498
1.5 in.
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*7–36. The beam is fabricated from two equivalent
structural tees and two plates. Each plate has a height of
6 in. and a thickness of 0.5 in. If a shear of V = 50 kip is
applied to the cross section, determine the maximum spacing
of the bolts. Each bolt can resist a shear force of 15 kip.
0.5 in.
s
3 in.
1 in.
A
Section Properties:
INA =
V
6 in.
1
1
(3) A 93 B (2.5) A 83 B
12
12
0.5 in.
N
1
1
(0.5) A 23 B +
(1) A 63 B
12
12
3 in.
= 93.25 in4
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(15)
30
.
=
q =
s
s
VQ
q =
I
50(10.125)
30
=
s
93.25
s = 5.53 in.
Ans.
•7–37.
The beam is fabricated from two equivalent
structural tees and two plates. Each plate has a height of
6 in. and a thickness of 0.5 in. If the bolts are spaced at
s = 8 in., determine the maximum shear force V that can
be applied to the cross section. Each bolt can resist a
shear force of 15 kip.
0.5 in.
s
3 in.
1 in.
A
Section Properties:
INA
-
1
1
(0.5) A 23 B +
(1) A 63 B
12
12
3 in.
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(15)
= 3.75 kip>in.
q =
8
3.75 =
0.5 in.
N
= 93.25 in4
q =
V
6 in.
1
1
=
(3) A 93 B (2.5) A 83 B
12
12
VQ
I
V(10.125)
93.25
y = 34.5 kip
Ans.
499
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7–38. The beam is subjected to a shear of V = 2 kN.
Determine the average shear stress developed in each nail
if the nails are spaced 75 mm apart on each side of the
beam. Each nail has a diameter of 4 mm.
The neutral axis passes through centroid C of the cross-section as shown in Fig. a.
'
0.175(0.05)(0.2) + 0.1(0.2)(0.05)
© y A
y =
=
= 0.1375 m
©A
0.05(0.2) + 0.2(0.05)
200 mm
25 mm
75 mm
50 mm 75 mm
V
200 mm
Thus,
I =
1
(0.2)(0.053) + 0.2 (0.05)(0.175 - 0.1375)2
12
+
25 mm
1
(0.05)(0.23) + 0.05(0.2)(0.1375 - 0.1)2
12
= 63.5417(10 - 6) m4
Q for the shaded area shown in Fig. b is
Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10 - 3) m3
Since there are two rows of nails q = 2a
q =
VQ
;
I
26.67 F =
F
2F
b =
= (26.67 F) N>m.
s
0.075
2000 C 0.375 (10 - 3) D
63.5417 (10 - 6)
F = 442.62 N
Thus, the shear stress developed in the nail is
tn =
F
442.62
=
= 35.22(106)Pa = 35.2 MPa
p
A
2
(0.004 )
4
Ans.
500
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7–39. A beam is constructed from three boards bolted
together as shown. Determine the shear force developed
in each bolt if the bolts are spaced s = 250 mm apart and the
applied shear is V = 35 kN.
25 mm
25 mm
100 mm 250 mm
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025)
y =
= 0.18676 m
2 (0.25)(0.025) + 0.35 (0.025)
I = (2)a
+
1
b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2
12
V
1
(0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2
12
350 mm
s = 250 mm
= 0.270236 (10 - 3) m4
25 mm
-3
3
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10 ) m
q =
35 (0.386)(10 - 3)
VQ
= 49.997 kN>m
=
I
0.270236 (10 - 3)
F = q(s) = 49.997 (0.25) = 12.5 kN
Ans.
*7–40. The double-web girder is constructed from two
plywood sheets that are secured to wood members at its top
and bottom. If each fastener can support 600 lb in single
shear, determine the required spacing s of the fasteners
needed to support the loading P = 3000 lb. Assume A is
pinned and B is a roller.
2 in.
2 in.
s
10 in.
A
4 ft
2 in.
2 in.
6 in.
0.5 in.
0.5 in.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, Vmax = 1500 lb.
Section Properties:
INA =
P
1
1
(7) A 183 B (6) A 103 B = 2902 in4
12
12
Q = y¿A¿ = 7(4)(6) = 168 in3
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
2(600)
1200
=
q =
.
s
s
VQ
q =
I
1500(168)
1200
=
s
2902
s = 13.8 in.
Ans.
501
4 ft
B
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•7–41.
The double-web girder is constructed from two
plywood sheets that are secured to wood members at its top
and bottom. The allowable bending stress for the wood is
sallow = 8 ksi and the allowable shear stress is tallow = 3 ksi.
If the fasteners are spaced s = 6 in. and each fastener can
support 600 lb in single shear, determine the maximum load
P that can be applied to the beam.
2 in.
2 in.
s
10 in.
A
4 ft
2 in.
2 in.
6 in.
0.5 in.
0.5 in.
Support Reactions: As shown on FBD.
