Chapter 07 in som

472
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The moment of inertia of the cross-section about the neutral axis is
From Fig. a,
Applying the shear formula,
Ans.
The shear stress component at A is represented by the volume element shown in
Fig. b.
= 2.559(106
) Pa = 2.56 MPa
tA =
VQA
It
=
20(103
)[0.64(10-3
)]
0.2501(10-3
)(0.02)
QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10-3
) m3
I =
1
12
(0.2)(0.343
) -
1
12
(0.18)(0.33
) = 0.2501(10-3
) m4
•7–1. If the wide-flange beam is subjected to a shear of
determine the shear stress on the web at A.
Indicate the shear-stress components on a volume element
located at this point.
V = 20 kN,
A
B
V
20 mm
20 mm
20 mm
300 mm
200 mm
200 mm
07 Solutions 46060 5/26/10 2:04 PM Page 472

473
From Fig. a.
The maximum shear stress occurs at the points along neutral axis since Q is
maximum and thicknest t is the smallest.
Ans.= 3.459(106
) Pa = 3.46 MPa
tmax =
VQmax
It
=
20(103
) [0.865(10-3
)]
0.2501(10-3
) (0.02)
Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10-3
) m3
I =
1
12
(0.2)(0.343
) -
1
12
(0.18)(0.33
) = 0.2501(10-3
) m4
7–2. If the wide-flange beam is subjected to a shear of
determine the maximum shear stress in the beam.V = 20 kN,
A
B
V
20 mm
20 mm
20 mm
300 mm
200 mm
200 mm

For , Fig. a, Q as a function of y is
For , .Thus.
The sheer force resisted by the web is,
Ans.= 18.95 (103
) N = 19.0 kN
Vw = 2
L
0.15 m
0
tdA = 2
L
0.15 m
0
C3.459(10
6
) - 39.99(10
6
) y
2
D (0.02 dy)
= E3.459(106
) - 39.99(106
) y2
F Pa.
t =
VQ
It
=
20(103
) C0.865(10-3
) - 0.01y2
D
0.2501(10-3
) (0.02)
t = 0.02 m0 … y 6 0.15 m
= 0.865(10-3
) - 0.01y2
Q = ©y¿A¿ = 0.16 (0.02)(0.2) +
1
2
(y + 0.15)(0.15 - y)(0.02)
0 … y 6 0.15 m
I =
1
12
(0.2)(0.343
) -
1
12
(0.18)(0.33
) = 0.2501(10-3
) m4
determine the shear force resisted by the web
of the beam.
V = 20 kN,
474
A
B
V
20 mm
20 mm
20 mm
300 mm
200 mm
200 mm

475
Section Properties:
Shear Stress: Applying the shear formula
Ans.
Ans.
Ans.(tAB)W =
VQAB
I tW
=
12(64.8)
390.60(4)
= 0.498 ksi
(tAB)f =
VQAB
Itf
=
12(64.8)
390.60(12)
= 0.166 ksi
tmax =
VQmax
It
=
12(64.98)
390.60(4)
= 0.499 ksi
t =
VQ
It
QAB = yœ
2 A¿ = 1.8(3)(12) = 64.8 in3
Qmax = yœ
1 A¿ = 2.85(5.7)(4) = 64.98 in3
= 390.60 in4
INA =
1
12
(12)A33
B + 12(3)(3.30 - 1.5)2
+
1
12
(4)A63
B + 4(6)(6 - 3.30)2
y =
©yA
©A
=
1.5(12)(3) + 6(4)(6)
12(3) + 4(6)
= 3.30 in.
*7–4. If the T-beam is subjected to a vertical shear of
determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flange-
web junction AB. Sketch the variation of the shear-stress
intensity over the entire cross section.
V = 12 kip,
BB
V ϭ 12 kip
6 in.
3 in.
4 in.
4 in.
4 in.
A

Section Properties:
Shear Stress: Applying the shear formula
Resultant Shear Force: For the flange
Ans.= 3.82 kip
=
L
3.3 in
0.3 in
A0.16728 - 0.01536y2
B(12dy)
Vf =
LA
tdA
= 0.16728 - 0.01536y2
t =
VQ
It
=
12(65.34 - 6y2
)
390.60(12)
Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2
= 390.60 in4
INA =
1
12
(12)A33
B + 12(3)(3.30 - 1.5)2
+
1
12
(4)A63
B + 6(4)(6 - 3.30)2
y =
©yA
©A
=
1.5(12)(3) + 6(4)(6)
12(3) + 4(6)
= 3.30 in.
•7–5. If the T-beam is subjected to a vertical shear of
determine the vertical shear force resisted by
the flange.
V = 12 kip,
476
BB
V ϭ 12 kip
6 in.
3 in.
4 in.
4 in.
4 in.
A

477
Ans.
Ans.tB =
VQB
I t
=
15(103
)(0.59883)(10-3
)
0.218182(10-3
)0.025)
= 1.65 MPa
tA =
VQA
I t
=
15(103
)(0.7219)(10-3
)
0.218182(10-3
)(0.025)
= 1.99 MPa
QB = yAœ
B = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10-3
) m3
QA = yAœ
A = (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10-3
) m3
+
1
12
(0.2)(0.033
) + 0.2(0.03)(0.295 - 0.1747)2
= 0.218182(10-3
) m4
+
1
12
(0.025)(0.253
) + 0.25(0.025)(0.1747 - 0.155)2
I =
1
12
(0.125)(0.033
) + 0.125(0.03)(0.1747 - 0.015)2
y =
(0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)
0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)
= 0.1747 m
7–6. If the beam is subjected to a shear of
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
from the bottom and INA = 0.2182110-3
2 m4
.y = 0.1747 m
V = 15 kN,
A
B
V
30 mm
25 mm
30 mm
250 mm
200 mm
125 mm
A
B
V
30 mm
25 mm
30 mm
250 mm
200 mm
200 mm
Section Properties:
Ans.tmax =
VQ
I t
=
30(10)3
(1.0353)(10)-3
268.652(10)-6
(0.025)
= 4.62 MPa
Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10)-3
m3
I =
1
12
(0.2)(0.310)3
-
1
12
(0.175)(0.250)3
= 268.652(10)-6
m4
determine the maximum shear stress in the beam.V = 30 kN,

478
Ans.Vw = 30 - 2(1.457) = 27.1 kN
Vf = 1.457 kN
= 11.1669(10)6
[ 0.024025y -
1
2
y3
]
0.155
0.125
Vf =
L
tf dA = 55.8343(10)6
L
0.155
0.125
(0.024025 - y2
)(0.2 dy)
tf =
30(10)3
(0.1)(0.024025 - y2
)
268.652(10)-6
(0.2)
Q = a
0.155 + y
2
b(0.155 - y)(0.2) = 0.1(0.024025 - y2
)
I =
1
12
(0.2)(0.310)3
-
1
12
(0.175)(0.250)3
= 268.652(10)-6
m4
*7–8. If the wide-flange beam is subjected to a shear of
determine the shear force resisted by the web
of the beam.
V = 30 kN,
A
B
V
30 mm
25 mm
30 mm
250 mm
200 mm
200 mm
Ans.V = 32132 lb = 32.1 kip
8(103
) = -
V(3.3611)
6.75(2)(1)
tmax = tallow =
VQmax
I t
Qmax = ©y¿A¿ = 2 (0.91665)(1.8333)(1) = 3.3611 in3
+ 2 a
1
12
b(1)(23
) + 2(1)(2)(2 - 1.1667)2
= 6.75 in4
I =
1
12
(5)(13
) + 5 (1)(1.1667 - 0.5)2
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
1 (5) + 2(1)(2)
= 1.1667 in.
•7–9. Determine the largest shear force V that the member
can sustain if the allowable shear stress is tallow = 8 ksi.
V
3 in. 1 in.
1 in.
1 in.
3 in.

479
Ans.tmax =
VQmax
I t
=
18(3.3611)
6.75(2)(1)
= 4.48 ksi
Qmax = ©y¿A¿ = 2(0.91665)(1.8333)(1) = 3.3611 in3
+ 2 a
1
12
b(1)(23
) + 2(1)(2)(2 - 1.1667) = 6.75 in4
I =
1
12
(5)(13
) + 5(1)(1.1667 - 0.5)2
y =
(0.5)(1)(5) + 2 [(2)(1)(2)]
1 (5) + 2(1)(2)
= 1.1667 in.
7–10. If the applied shear force determine the
maximum shear stress in the member.
V = 18 kip,
V
3 in. 1 in.
1 in.
1 in.
3 in.
Ans.V = 100 kN
7(106
) =
V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)]
125(10-6
)(0.1)
tallow =
VQmax
It
I =
1
12
(0.2)(0.2)3
-
1
12
(0.1)(0.1)3
= 125(10-6
) m4
7–11. The wood beam has an allowable shear stress of
Determine the maximum shear force V that
can be applied to the cross section.
tallow = 7 MPa.
50 mm
50 mm
200 mm
100 mm
50 mm
V
50 mm

480
Section Properties The moment of inertia of the cross-section about the neutral axis is
Q as the function of y, Fig. a,
Qmax occurs when .Thus,
The maximum shear stress occurs of points along the neutral axis since Q is
maximum and the thickness is constant.
Ans.
Thus, the shear stress distribution as a function of y is
= E5.56 (36 - y2
)F psi
t =
VQ
It
=
12.8(103
)C4(36 - y2
)D
1152 (8)
V = 12800 16 = 12.8 kip
tallow =
VQmax
It
; 200 =
V(144)
1152(8)
t = 8 in.
Qmax = 4(36 - 02
) = 144 in3
y = 0
Q =
1
2
(y + 6)(6 - y)(8) = 4(36 - y2
)
I =
1
12
(8)(123
) = 1152 in4
*7–12. The beam has a rectangular cross section and is
made of wood having an allowable shear stress of
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.
tallow =
V
12 in.
8 in.

481
Section Properties:
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Applying the shear formula
Ans.= 4 22 MPa
=
20(103
)(87.84)(10-6
)
5.20704(10-6
)(0.08)
tmax =
VQmax
It
= 87.84A10-6
B m3
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)
Qmax = ©y¿A¿
= 5.20704 A10-6
B m4
INA =
1
12
(0.12)A0.0843
B -
1
12
(0.04)A0.063
B
7–13. Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 20 kN.
V
60 mm
12 mm
20 mm
20 mm
80 mm
12 mm
Section Properties:
Allowable shear stress: Maximum shear stress occurs at the point where the neutral
axis passes through the section.
Ans.V = 189 692 N = 190 kN
40A 106
B =
V(87.84)(10-6
)
5.20704(10-6
)(0.08)
tmax = tallow =
VQmax
It
= 87.84A 10-6
B m3
= 0.015(0.08)(0.03) + 0.036(0.012)(0.12)
Qmax = ©y¿A¿
= 5.20704A 10-6
B m4
INA =
1
12
(0.12)A0.0843
B -
1
12
(0.04)A0.063
B
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.
V
60 mm
12 mm
20 mm
20 mm
80 mm
12 mm

482
The maximum shear stress occur when
Ans.The faector =
tmax
tavg
=
4V
3 pc2
V
pc2
=
4
3
tavg =
V
A
=
V
p c2
tmax =
4V
3 pc2
y = 0
t =
VQ
I t
=
V[2
3 (c2
- y2
)
3
2]
(p
4 c4
)(22c2
- y2
)
=
4V
3pc4
[c2
- y2
)
Q =
L
x
y
2y2c2
- y2
dy = -
2
3
(c2
- y2
)
3
2 |
x
y =
2
3
(c2
- y2
)
2
3
dQ = ydA = 2y2c2
- y2
dy
dA = 2xdy = 22c2
- y2
dy
t = 2x = 22c2
- y2
x = 2c2
- y2
; I =
p
4
c4
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
V
y

483
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.
tmax =
3V
ah
tmax =
24V
a2
h
a
a
4
b a1 -
2
a
a
a
4
b b
y =
2h
a
a
a
4
b =
h
2
At x =
a
4
dt
dx
=
24V
a2
h2
a1 -
4
a
xb = 0
t =
24V(x - 2
a x2
)
a2
h
t =
VQ
It
=
V(4h2
>3a)(x2
)(1 - 2x
a )
((1>36)(a)(h3
))(2x)
t = 2x
Q = a
4h2
3a
b(x2
)a1 -
2x
a
b
Q =
LA¿
y dA = 2c a
1
2
b(x)(y)a
2
3
h -
2
3
yb d
y
x
=
h
a>2
; y =
2h
a
x
I =
1
36
(a)(h)3
*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. V
a
h

484
From Fig. a,
The maximum shear stress occurs at the points along the neutral axis since Q is
maximum and thickness is the smallest.
Ans.= 37.36(106
) Pa = 37.4 MPa
tmax =
VQmax
It
=
600(103
)[1.09125(10-3
)]
0.175275(10-3
) (0.1)
t = 0.1 m
= 1.09125(10-3
) m3
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375(0.075)(0.1)
I =
1
12
(0.3)(0.213
) -
1
12
(0.2)(0.153
) = 0.175275(10-3
) m4
•7–17. Determine the maximum shear stress in the strut if
it is subjected to a shear force of V = 600 kN.
V
150 mm
30 mm
100 mm
100 mm
100 mm
30 mm