Internal Shear Force and Moment: As shown on shear and moment diagram,
Vmax = 0.500P and Mmax = 2.00P.
Section Properties:
INA =
P
1
1
(7) A 183 B (6) A 103 B = 2902 in4
12
12
Q = y2œ A¿ = 7(4)(6) = 168 in3
Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3
Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the
2(600)
= 200 lb>in.
allowable shear flow is q =
6
VQ
q =
I
0.500P(168)
200 =
2902
P = 6910 lb = 6.91 kip (Controls !)
Ans.
Shear Stress: Assume failure due to shear stress.
VQmax
It
0.500P(208.5)
3000 =
2902(1)
tmax = tallow =
P = 22270 lb = 83.5 kip
Bending Stress: Assume failure due to bending stress.
Mc
I
2.00P(12)(9)
8(103) =
2902
smax = sallow =
P = 107 ksi
502
4 ft
B
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7–42. The T-beam is nailed together as shown. If the nails
can each support a shear force of 950 lb, determine the
maximum shear force V that the beam can support and the
corresponding maximum nail spacing s to the nearest 18 in.
The allowable shear stress for the wood is tallow = 450 psi.
2 in.
s
The neutral axis passes through the centroid c of the cross-section as shown in Fig. a.
'
13(2)(12) + 6(12)(2)
© y A
y =
=
= 9.5 in.
©A
2(12) + 12(2)
I =
1
(2)(123) + 2(12)(9.5 - 6)2
12
+
= 884 in4
Refering to Fig. a, Qmax and QA are
Qmax = y1œ A1œ = 4.75(9.5)(2) = 90.25 in3
QA = y2œ A2œ = 3.5 (2)(12) = 84 in3
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and t = 2 in.
VQmax
;
It
450 =
V (90.25)
884 (2)
V = 8815.51 lb = 8.82 kip
Here, qallow =
F
950
=
lb>in. Then
s
s
VQA
;
qallow =
I
Ans.
8815.51(84)
950
=
s
884
s = 1.134 in = 1
12 in.
V
2 in.
1
(12)(23) + 12(2)(13 - 9.5)2
12
tallow =
s
12 in.
1
in
8
Ans.
503
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7–43. Determine the average shear stress developed in the
nails within region AB of the beam. The nails are located on
each side of the beam and are spaced 100 mm apart. Each
nail has a diameter of 4 mm. Take P = 2 kN.
P
2 kN/m
A
B
C
1.5 m
The FBD is shown in Fig. a.
As indicated in Fig. b, the internal shear force on the cross-section within region AB
is constant that is VAB = 5 kN.
1.5 m
100 mm
The neutral axis passes through centroid C of the cross section as shown in Fig. c.
'
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
© y A
=
y =
©A
0.04(0.2) + 0.2(0.04)
40 mm
= 0.14 m
200 mm
1
I =
(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2
12
1
+
(0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2
12
200 mm 20 mm
20 mm
= 53.333(10 - 6) m4
Q for the shaded area shown in Fig. d is
Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3
Since there are two rows of nail, q = 2 a
q =
VAB Q
;
I
20F =
F
F
b = 2a
b = 20F N>m.
s
0.1
5(103) C 0.32(10 - 3) D
53.333(10 - 6)
F = 1500 N
Thus, the average shear stress developed in each nail is
A tnail B avg =
F
1500
=
= 119.37(106)Pa = 119 MPa
p
Anail
2
(0.004 )
4
504
Ans.
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*7–44. The nails are on both sides of the beam and each
can resist a shear of 2 kN. In addition to the distributed
loading, determine the maximum load P that can be applied
to the end of the beam. The nails are spaced 100 mm apart
and the allowable shear stress for the wood is tallow = 3 MPa.
P
2 kN/m
A
B
C
1.5 m
1.5 m
100 mm
The FBD is shown in Fig. a.
40 mm
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of
Constant value, Vmax = (P + 3) kN.
The neutral axis passes through Centroid C of the cross-section as shown in Fig. c.
'
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
© y A
=
y =
©A
0.04(0.2) + 0.2(0.04)
= 0.14 m
I =
1
(0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2
12
1
+
(0.2)(0.043) + 0.2(0.04)(0.18 - 0.142)
12
Refering to Fig. d,
Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10 - 3) m3
QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3
The maximum shear stress occurs at the points on Neutral axis where Q is maximum
and t = 0.04 m.
Vmax Qmax
;
It
3(106) =
(P + 3)(103) C 0.392(10 - 3) D
53.333(10 - 6)(0.04)
P = 13.33 kN
Since there are two rows of nails qallow = 2 a
qallow
Vmax QA
=
;
I
40 000 =
200 mm 20 mm
20 mm
= 53.333(10 - 6) m4
tallow =
200 mm
2(103)
F
d = 40 000 N>m.
b = 2c
s
0.1
(P + 3)(103) C 0.32(10 - 3) D
53.333(10 - 6)
P = 3.67 kN (Controls!)