485
From Fig. a
The maximum shear stress occeurs at the points along the neutral axis since Q is
maximum and thickness is the smallest.
Ans.V = 722.78(103
)N = 723 kN
tallow =
VQmax
It
; 45(106
) =
V C1.09125(10-3
)D
0.175275(10-3
)(0.1)
t = 0.1 m
= 1.09125 (10-3
) m3
Qmax = ©y¿A¿ = 0.09(0.03)(0.3) + 0.0375 (0.075)(0.1)
I =
1
12
(0.3)(0.213
) -
1
12
(0.2)(0.153
) = 0.175275 (10-3
) m4
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.
V
150 mm
30 mm
100 mm
100 mm
100 mm
30 mm

486
For , Fig. a, Q as a function of y is
For , Fig. b, Q as a function of y is0 … y 6 0.075 m
Q = y¿A¿ =
1
2
(0.105 + y) (0.105 - y)(0.3) = 1.65375(10-3
) - 0.15y2
0.075 m 6 y … 0.105 m
I =
1
12
(0.3)(0.213
) -
1
12
(0.2)(0.153
) = 0.175275(10-3
) m4
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.
V
150 mm
30 mm
100 mm
100 mm
100 mm
30 mm
Q = ©y¿A¿ = 0.09 (0.03)(0.3) +
1
2
(0.075 + y)(0.075 - y)(0.1) = 1.09125(10-3
) - 0.05 y2
For , .Thus,
At and ,
For , .Thus,
At and ,
The plot shear stress distribution over the cross-section is shown in Fig. c.
t|y=0 = 37.4 MPa ty=0.075 m = 27.7 MPa
y = 0.075 my = 0
t =
VQ
It
=
600(103
) [1.09125(10-3
) - 0.05y2
]
0.175275(10-3
) (0.1)
= (37.3556 - 1711.60 y2
) MPa
t = 0.1 m0 … y 6 0.075 m
t|y=0.015 m = 9.24 MPa ty=0.105 m = 0
y = 0.105 my = 0.075 m
t =
VQ
It
=
600(103
) C1.65375(10-3
) - 0.15y2
D
0.175275(10-3
) (0.3)
= (18.8703 - 1711.60y2
) MPa
t = 0.3 m0.075 m 6 y … 0.105 m

487
The moment of inertia of the ciralor cross-section about the neutral axis (x axis) is
Q for the differential area shown shaded in Fig. a is
However, from the equation of the circle, ,Then
Thus, Q for the area above y is
Here, .Thus
By inspecting this equation, at .Thus
Ans.tmax¿ =
20
2p
=
10
p
= 3.18 ksi
y = 0t = tmax
t =
5
2p
(4 - y2
) ksi
t =
VQ
It
=
30C2
3 (4 - y2
)
3
2
D
4p C2(4 - y2
)
1
2
D
t = 2x = 2(4 - y2
)
1
2
=
2
3
(4 - y2
)
3
2
= -
2
3
(4 - y2
)
3
2
Η
2 in
y
Q =
L
2 in
y
2y (4 - y2
)
1
2 dy
dQ = 2y(4 - y2
)
1
2 dy
x = (4 - y2
)
1
2
dQ = ydA = y(2xdy) = 2xy dy
I =
p
4
r4
=
p
4
(24
) = 4 p in4
*7–20. The steel rod is subjected to a shear of 30 kip.
Determine the maximum shear stress in the rod.
30 kip
2 in.
1 in.
A

The moment of inertia of the circular cross-section about the neutral axis (x axis) is
Q for the differential area shown in Fig. a is
However, from the equation of the circle, ,Then
Thus, Q for the area above y is
Here .Thus,
For point A, .Thus
Ans.
The state of shear stress at point A can be represented by the volume element
shown in Fig. b.
tA =
5
2p
(4 - 12
) = 2.39 ksi
y = 1 in
t =
5
2p
(4 - y2
) ksi
t =
VQ
It
=
30 C2
3 (4 - y2
)
3
2
D
4p C2(4 - y2
)
1
2
D
t = 2x = 2(4 - y2
)
1
2
= -
2
3
(4 - y2
)
3
2
`
2 in.
y
=
2
3
(4 - y2
)
3
2
Q =
L
2 in.
y
2y (4 - y2
)
1
2 dy
dQ = 2y (4 - y2
)
1
2 dy
x = (4 - y2
)
1
2
dQ = ydA = y (2xdy) = 2xy dy
I =
p
4
r4
=
p
4
(24
) = 4p in4
•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.
488
30 kip
2 in.
1 in.
A

489
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
(0.05)(0.02) + (0.07)(0.02)
= 0.03625 m
7–22. Determine the shear stress at point B on the web of
the cantilevered strut at section a–a.
a
a
2 kN 4 kN
250 mm 250 mm 300 mm
20 mm
50 mm
70 mm
20 mm
B
Ans.= 4.41 MPa
tB =
VQB
I t
=
6(103
)(26.25)(10-6
)
1.78622(10-6
)(0.02)
QB = (0.02)(0.05)(0.02625) = 26.25(10-6
) m3
yœ
B = 0.03625 - 0.01 = 0.02625 m
+
1
12
(0.02)(0.073
) + (0.02)(0.07)(0.055 - 0.03625)2
= 1.78625(10-6
) m4
I =
1
12
(0.05)(0.023
) + (0.05)(0.02)(0.03625 - 0.01)2
Ans.= 4.85 MPa
tmax =
VQmax
It
=
6(103
)(28.8906)(10-6
)
1.78625(10-6
)(0.02)
Qmax = y¿A¿ = (0.026875)(0.05375)(0.02) = 28.8906(10-6
) m3
+
1
12
(0.02)(0.073
) + (0.02)(0.07)(0.055 - 0.03625)2
= 1.78625(10-6
) m4
I =
1
12
(0.05)(0.023
) + (0.05)(0.02)(0.03625 - 0.01)2
y =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
(0.05)(0.02) + (0.07)(0.02)
= 0.03625 m
7–23. Determine the maximum shear stress acting at
section a–a of the cantilevered strut.
a
a
2 kN 4 kN
250 mm 250 mm 300 mm
20 mm
50 mm
70 mm
20 mm
B

490
*7–24. Determine the maximum shear stress in the T-beam
at the critical section where the internal shear force is
maximum.
3 m 1.5 m1.5 m
10 kN/m
A
150 mm
150 mm 30 mm
30 mm
B
C
The FBD of the beam is shown in Fig. a,
The shear diagram is shown in Fig. b.As indicated,
The neutral axis passes through centroid c of the cross-section, Fig. c.
From Fig. d,
The maximum shear stress occurs at points on the neutral axis since Q is maximum
and thickness is the smallest.
Ans.= 7.33 MPa
= 7.333(106
) Pa
tmax =
VmaxQmax
It
=
27.5(103
)C0.216(10-3
)D
27.0(10-6
)(0.03)
t = 0.03 m
= 0.216 (10-3
) m3
Qmax = y¿A¿ = 0.06(0.12)(0.03)
= 27.0(10-6
) m4
+
1
12
(0.15)(0.033
) + 0.15(0.03)(0.165 - 0.12)2
I =
1
12
(0.03)(0.153
) + 0.03(0.15)(0.12 - 0.075)2
= 0.12 m
y =
©
'
y A
©A
=
0.075(0.15)(0.03) + 0.165(0.03)(0.15)
0.15(0.03) + 0.03(0.15)
Vmax = 27.5 kN

491
using the method of sections,
The neutral axis passes through centroid C of the cross-section,
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.
Ans.= 3.667(106
) Pa = 3.67 MPa
tmax =
VC Qmax
It
=
13.75(103
) C0.216(10-3
)D
27.0(10-6
) (0.03)
= 0.216 (10-3
) m3
Qmax = y¿A¿ = 0.06 (0.12)(0.03)
= 27.0 (10-6
) m4
+
1
12
(0.15)(0.033
) + 0.15(0.03)(0.165 - 0.12)2
I =
1
12
(0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2
= 0.12 m
y =
©yA
©A
=
0.075 (0.15)(0.03) + 0.165(0.03)(0.15)
0.15(0.03) + 0.03(0.15)
VC = -13.75 kN
+ c©Fy = 0; VC + 17.5 -
1
2
(5)(1.5) = 0
•7–25. Determine the maximum shear stress in the
T-beam at point C. Show the result on a volume element
at this point.
3 m 1.5 m1.5 m
10 kN/m
A
150 mm
150 mm 30 mm
30 mm
B
C

492
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram, .
Section Properties:
Maximum Shear Stress: Maximum shear stress occurs at the point where the
neutral axis passes through the section.
Ans.=
878.57(12.375)
77.625(0.5)
= 280 psi
tmax =
VQmax
It
= 3.375(4)(0.75) + 1.5(3)(0.5) = 12.375 in3
Qmax = ©y¿A¿
INA =
1
12
(4) A7.53
B -
1
12
(3.5) A63
B = 77.625 in4
Vmax = 878.57 lb
7–26. Determine the maximum shear stress acting in the
fiberglass beam at the section where the internal shear
force is maximum.
A
150 lb/ft
D
0.75 in.
0.75 in.4 in.
6 in.
2 ft
6 ft6 ft
0.5 in.
200 lb/ft
4 in.

493
The FBD is shown in Fig. a.
Using the method of sections, Fig. b,
The moment of inertia of the beam’s cross section about the neutral axis is
QC and QD can be computed by refering to Fig. c.
Shear Stress. since points C and D are on the web, .
Ans.
Ans.tD =
VQD
It
=
9.00 (27)
276 (0.75)
= 1.17 ksi
tC =
VQC
It
=
9.00 (33)
276 (0.75)
= 1.43 ksi
t = 0.75 in
QD = yœ
3A¿ = 4.5 (1)(6) = 27 in3
= 33 in3
QC = ©y¿A¿ = 4.5 (1)(6) + 2 (4)(0.75)
I =
1
12
(6)(103
) -
1
12
(5.25)(83
) = 276 in4
V = 9.00 kip.
+ c©Fy = 0; 18 -
1
2
(3)(6) - V = 0
7–27. Determine the shear stress at points C and D
located on the web of the beam.
A
3 kip/ft
D
D
C
C
B
1 in.
1 in.6 in.
4 in.
4 in.
6 ft 6 ft6 ft
0.75 in.
6 in.

494
The shear diagram is shown in Fig. b, .
The moment of inertia of the beam’s cross-section about the neutral axis is
From Fig. c
The maximum shear stress occurs at points on the neutral axis since Q is the
maximum and thickness is the smallest
Ans.tmax =
Vmax Qmax
It
=
18.0 (33)
276 (0.75)
= 2.87 ksi
t = 0.75 in
= 33 in3
Qmax = ©y¿A¿ = 4.5 (1)(6) + 2(4)(0.75)
= 276 in4
I =
1
12
(6)(103
) -
1
12
(5.25)(83
)
Vmax = 18.0 kip
*7–28. Determine the maximum shear stress acting in the
beam at the critical section where the internal shear force
is maximum.
A
3 kip/ft
D
D
C
C
B
1 in.
1 in.6 in.
4 in.
4 in.
6 ft 6 ft6 ft
0.75 in.
6 in.

495
Force Equilibrium: The shaded area indicares the plastic zone. Isolate an element in
the plastic zone and write the equation of equilibrium.
This proves that the longitudinal shear stress. , is equal to zero. Hence the
corresponding transverse stress, , is also equal to zero in the plastic zone.
Therefore, the shear force is carried by the malerial only in the elastic zone.
Section Properties:
Maximum Shear Stress: Applying the shear formula
However, hence
‚ (Q.E.D.)tmax =
3P
2A¿
A¿ = 2by¿
tmax =
VQmax
It
=
VAy¿3
b
2 B
A2
3 by¿3
B(b)
=
3P
4by¿
Qmax = y¿ A¿ =
y¿
2
(y¿)(b) =
y¿2
b
2
INA =
1
12
(b)(2y¿)3
=
2
3
b y¿3
V = P
tmax
tlong
tlong = 0
;©Fx = 0; tlong A2 + sgA1 - sg A1 = 0
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 at the fixed support. If the
material is elastic-plastic, then at a distance the
moment creates a region of plastic yielding with
an associated elastic core having a height 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 where the cross-
sectional area of the elastic core.
A¿ = 2y¿b,tmax = 3
21P>A¿2,
2y¿.
M = Px
x 6 L
Mp = PL
h
b L
P
x
Plastic region
2y¿
Elastic region