Ans.
505
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7–44.
Continued
506
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•7–45.
The beam is constructed from four boards which
are nailed together. If the nails are on both sides of the beam
and each can resist a shear of 3 kN, determine the maximum
load P that can be applied to the end of the beam.
3 kN
A
P
B
C
2m
2m
100 mm
30 mm
150 mm
30 mm
250 mm 30 mm
30 mm
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN.
Section Properties:
INA =
1
1
(0.31) A 0.153 B (0.25) A 0.093 B
12
12
= 72.0 A 10 - 6 B m4
Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A 10 - 3 B m3
Shear Flow: There are two rows of nails. Hence the allowable shear flow is
3(2)
= 60.0 kN>m.
q =
0.1
VQ
q =
I
(P
+ 3)(103)0.450(10 - 3)
60.0 A 103 B =
72.0(10 - 6)
P = 6.60 kN
Ans.
507
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7–47. The beam is made from four boards nailed together
as shown. If the nails can each support a shear force of
100 lb., determine their required spacing s and s if the beam
is subjected to a shear of V = 700 lb.
D
1 in.
1 in.
2 in.
s¿
s¿
s
A
C
s
10 in.
1 in.
10 in.
V
B
1.5 in.
Section Properties:
y =
©yA
0.5(10)(1) + 1.5(2)(3) + 6(1.5)(10)
=
©A
10(1) + 2(3) + 1.5(10)
= 3.3548 in
INA =
1
(10) A 13 B + 10(1)(3.3548 - 0.5)2
12
1
+
(2) A 33 B + 2(3)(3.3548 - 1.5)2
12
= 337.43 in4
QC = y1 ¿A¿ = 1.8548(3)(1) = 5.5645 in3
QD = y2 ¿A¿ = (3.3548 - 0.5)(10)(1) + 2 C (3.3548 - 1.5)(3)(1) D = 39.6774 in3
Shear Flow: The allowable shear flow at points C and D is qC =
100
, respectively.
qB =
s¿
VQC
qC =
I
700(5.5645)
100
=
s
337.43
s = 8.66 in.
VQD
qD =
I
700(39.6774)
100
=
s¿
337.43
100
and
s
Ans.
s¿ = 1.21 in.
Ans.
508
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*7–48. The box beam is constructed from four boards that
are fastened together using nails spaced along the beam
every 2 in. If each nail can resist a shear of 50 lb, determine
the greatest shear V that can be applied to the beam without
causing failure of the nails.
1 in.
12 in.
5 in.
V
2 in.
1 in.
6 in.
1 in.
y =
©yA
0.5 (12)(1) + 2 (4)(6)(1) + (6.5)(6)(1)
=
= 3.1 in.
©A
12(1) + 2(6)(1) + (6)(1)
I =
1
(12)(13) + 12(1)(3.1 - 0.5)2
12
+ 2a
+
1
b (1)(63) + 2(1)(6)(4 - 3.1)2
12
1
(6)(13) + 6(1)(6.5 - 3.1)2 = 197.7 in4
12
QB = y1œ A¿ = 2.6(12)(1) = 31.2 in3
qB =
V(31.2)
1 VQB
a
b =
= 0.0789 V
2
I
2(197.7)
qB s = 0.0789V(2) = 50
V = 317 lb (controls)
Ans.
QA = y2œ A¿ = 3.4(6)(1) = 20.4 in3
qA =
V(20.4)
1 VQA
a
b =
= 0.0516 V
2
I
2(197.7)
qA s = 0.0516V(2) = 50
V = 485 lb
509
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7–50. A shear force of V = 300 kN is applied to the box
girder. Determine the shear flow at points A and B.
90 mm
90 mm
C
A
D
200 mm
B
190 mm
V
200 mm
10 mm
180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
Refering to Fig. a Fig. b,
QA = y1œ A1œ = 0.195 (0.01)(0.19) = 0.3705 (10 - 3) m3
QB = 2yzœ A2œ + y3œ A3œ = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10 - 3) m3
Due to symmety, the shear flow at points A and A¿ , Fig. a, and at points B and B¿ ,
Fig. b, are the same. Thus
qA
3
-3
1 300(10 ) C 0.3705(10 ) D
1 VQA
s
b = c
= a
2
I
2
0.24359(10 - 3)
= 228.15(103) N>m = 228 kN>m
qB =
Ans.
3
-3
1 VQB
1 300(10 ) C 0.751(10 ) D
s
a
b = c
2
I
2
0.24359(10 - 3)
= 462.46(103) N>m = 462 kN>m
Ans.
510
100 mm
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7–51. A shear force of V = 450 kN is applied to the box
girder. Determine the shear flow at points C and D.
90 mm
90 mm
C
A
D
200 mm
B
190 mm
V
200 mm
10 mm
180 mm
10 mm
The moment of inertia of the cross-section about the neutral axis is
I =
1
1
(0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4
12
12
Refering to Fig. a, due to symmetry ACœ = 0. Thus
QC = 0
Then refering to Fig. b,
QD = y1œ A1œ + y2œ A2œ = 0.195 (0.01)(0.09) + 0.15(0.1)(0.01)
= 0.3255(10 - 3) m3
Thus,
qC =
qD =
VQC
= 0
I
Ans.
450(103) C 0.3255(10 - 3) D
VQD
=
I
0.24359(10 - 3)
= 601.33(103) N>m = 601 kN>m
Ans.
100 mm
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*7–52. A shear force of V = 18 kN is applied to the
symmetric box girder. Determine the shear flow at A and B.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
150 mm
10 mm
10 mm
V
150 mm
10 mm 125 mm
10 mm
Section Properties:
INA =
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+ 2c
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
= 125.17 A 10 - 6 B m4
QA = y2œ A¿ = 0.145(0.125)(0.01) = 0.18125 A 10 - 3 B m3
QB = y1œ A¿ = 0.105(0.125)(0.01) = 0.13125 A 10 - 3 B m3
Shear Flow:
qA =
=
1 VQA
c
d
2
I
1 18(103)(0.18125)(10 - 3)
d
c
2
125.17(10 - 6)
= 13033 N>m = 13.0 kN>m
qB =
=
Ans.
1 VQB
c
d
2
I
1 18(103)(0.13125)(10 - 3)
d
c
2
125.17(10 - 6)
= 9437 N>m = 9.44 kN>m
Ans.
512
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A shear force of V = 18 kN is applied to the box
girder. Determine the shear flow at C.
•7–53.
10 mm
30 mm
10 mm
A
100 mm
C
B
100 mm
150 mm
10 mm
10 mm
V
150 mm
10 mm 125 mm
10 mm
Section Properties:
INA =
1
1
(0.145) A 0.33 B (0.125) A 0.283 B
12
12
+2 c
1
(0.125) A 0.013 B + 0.125(0.01) A 0.1052 B d
12
= 125.17 A 10 - 6 B m4
QC = ©y¿A¿
= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02)
= 0.5375 A 10 - 3 B m3
Shear Flow:
qC =
=
1 VQC
c
d
2
I
1 18(103)(0.5375)(10 - 3)
d
c
2
125.17(10 - 4)
= 38648 N>m = 38.6 kN>m
Ans.
513
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7–54. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of, V = 150 N,
determine the shear flow at points A and B.
10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
= 0.027727 m
y =
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
I = 2c
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
+ 2c
+
40 mm
10 mm
30 mm
1
(0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
B
A
V
40 mm
10 mm
30 mm
10 mm
1
(0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 = 0.98197(10 - 6) m4
12
yB ¿ = 0.055 - 0.027727 = 0.027272 m
yA ¿ = 0.027727 - 0.005 = 0.022727 m
QA = yA ¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10 - 6) m3
QB = yB ¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10 - 6) m3
qA =
VQA
150(9.0909)(10 - 6)
= 1.39 kN>m
=
I
0.98197(10 - 6)
Ans.
qB =
VQB
150(8.1818)(10 - 6)
= 1.25 kN>m
=
I
0.98197(10 - 6)
Ans.
7–55. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of V = 150 N,
determine the maximum shear flow in the strut.
y =
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01)
2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
10 mm
40 mm
B
A
= 0.027727 m
I = 2c
10 mm
1
(0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d
12
30 mm
1
+ 2 c (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d
12
+
10 mm
1
(0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2
12
= 0.98197(10 - 6) m4
Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a
0.06 - 0.0277
b
2
= 21.3(10 - 6) m3
qmax =
V
40 mm
1 150(21.3(10 - 6))
1 VQmax
b = 1.63 kN>m
a
b = a
2
I
2 0.98197(10 - 6)
Ans.
514
30 mm
10 mm
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*7–56. The beam is subjected to a shear force of
V = 5 kip. Determine the shear flow at points A and B.
0.5 in.
C
5 in.
5 in. 0.5 in.
0.5 in.
2 in.
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
©yA
y =
=
= 3.70946 in.
©A
11(0.5) + 2(8)(0.5) + 10(0.5)
A
D
8 in.
1
1
(11)(0.53) + 11(0.5)(3.70946 - 0.25)2 + 2c (0.5)(83) + 0.5(8)(4.5 - 3.70946)2 d
12
12
I =
+
0.5 in.
V
B
1
(10)(0.53) + 10(0.5)(6.25 - 3.70946)2
12
= 145.98 in4
œ
= 3.70946 - 0.25 = 3.45946 in.
yA
yBœ = 6.25 - 3.70946 = 2.54054 in.
œ
QA = yA
A¿ = 3.45946(11)(0.5) = 19.02703 in3
QB = yBœ A¿ = 2.54054(10)(0.5) = 12.7027 in3
qA =
1 VQA
1 5(103)(19.02703)
a
b = a
b = 326 lb>in.