496
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
Thus no shear stress is developed on the longitudinal or transverse plane of the
element. (Q. E. D.)
tlong = 0
; ©Fx = 0; sg A1 + tlong A2 - sg A1 = 0
7–31. The beam in Fig. 6–48f is subjected to a fully plastic
moment 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.
Mp.
Section Properties:
Shear Flow: There are two rows of nails. Hence, the allowable shear flow
.
Ans.V = 444 lb
166.67 =
V(12.0)
32.0
q =
VQ
I
q =
2(500)
6
= 166.67 lb>in
Q = y¿A¿ = 1(6)(2) = 12.0 in4
I =
1
12
(6)A43
B = 32.0 in4
*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.
V
2 in.
6 in.
6 in.
6 in.
2 in.
h
b L
P
x
Plastic region
2y¿
Elastic region

497
Section Properties:
Shear Flow:
There are two rows of nails. Hence, the shear force resisted by each nail is
Ans.F = a
q
2
bs = a
225 lb>in.
2
b(6 in.) = 675 lb
q =
VQ
I
=
600(12.0)
32.0
= 225 lb>in.
Q = y¿A¿ = 1(6)(2) = 12.0 in4
I =
1
12
(6)A43
B = 32.0 in4
•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
is applied to the boards, determine the shear
force resisted by each nail.
V = 600 lb
V
2 in.
6 in.
6 in.
6 in.
2 in.
7–34. The beam is constructed from two boards fastened
together with three rows of nails spaced 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 = 2 in. apart.
V
1.5 in.
s
s
6 in.
1.5 in.The moment of inertia of the cross-section about the neutral axis is
Refering to Fig. a,
The maximum shear stress occurs at the points on the neutral axis where Q is
maximum and .
Shear Flow: Since there are three rows of nails,
Ans.V = 1350 lb = 1.35 kip
qallow =
VQA
I
; 675 =
V(6.75)
13.5
qallow = 3a
F
s
b = 3a
450
2
b = 675 lb>in.
V = 3600 lb = 3.60 kips
tallow =
VQmax
It
; 300 =
V(6.75)
13.5(6)
t = 6 in
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
I =
1
12
(6)(33
) = 13.5 in4

498
Refering to Fig. a,
maximum and .
Ans.
Since there are three rows of nails,
Ans.s = 2.167 in = 2
1
8
in
qallow =
VQA
I
;
1950
s
=
1800(6.75)
13.5
qallow = 3a
F
s
b = 3¢
650
s
≤ = a
1950
s
b
lb
in.
V = 1800 lb = 1.80 kip
tallow =
VQmax
It
; 150 =
V(6.75)
13.5(6)
t = 6 in
QA = Qmax = y¿A¿ = 0.75(1.5)(6) = 6.75 in3
I =
1
12
(6)(33
) = 13.5 in4
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 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.
tallow = 150 psi,
V
1.5 in.
s
s
6 in.
1.5 in.

499
Section Properties:
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is
.
Ans.s = 5.53 in.
30
s
=
50(10.125)
93.25
q =
VQ
I
q =
2(15)
s
=
30
s
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
= 93.25 in4
-
1
12
(0.5)A23
B +
1
12
(1)A63
B
INA =
1
12
(3)A93
B -
1
12
(2.5)A83
B
*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 is
applied to the cross section, determine the maximum spacing
of the bolts. Each bolt can resist a shear force of 15 kip.
V = 50 kip
3 in.
3 in.
A
V
0.5 in.
1 in.
0.5 in.
6 in.
ss
NN
Section Properties:
.
Ans.y = 34.5 kip
3.75 =
V(10.125)
93.25
q =
VQ
I
q =
2(15)
8
= 3.75 kip>in
Q = ©y¿A¿ = 2.5(3)(0.5) + 4.25(3)(0.5) = 10.125 in3
= 93.25 in4
-
1
12
(0.5)A23
B +
1
12
(1)A63
B
INA =
1
12
(3)A93
B -
1
12
(2.5)A83
B
•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
determine the maximum shear force V that can
be applied to the cross section. Each bolt can resist a
shear force of 15 kip.
s = 8 in.,
3 in.
3 in.
A
V
0.5 in.
1 in.
0.5 in.
6 in.
ss
NN

500
The neutral axis passes through centroid C of the cross-section as shown in Fig. a.
Thus,
Q for the shaded area shown in Fig. b is
Since there are two rows of nails .
Thus, the shear stress developed in the nail is
Ans.tn =
F
A
=
442.62
p
4
(0.0042
)
= 35.22(106
)Pa = 35.2 MPa
F = 442.62 N
q =
VQ
I
; 26.67 F =
2000 C0.375 (10-3
)D
63.5417 (10-6
)
q = 2a
F
s
b =
2F
0.075
= (26.67 F) N>m
Q = y¿A¿ = 0.0375 (0.05)(0.2) = 0.375(10-3
) m3
= 63.5417(10-6
) m4
+
1
12
(0.05)(0.23
) + 0.05(0.2)(0.1375 - 0.1)2
I =
1
12
(0.2)(0.053
) + 0.2 (0.05)(0.175 - 0.1375)2
y =
©
'
y A
©A
=
0.175(0.05)(0.2) + 0.1(0.2)(0.05)
0.05(0.2) + 0.2(0.05)
= 0.1375 m
7–38. The beam is subjected to a shear of
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.
V = 2 kN.
75 mm
75 mm
50 mm
25 mm
200 mm
200 mm
25 mm V

501
Ans.F = q(s) = 49.997 (0.25) = 12.5 kN
q =
VQ
I
=
35 (0.386)(10-3
)
0.270236 (10-3
)
= 49.997 kN>m
Q = y¿A¿ = 0.06176(0.025)(0.25) = 0.386(10-3
) m3
= 0.270236 (10-3
) m4
+
1
12
(0.025)(0.35)3
+ (0.025)(0.35)(0.275 - 0.18676)2
I = (2)a
1
12
b(0.025)(0.253
) + 2 (0.025)(0.25)(0.18676 - 0.125)2
y =
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025)
2 (0.25)(0.025) + 0.35 (0.025)
= 0.18676 m
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 apart and the
applied shear is V = 35 kN.
s = 250 mm
s = 250 mm
250 mm100 mm
25 mm
25 mm
25 mm
350 mm
V
Section Properties:
.
Ans.s = 13.8 in.
1200
s
=
1500(168)
2902
q =
VQ
I
q =
2(600)
s
=
1200
s
Q = y¿A¿ = 7(4)(6) = 168 in3
INA =
1
12
(7)A183
B -
1
12
(6)A103
B = 2902 in4
Vmax = 1500 lb
*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 Assume A is
pinned and B is a roller.
P = 3000 lb.
P
B
s
A
2 in.
2 in.
10 in.
6 in.
0.5 in. 0.5 in.
2 in.
2 in.
4 ft 4 ft