2
I
2
145.98
Ans.
qB =
1 VQB
1 5(103)(12.7027)
a
b = a
b = 218 lb>in.
2
I
2
145.98
Ans.
•7–57.
The beam is constructed from four plates and is
subjected to a shear force of V = 5 kip. Determine the
maximum shear flow in the cross section.
0.5 in.
C
5 in.
5 in. 0.5 in.
y =
©yA
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
=
= 3.70946 in.
©A
11(0.5) + 2(8)(0.5) + 10(0.5)
I =
1
1
(11)(0.53) + 11(0.5)(3.45952) + 2 c (0.5)(83) + 0.5(8)(0.79052) d
12
12
0.5 in.
2 in.
A
D
8 in.
V
+
1
(10)(0.53) + 10(0.5)(2.54052)
12
= 145.98 in4
Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)]
= 24.177 in3
qmax =
1 VQmax
1 5(103)(24.177)
a
b = a
b
2
I
2
145.98
= 414 lb>in.
Ans.
515
0.5 in.
B
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7–58. The channel is subjected to a shear of V = 75 kN.
Determine the shear flow developed at point A.
30 mm
400 mm
A
200 mm
30 mm
V ⫽ 75 kN
30 mm
y =
©yA
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
=
= 0.0725 m
©A
0.4(0.03) + 2(0.2)(0.03)
I =
1
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
+ 2c
1
(0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d = 0.12025(10 - 3) m4
12
œ
A¿ = 0.0575(0.2)(0.03) = 0.3450(10 - 3) m3
QA = yA
q =
qA =
VQ
I
75(103)(0.3450)(10 - 3)
0.12025(10 - 3)
= 215 kN>m
Ans.
7–59. The channel is subjected to a shear of V = 75 kN.
Determine the maximum shear flow in the channel.
30 mm
400 mm
A
V ⫽ 75 kN
30 mm
y =
©yA
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
=
©A
0.4(0.03) + 2(0.2)(0.03)
= 0.0725 m
1
I =
(0.4)(0.033) + 0.4(0.03)(0.0725 - 0.015)2
12
1
+ 2c (0.03)(0.23) + 0.03(0.2)(0.13 - 0.0725)2 d
12
= 0.12025(10 - 3) m4
Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10 - 3) m3
qmax =
75(103)(0.37209)(10 - 3)
0.12025(10 - 3)
= 232 kN>m
Ans.
516
200 mm
30 mm
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*7–60. The angle is subjected to a shear of V = 2 kip.
Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
A
5 in.
5 in.
45⬚ 45⬚
0.25 in.
Section Properties:
b =
0.25
= 0.35355 in.
sin 45°
h = 5 cos 45° = 3.53553 in.
INA = 2c
1
(0.35355) A 3.535533 B d = 2.604167 in4
12
Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 -
y
b(0.25)
sin 45°
= 0.55243 - 0.17678y2
Shear Flow:
VQ
I
2(103)(0.55243 - 0.17678y2)
=
2.604167
q =
= {424 - 136y2} lb>in.
At y = 0,
Ans.
q = qmax = 424 lb>in.
Ans.
517
B
V
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•7–61.
The assembly is subjected to a vertical shear of
V = 7 kip. Determine the shear flow at points A and B and
the maximum shear flow in the cross section.
A
0.5 in.
B
V
2 in.
0.5 in.
0.5 in.
6 in.
6 in.
2 in.
0.5 in.
y =
©yA
(0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5)
=
= 2.8362 in.
©A
0.5(11) + 2(0.5)(5.5) + 7(0.5)
I =
1
1
(11)(0.53) + 11(0.5)(2.8362 - 0.25)2 + 2a b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2
12
12
+
1
(7)(0.53) + (0.5)(7)(6.25 - 2.8362)2 = 92.569 in4
12
QA = y1 ¿A1 ¿ = (2.5862)(2)(0.5) = 2.5862 in3
QB = y2 ¿A2 ¿ = (3.4138)(7)(0.5) = 11.9483 in3
Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3
q =
VQ
I
7(103)(2.5862)
= 196 lb>in.
92.569
1 7(103)(11.9483)
qB = a
b = 452 lb>in.
2
92.569
1 7(103)(16.9531)
b = 641 lb>in.
qmax = a
2
92.569
qA =
Ans.
Ans.
Ans.
518
0.5 in.
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7–62. Determine the shear-stress variation over the cross
section of the thin-walled tube as a function of elevation y and
show that t max = 2V>A, where A = 2prt. Hint: Choose a
differential area element dA = Rt du. Using dQ = y dA,
formulate Q for a circular section from u to (p - u) and show
that Q = 2R2t cos u, where cos u = 2R2 - y2>R.
ds
du
y
u
t
dA = R t du
dQ = y dA = yR t du
Here y = R sin u
Therefore dQ = R2 t sin u du
p-u
Q =
p-u
R2 t sin u du = R2 t(-cos u) |
Lu
u
2
= R t [-cos (p - u) - (-cos u)] = 2R2 t cos u
dI = y2 dA = y2 R t du = R3 t sin2 u du
2p
I =
L0
2p
R3 t sin2 u du = R3 t
2p
=
t =
sin 2u
R3 t
[u ]
2
2 0
R3 t
[2p - 0] = pR3 t
2
VQ
V(2R2t cos u)
V cos u
=
=
3
It
pR t
pR t(2t)
Here cos u =
t =
=
L0
(1 - cos 2u)
du
2
2R2 - y2
R
V
2R2 - y2
pR2t
Ans.
tmax occurs at y = 0; therefore
tmax =
V
pR t
A = 2pRt; therefore
tmax =
2V
A
QED
519
R
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7–63. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown where b2 7 b1. The member segments have
the same thickness t.
t
h
e
b2
Section Properties:
I =
1
h 2
t h2
t h3 + 2c(b1 + b2)ta b d =
C h + 6(b1 + b2) D
12
2
12
Q1 = y¿A¿ =
h
ht
(x )t =
x
2 1
2 1
Q2 = y¿A¿ =
h
ht
(x )t =
x
2 2
2 2
Shear Flow Resultant:
VQ1
q1 =
=
I
q2 =
VQ2
=
I
P A ht2 x1 B
P A ht2 x2 B
h C h + 6(b1 + b2) D
h C h + 6(b1 + b2) D
6P
t h2
12
C h + 6(b1 + b2) D
=
t h2
12
C h + 6(b1 + b2) D
=
6P
b1
(Ff)1 =
L0
q1 dx1 =
6P
x1
x2
b1
h C h + 6(b1 + b2) D L0
x1 dx1
3Pb21
=
b2
(Ff)2 =
L0
q2 dx2 =
h C h + 6(b1 + b2) D
6P
b2
h C h + 6(b1 + b2) D L0
x2 dx2
3Pb22
=
h C h + 6(b1 + b2) D
Shear Center: Summing moment about point A.
Pe = A Ff B 2 h - A Ff B 1 h
Pe =
e =
3Pb22
h C h + 6(b1 + b2) D
3(b22 - b21)
h + 6(b1 + b2)
(h) -
3Pb21
h C h + 6(b1 + b2) D
(h)
Ans.
Note that if b2 = b1, e = 0 (I shape).
520
b1
O
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*7–64. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
b
d
45⬚
O
e
Section Properties:
I =
=
t
1
a
b(2d sin 45°)3 + 2 C bt(d sin 45°)2 D
12 sin 45°
td2
(d + 3b)
3
Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x
Shear Flow Resultant:
qf =
P(td sin 45°)x
VQ
3P sin 45°
=
=
x
td2
I
d(d + 3b)
(d
+
3b)
3
b
Ff =
L0
b
qfdx =
2
3P sin 45°
3b sin 45°
P
xdx =
d(d + 3b) L0
2d(d + 3b)
Shear Center: Summing moments about point A,
Pe = Ff(2d sin 45°)
Pe = c
e =
3b2 sin 45°
P d(2d sin 45°)
2d(d + 3b)
3b2
2(d + 3b)
Ans.
521
45⬚
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•7–65.
Determine the location e of the shear center,
point O, for the thin-walled member having a slit along its
side. Each element has a constant thickness t.
a
e
a
t
a
Section Properties:
I =
1
10 3
(2t)(2a)3 + 2 C at A a2 B D =
a t
12
3
Q1 = y1œ A¿ =
y
t
(yt) = y2
2
2
Q2 = ©y¿A¿ =
a
at
(at) + a(xt) =
(a + 2x)
2
2
Shear Flow Resultant:
q1 =
P A 12 y2 B
VQ1
3P 2
= 10 3 =
y
3
I
20a
a
t
3
P C at2 (a + 2x) D
VQ2
3P
=
=
(a + 2x)
q2 =
10 3
2
I
20a
a
t
3
a
(Fw)1 =
L0
a
q1 dy =
a
Ff =
L0
3P
P
y2 dy =
20
20a3 L0
a
q2 dx =
3P
3
(a + 2x)dx =
P
2
10
20a L0
Shear Center: Summing moments about point A.
Pe = 2(Fw)1 (a) + Ff(2a)
Pe = 2 a
e =
3
P
b a + a Pb 2a
20
10
7
a
10
Ans.
522
O
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7–66. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown.
a
60⬚
O
a
60⬚
a
e
Summing moments about A.
Pe = F2 a
I =
13
ab
2
t
1
1
1
(t)(a)3 +
a
b(a)3 = t a3
12
12 sin 30°
4
q1 =
V(a)(t)(a>4)
1
4
q2 = q1 +
F2 =
=
3
ta
V
a
V(a>2)(t)(a>4)
1
4
ta
3
= q1 +
V
2a
V
4V
2 V
(a) + a b (a) =
a
3 2a
3
e =
223
a
3
Ans.