502
Internal Shear Force and Moment: As shown on shear and moment diagram,
and .
Section Properties:
Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the
allowable shear flow is .
kip (Controls !) Ans.
Shear Stress: Assume failure due to shear stress.
Bending Stress: Assume failure due to bending stress.
P = 107 ksi
8(103
) =
2.00P(12)(9)
2902
smax = sallow =
Mc
I
P = 22270 lb = 83.5 kip
3000 =
0.500P(208.5)
2902(1)
tmax = tallow =
VQmax
It
P = 6910 lb = 6.91
200 =
0.500P(168)
2902
q =
VQ
I
q =
2(600)
6
= 200 lb>in
Qmax = ©y¿A¿ = 7(4)(6) + 4.5(9)(1) = 208.5 in3
Q = yœ
2A¿ = 7(4)(6) = 168 in3
INA =
1
12
(7)A183
B -
1
12
(6)A103
B = 2902 in4
Mmax = 2.00PVmax = 0.500P
•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
and the allowable shear stress is
If the fasteners are spaced and each fastener can
support 600 lb in single shear, determine the maximum load
P that can be applied to the beam.
s = 6 in.
tallow = 3 ksi.sallow = 8 ksi
P
B
s
A
2 in.
2 in.
10 in.
6 in.
0.5 in. 0.5 in.
2 in.
2 in.
4 ft 4 ft

503
The neutral axis passes through the centroid c of the cross-section as shown in Fig. a.
Refering to Fig. a, Qmax and QA are
maximum and .
Ans.
Here, .Then
Ans.s = 1.134 in = 1
1
8
in
qallow =
VQA
I
;
950
s
=
8815.51(84)
884
qallow =
F
s
=
950
s
lb>in
V = 8815.51 lb = 8.82 kip
tallow =
VQmax
It
; 450 =
V (90.25)
884 (2)
t = 2 in
QA = y2
œ
A2
œ
= 3.5 (2)(12) = 84 in3
Qmax = y1
œ
A1
œ
= 4.75(9.5)(2) = 90.25 in3
= 884 in4
+
1
12
(12)(23
) + 12(2)(13 - 9.5)2
I =
1
12
(2)(123
) + 2(12)(9.5 - 6)2
y =
©
'
y A
©A
=
13(2)(12) + 6(12)(2)
2(12) + 12(2)
= 9.5 in.
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 in.
The allowable shear stress for the wood is .tallow = 450 psi
1
8
12 in.
12 in.2 in.
2 in.
V
s
s

504
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.
As indicated in Fig. b, the internal shear force on the cross-section within region AB
is constant that is .
The neutral axis passes through centroid C of the cross section as shown in Fig. c.
Q for the shaded area shown in Fig. d is
Since there are two rows of nail, .
Thus, the average shear stress developed in each nail is
Ans.AtnailBavg =
F
Anail
=
1500
p
4
(0.0042
)
= 119.37(106
)Pa = 119 MPa
F = 1500 N
q =
VAB Q
I
; 20F =
5(103
)C0.32(10-3
)D
53.333(10-6
)
q = 2 a
F
s
b = 2 a
F
0.1
b = 20F N>m
Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10-3
) m3
= 53.333(10-6
) m4
+
1
12
(0.2)(0.043
) + 0.2(0.04)(0.18 - 0.14)2
I =
1
12
(0.04)(0.23
) + 0.04(0.2)(0.14 - 0.1)2
= 0.14 m
y =
©
'
y A
©A
=
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
0.04(0.2) + 0.2(0.04)
VAB = 5 kN
P
1.5 m 1.5 m
B CA
40 mm
20 mm
20 mm
100 mm
200 mm
200 mm
2 kN/m

505
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of
Constant value, .
The neutral axis passes through Centroid C of the cross-section as shown in Fig. c.
Refering to Fig. d,
The maximum shear stress occurs at the points on Neutral axis where Q is maximum
and .
Since there are two rows of nails .
Ans.P = 3.67 kN (Controls!)
qallow =
Vmax QA
I
; 40 000 =
(P + 3)(103
)C0.32(10-3
)D
53.333(10-6
)
qallow = 2 a
F
s
b = 2 c
2(103
)
0.1
d = 40 000 N>m
P = 13.33 kN
tallow =
Vmax Qmax
It
; 3(106
) =
(P + 3)(103
)C0.392(10-3
)D
53.333(10-6
)(0.04)
t = 0.04 m
QA = y2
œ
A2
œ
= 0.04(0.04)(0.2) = 0.32(10-3
) m3
Qmax = y1
œ
A1
œ
= 0.07(0.14)(0.04) = 0.392(10-3
) m3
= 53.333(10-6
) m4
+
1
12
(0.2)(0.043
) + 0.2(0.04)(0.18 - 0.142
)
I =
1
12
(0.04)(0.23
) + 0.04(0.2)(0.14 - 0.1)2
= 0.14 m
y =
©
'
y A
©A
=
0.18(0.04)(0.2) + 0.1(0.2)(0.04)
0.04(0.2) + 0.2(0.04)
Vmax = (P + 3) kN
*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
1.5 m 1.5 m
B CA
40 mm
20 mm
20 mm
100 mm
200 mm
200 mm
2 kN/m

506
7–44. Continued

507
Section Properties:
Shear Flow: There are two rows of nails. Hence the allowable shear flow is
.
Ans.P = 6.60 kN
60.0A103
B =
(P + 3)(103
)0.450(10-3
)
72.0(10-6
)
q =
VQ
I
q =
3(2)
0.1
= 60.0 kN>m
Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450A10-3
B m3
= 72.0A10-6
B m4
INA =
1
12
(0.31)A0.153
B -
1
12
(0.25)A0.093
B
VAB = (P + 3) kN
•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.
P
2 m 2 m
3 kN
B CA
30 mm
30 mm
30 mm
100 mm
250 mm30 mm
150 mm

508
Section Properties:
Shear Flow: The allowable shear flow at points C and D is and
, respectively.
Ans.
Ans.s¿ = 1.21 in.
100
s¿
=
700(39.6774)
337.43
qD =
VQD
I
s = 8.66 in.
100
s
=
700(5.5645)
337.43
qC =
VQC
I
qB =
100
s¿
qC =
100
s
QD = y2¿A¿ = (3.3548 - 0.5)(10)(1) + 2 C(3.3548 - 1.5)(3)(1)D = 39.6774 in3
QC = y1¿A¿ = 1.8548(3)(1) = 5.5645 in3
= 337.43 in4
+
1
12
(2)A33
B + 2(3)(3.3548 - 1.5)2
INA =
1
12
(10)A13
B + 10(1)(3.3548 - 0.5)2
= 3.3548 in
y =
©yA
©A
=
0.5(10)(1) + 1.5(2)(3) + 6(1.5)(10)
10(1) + 2(3) + 1.5(10)
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 sЈ and s if the beam
is subjected to a shear of .V = 700 lb
1.5 in.
10 in.
2 in.
B
V
1 in.
10 in.
1 in.
A
1 in.
C
s
s
D
s¿
s¿

509
Ans.
V = 485 lb
qA s = 0.0516V(2) = 50
qA =
1
2
a
VQA
I
b =
V(20.4)
2(197.7)
= 0.0516 V
QA = y2
œ
A¿ = 3.4(6)(1) = 20.4 in3
V = 317 lb (controls)
qB s = 0.0789V(2) = 50
qB =
1
2
a
VQB
I
b =
V(31.2)
2(197.7)
= 0.0789 V
QB = y1
œ
A¿ = 2.6(12)(1) = 31.2 in3
+
1
12
(6)(13
) + 6(1)(6.5 - 3.1)2
= 197.7 in4
+ 2a
1
12
b(1)(63
) + 2(1)(6)(4 - 3.1)2
I =
1
12
(12)(13
) + 12(1)(3.1 - 0.5)2
y =
©yA
©A
=
0.5 (12)(1) + 2 (4)(6)(1) + (6.5)(6)(1)
12(1) + 2(6)(1) + (6)(1)
= 3.1 in.
*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.
6 in.
1 in.
1 in.
1 in.
5 in.
V
12 in.
2 in.

510510
Refering to Fig. a Fig. b,
Due to symmety, the shear flow at points A and , Fig. a, and at points B and ,
Fig. b, are the same.Thus
Ans.
Ans.= 462.46(103
) N>m = 462 kN>m
qB =
1
2
a
VQB
I
b =
1
2
c
300(103
) C0.751(10-3
)D
0.24359(10-3
)
s
= 228.15(103
) N>m = 228 kN>m
qA =
1
2
a
VQA
I
b =
1
2
c
300(103
) C0.3705(10-3
)D
0.24359(10-3
)
s
B¿A¿
QB = 2yœ
zA2
œ
+ y3
œ
A3
œ
= 2[0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10-3
) m3
QA = y1
œ
A1
œ
= 0.195 (0.01)(0.19) = 0.3705 (10-3
) m3
I =
1
12
(0.2)(0.43
) -
1
12
(0.18)(0.383
) = 0.24359(10-3
) m4
7–50. A shear force of is applied to the box
girder. Determine the shear flow at points A and B.
V = 300 kN
100 mm
90 mm90 mm
200 mm
200 mm
180 mm
190 mm
10 mm
10 mm
V
A D
C
B

Refering to Fig. a, due to symmetry .Thus
Then refering to Fig. b,
Thus,
Ans.
Ans.= 601.33(103
) N>m = 601 kN>m
qD =
VQD
I
=
450(103
) C0.3255(10-3
)D
0.24359(10-3
)
qC =
VQC
I
= 0
= 0.3255(10-3
) m3
QD = y1
œ
A1
œ
+ y2
œ
A2
œ
= 0.195 (0.01)(0.09) + 0.15(0.1)(0.01)
QC = 0
AC
œ
= 0
I =
1
12
(0.2)(0.43
) -
1
12
(0.18)(0.383
) = 0.24359(10-3
) m4
7–51. A shear force of is applied to the box
girder. Determine the shear flow at points C and D.
V = 450 kN
100 mm
90 mm90 mm
200 mm
200 mm
180 mm
190 mm
10 mm
10 mm
V
A D
C
B

512
Section Properties:
Shear Flow:
Ans.
Ans.= 9437 N>m = 9.44 kN>m
=
1
2
c
18(103
)(0.13125)(10-3
)
125.17(10-6
)
d
qB =
1
2
c
VQB
I
d
= 13033 N>m = 13.0 kN>m
=
1
2
c
18(103
)(0.18125)(10-3
)
125.17(10-6
)
d
qA =
1
2
c
VQA
I
d
QB = y1
œ
A¿ = 0.105(0.125)(0.01) = 0.13125A10-3
B m3
QA = y2
œ
A¿ = 0.145(0.125)(0.01) = 0.18125A10-3
B m3
= 125.17A10-6
B m4
+ 2c
1
12
(0.125)A0.013
B + 0.125(0.01)A0.1052
B d
INA =
1
12
(0.145)A0.33
B -
1
12
(0.125)A0.283
B
*7–52. A shear force of is applied to the
symmetric box girder. Determine the shear flow at A and B.
V = 18 kN
C
A
150 mm
10 mm
10 mm
100 mm
100 mm
10 mm
10 mm
125 mm
150 mm
10 mm
30 mm
B
V10 mm

513
Section Properties:
Shear Flow:
Ans.= 38648 N>m = 38.6 kN>m
=
1
2
c
18(103
)(0.5375)(10-3
)
125.17(10-4
)
d
qC =
1
2
c
VQC
I
d
= 0.5375A10-3
B m3
= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02)
QC = ©y¿A¿
= 125.17A10-6
B m4
+2c
1
12
(0.125)A0.013
B + 0.125(0.01)A0.1052
B d
INA =
1
12
(0.145)A0.33
B -
1
12
(0.125)A0.283
B
•7–53. A shear force of is applied to the box
girder. Determine the shear flow at C.
V = 18 kN
C
A
150 mm
10 mm
10 mm
100 mm
100 mm
10 mm
10 mm
125 mm
150 mm
10 mm
30 mm
B
V10 mm