523
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7–67. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
b
t
h
2
O
e
h
2
b
Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3
on FBD (b). Hence, The horizontal force equilibrium is not satisfied (©Fx Z 0). In
order to satisfy this equilibrium requirement. F1 and F2 must be equal to zero.
Shear Center: Summing moments about point A.
Pe = F2(0)
e = 0
Ans.
Also,
The shear flows through the section as indicated by F1, F2, F3.
+ ©F Z 0
However, :
x
To satisfy this equation, the section must tip so that the resultant of
:
:
:
:
F1 + F2 + F3 = P
Also, due to the geometry, for calculating F1 and F3, we require F1 = F3.
Hence, e = 0
Ans.
524
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*7–68. Determine the location e of the shear center,
point O, for the beam having the cross section shown. The
thickness is t.
1
—
r
2
e
r
O
I = (2)c
1
r 2
(t)(r>2)3 + (r>2)(t)ar + b d + Isemi-circle
12
4
= 1.583333t r3 + Isemi-circle
p>2
Isemi-circle =
p>2
2
L-p>2
(r sin u) t r du = t r3
L-p>2
sin2 u du
p
Isemi-circle = t r3 a b
2
Thus,
p
I = 1.583333t r3 + t r3 a b = 3.15413t r3
2
r
r
Q = a b t a + rb +
2
4
Lu
p>2
r sin u (t r du)
Q = 0.625 t r2 + t r2 cos u
q =
VQ
P(0.625 + cos u)t r2
=
I
3.15413 t r3
Summing moments about A:
p>2
Pe =
L-p>2
(q r du)r
p>2
Pe =
e =
Pr
(0.625 + cos u)du
3.15413 L-p>2
r (1.9634 + 2)
3.15413
e = 1.26 r
Ans.
525
1
—
r
2
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•7–69.
Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown. The member segments have the same
thickness t.
h1
h
O
e
h1
b
Summing moments about A.
Pe = F(h) + 2V(b)
h 2
1
1
(t)(h3) + 2b(t)a b +
(t)[h3 - (h - 2h1)3]
12
2
12
I =
=
(1)
t(h - 2h1)3
bth2
th3
+
6
2
12
Q1 = y¿A¿ =
t(hy - 2h1 y + y2)
1
(h - 2h1 + y)yt =
2
2
VQ
Pt(hy - 2h1 y + y2)
=
I
2I
q1 =
V =
L
h1
Pt
Pt hh1 2
2
(hy - 2h1 y + y2)dy =
c
- h31 d
2I L0
2I
2
3
q1 dy =
Q2 = ©y¿A¿ =
1
1
h
(h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx]
2
2
2
VQ2
Pt
=
(h (h - h1) + hx)
I
2I 1
q2 =
b
F =
L
q2 dx =
Pt
Pt
hb2
[h1 (h - h1) + hx]dx =
ah1 hb - h21 b +
b
2I L0
2I
2
From Eq, (1).
Pe =
h2b2
4
Pt
[h1 h2b - h21 hb +
+ hh21 b - h31 b]
2I
2
3
I =
t
(2h3 + 6bh2 - (h - 2h1)3)
12
e =
b(6h1 h2 + 3h2b - 8h31)
t
(6h1 h2b + 3h2b2 - 8h1 3b) =
12I
2h3 + 6bh2 - (h - 2h1)3
526
Ans.
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7–70. Determine the location e of the shear center, point O,
for the thin-walled member having the cross section shown.
t
r
a
O
a
e
Summing moments about A.
Pe = r
dF
L
dA = t ds = t r du
(1)
y = r sin u
dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2 udu
p+a
I = r3 t
L
sin2 u du = r3 t
Lp - a
1 - cos 2u
du
2
=
sin 2u p + a
r3 t
(u )
2
2 p - a
=
sin 2(p + a)
sin 2(p - a)
r3 t
c ap + a b - ap - a bd
2
2
2
=
r3 t
r3 t
2 (2a - 2 sin a cos a) =
(2a - sin 2a)
2
2
dQ = y dA = r sin u(t r du) = r2 t sin u du
u
Q = r2 t
q =
L
u
sin u du = r2 t (-cos u)|
Lp-a
= r2 t(-cos u - cos a) = -r2 t(cos u + cos a)
p-a
P(-r2t)(cos u + cos a)
-2P(cos u + cos a)
VQ
=
=
r3t
I
r(2a - sin 2a)
2 (2a - sin 2a)
dF =
L
q ds =
L
q r du
p+p
L
=
dF =
2P r
-2P
(cos u + cos a) du =
(2a cos a - 2 sin a)
r(2a - sin 2a) Lp - a
2a - sin 2a
4P
(sin a - a cos a)
2a - sin 2a
4P
(sin a - a cos a) d
2a - sin 2a
4r (sin a - a cos a)
e =
2a - sin 2a
From Eq. (1); P e = r c
Ans.
527
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7–71. Sketch the intensity of the shear-stress distribution
acting over the beam’s cross-sectional area, and determine
the resultant shear force acting on the segment AB. The
shear acting at the section is V = 35 kip. Show that
INA = 872.49 in4.
C
V
8 in.
B
A
6 in.
Section Properties:
y =
4(8)(8) + 11(6)(2)
©yA
=
= 5.1053 in.
©A
8(8) + 6(2)
INA =
2 in.
1
(8) A 83 B + 8(8)(5.1053 - 4)2
12
+
1
(2) A 63 B + 2(6)(11 - 5.1053)2
12
= 872.49 in4 (Q.E.D)
Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8)
= 104.25 - 4y21
Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2)
= 79.12 - y22
Shear Stress: Applying the shear formula t =
tCB =
VQ
,
It
VQ1
35(103)(104.25 - 4y21)
=
It
872.49(8)
= {522.77 - 20.06y21} psi
At y1 = 0,
tCB = 523 psi
At y1 = -2.8947 in.
tCB = 355 psi
tAB =
VQ2
35(103)(79.12 - y22)
=
It
872.49(2)
= {1586.88 - 20.06y22} psi
At y2 = 2.8947 in.
tAB = 1419 psi
Resultant Shear Force: For segment AB.
VAB =
L
tAB dA
0.8947 in
=
L2.8947 in
0.8947 in
=
L2.8947 in
3 in.
3 in.
A 1586.88 - 20.06y22 B (2dy)
A 3173.76 - 40.12y22 B dy
= 9957 lb = 9.96 kip
Ans.
528
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*7–72. The beam is fabricated from four boards nailed
together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are
uniformly spaced at s = 3 in. The beam is subjected to a
shear of V = 4.5 kip.
1 in.
1 in.
3 in.
10 in.
A
1 in.
12 in.
V
B
Section Properties:
y =
0.5(10)(1) + 2(4)(2) + 7(12)(1)
© yA
=
= 3.50 in.
©A
10(1) + 4(2) + 12(1)
INA =
1
(10) A 13 B + (10)(1)(3.50 - 0.5)2
12
+
1
(2) A 43 B + 2(4)(3.50 - 2)2
12
1
+
(1) A 123 B + 1(12)(7 - 3.50)2
12
= 410.5 in4
QC = y1œ A¿ = 1.5(4)(1) = 6.00 in2
QD = y2œ A¿ = 3.50(12)(1) = 42.0 in2
Shear Flow:
qC =
VQC
4.5(103)(6.00)
=
= 65.773 lb>in.
I
410.5
qD =
VQD
4.5(103)(42.0)
=
= 460.41 lb>in.
I
410.5
Hence, the shear force resisted by each nail is
FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb
Ans.
FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip
Ans.
529
1 in.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•7–73.
The member is subjected to a shear force of
V = 2 kN. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
200 mm
B
100 mm
A C
V ⫽ 2 kN
Section Properties:
y =
=
© yA
©A
0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015)
0.2(0.015) + 0.115(0.03) + 0.3(0.015)
= 0.08798 m
1
(0.2) A 0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2
12
1
+
(0.03) A 0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2
12
1
+
(0.015) A 0.33 B + 0.015(0.3)(0.165 - 0.08798)2
12
INA =
= 86.93913 A 10 - 6 B m4
QA = 0
'
QB = y 1œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - 6 B m3
Ans.
QC = ©y¿A¿
= 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015)
= 0.16424 A 10 - 3 B m3
Shear Flow:
qA =
VQA
= 0
I
Ans.
qB =
VQB
2(103)(52.57705)(10 - 6)
= 1.21 kN>m
=
I
86.93913(10 - 6)
Ans.
qC =
VQC
2(103)(0.16424)(10 - 3)
= 3.78 kN>m
=
I
86.93913(10 - 6)
Ans.
530
300 mm
07 Solutions 46060
5/26/10
2:04 PM
Page 531
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–74. The beam is constructed from four boards glued
together at their seams. If the glue can withstand
75 lb>in., what is the maximum vertical shear V that the
beam can support?
3 in.
0.5 in.
Section Properties:
INA =
1
1
(1) A 103 B + 2c (4) A 0.53 B + 4(0.5) A 1.752 B d
12
12
3 in.
0.5 in.
= 95.667 in4
V
Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3
4 in.
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
2(75) = 150 lb>in.
q =
150 =
3 in.
0.5 in.
0.5 in.
VQ
I
V(3.50)
95.667
V = 4100 lb = 4.10 kip
Ans.
7–75. Solve Prob. 7–74 if the beam is rotated 90° from the
position shown.
3 in.
0.5 in.
3 in.
0.5 in.
V
3 in.
4 in.
0.5 in.
Section Properties:
INA =
1
1
(10) A 53 B (9) A 43 B = 56.167 in4
12
12
Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
2(75) = 150 lb>in.
q =
150 =
VQ
I
V(11.25)
56.167
V = 749 lb
Ans.
531
0.5 in.