514
Ans.
Ans.qB =
VQB
I
=
150(8.1818)(10-6
)
0.98197(10-6
)
= 1.25 kN>m
qA =
VQA
I
=
150(9.0909)(10-6
)
0.98197(10-6
)
= 1.39 kN>m
QB = yB¿A¿ = 0.027272(0.03)(0.01) = 8.1818(10-6
) m3
QA = yA¿A¿ = 0.022727(0.04)(0.01) = 9.0909(10-6
) m3
yA¿ = 0.027727 - 0.005 = 0.022727 m
yB¿ = 0.055 - 0.027727 = 0.027272 m
+
1
12
(0.04)(0.01)3
+ 0.04(0.01)(0.055 - 0.027727)2
= 0.98197(10-6
) m4
+ 2c
1
12
(0.01)(0.06)3
+ 0.01(0.06)(0.03 - 0.027727)2
d
I = 2c
1
12
(0.03)(0.01)3
+ 0.03(0.01)(0.027727 - 0.005)2
d
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)
= 0.027727 m
7–54. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of, ,
determine the shear flow at points A and B.
V = 150 N
30 mm
40 mm
30 mm
V
A
B40 mm
10 mm
10 mm
10 mm10 mm
= 0.98197(10-6
) m4
+
1
12
(0.04)(0.01)3
+ 0.04(0.01)(0.055 - 0.027727)2
+ 2c
1
12
(0.01)(0.06)3
+ 0.01(0.06)(0.03 - 0.027727)2
d
I = 2c
1
12
(0.03)(0.01)3
+ 0.03(0.01)(0.027727 - 0.005)2
d
= 0.027727 m
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)
7–55. The aluminum strut is 10 mm thick and has the cross
section shown. If it is subjected to a shear of ,
determine the maximum shear flow in the strut.
V = 150 N
30 mm
40 mm
30 mm
V
A
B40 mm
10 mm
10 mm
10 mm10 mm
Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a
0.06 - 0.0277
2
b
Ans.qmax =
1
2
a
VQmax
I
b =
1
2
a
150(21.3(10-6
))
0.98197(10-6
)
b = 1.63 kN>m
= 21.3(10-6
) m3

515
y =
©yA
©A
=
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
11(0.5) + 2(8)(0.5) + 10(0.5)
= 3.70946 in.
*7–56. The beam is subjected to a shear force of
. Determine the shear flow at points A and B.V = 5 kip
A
V
0.5 in.
0.5 in.
5 in.
5 in.
0.5 in.
2 in. 0.5 in.
8 in.
B
A
C
D
I =
1
12
(11)(0.53
) + 11(0.5)(3.70946 - 0.25)2
+ 2c
1
12
(0.5)(83
) + 0.5(8)(4.5 - 3.70946)2
d
Ans.
Ans.qB =
1
2
a
VQB
I
b =
1
2
a
5(103
)(12.7027)
145.98
b = 218 lb>in.
qA =
1
2
a
VQA
I
b =
1
2
a
5(103
)(19.02703)
145.98
b = 326 lb>in.
QB = yB
œ
A¿ = 2.54054(10)(0.5) = 12.7027 in3
QA = yA
œ
A¿ = 3.45946(11)(0.5) = 19.02703 in3
yB
œ
= 6.25 - 3.70946 = 2.54054 in.
yA
œ
= 3.70946 - 0.25 = 3.45946 in.
= 145.98 in4
+
1
12
(10)(0.53
) + 10(0.5)(6.25 - 3.70946)2
Ans.= 414 lb>in.
qmax =
1
2
a
VQmax
I
b =
1
2
a
5(103
)(24.177)
145.98
b
= 24.177 in3
Qmax = 3.4594 (11)(0.5) + 2[(1.6047)(0.5)(3.7094 - 0.5)]
= 145.98 in4
+
1
12
(10)(0.53
) + 10(0.5)(2.54052
)
I =
1
12
(11)(0.53
) + 11(0.5)(3.45952
) + 2c
1
12
(0.5)(83
) + 0.5(8)(0.79052
)d
y =
©yA
©A
=
0.25(11)(0.5) + 2[4.5(8)(0.5)] + 6.25(10)(0.5)
11(0.5) + 2(8)(0.5) + 10(0.5)
= 3.70946 in.
•7–57. The beam is constructed from four plates and is
subjected to a shear force of . Determine the
maximum shear flow in the cross section.
V = 5 kip
A
V
0.5 in.
0.5 in.
5 in.
5 in.
0.5 in.
2 in. 0.5 in.
8 in.
B
A
C
D

516
Ans.qA =
75(103
)(0.3450)(10-3
)
0.12025(10-3
)
= 215 kN>m
q =
VQ
I
QA = yA
œ
A¿ = 0.0575(0.2)(0.03) = 0.3450(10-3
) m3
+ 2c
1
12
(0.03)(0.23
) + 0.03(0.2)(0.13 - 0.0725)2
d = 0.12025(10-3
) m4
I =
1
12
(0.4)(0.033
) + 0.4(0.03)(0.0725 - 0.015)2
y =
©yA
©A
=
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
0.4(0.03) + 2(0.2)(0.03)
= 0.0725 m
7–58. The channel is subjected to a shear of .
Determine the shear flow developed at point A.
V = 75 kN
200 mm
30 mm
30 mm
V ϭ 75 kN
30 mm
400 mm
A
Ans.qmax =
75(103
)(0.37209)(10-3
)
0.12025(10-3
)
= 232 kN>m
Qmax = y¿A¿ = 0.07875(0.1575)(0.03) = 0.37209(10-3
) m3
= 0.12025(10-3
) m4
+ 2c
1
12
(0.03)(0.23
) + 0.03(0.2)(0.13 - 0.0725)2
d
I =
1
12
(0.4)(0.033
) + 0.4(0.03)(0.0725 - 0.015)2
= 0.0725 m
y =
©yA
©A
=
0.015(0.4)(0.03) + 2[0.13(0.2)(0.03)]
0.4(0.03) + 2(0.2)(0.03)
7–59. The channel is subjected to a shear of .
Determine the maximum shear flow in the channel.
V = 75 kN
200 mm
30 mm
30 mm
V ϭ 75 kN
30 mm
400 mm
A

517
Section Properties:
Shear Flow:
Ans.
At Ans.y = 0, q = qmax = 424 lb>in.
= {424 - 136y2
} lb>in.
=
2(103
)(0.55243 - 0.17678y2
)
2.604167
q =
VQ
I
= 0.55243 - 0.17678y2
Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 -
y
sin 45°
b(0.25)
INA = 2c
1
12
(0.35355)A3.535533
B d = 2.604167 in4
h = 5 cos 45° = 3.53553 in.
b =
0.25
sin 45°
= 0.35355 in.
*7–60. The angle is subjected to a shear of .
Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
V = 2 kip
45Њ 45Њ
V
A
B
5 in.
5 in.
0.25 in.

518
y =
©yA
©A
=
(0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5)
0.5(11) + 2(0.5)(5.5) + 7(0.5)
= 2.8362 in.
•7–61. The assembly is subjected to a vertical shear of
. Determine the shear flow at points A and B and
the maximum shear flow in the cross section.
V = 7 kip
6 in.
0.5 in.
2 in.
2 in.
6 in.
0.5 in.
V
A
B
0.5 in.
0.5 in.
0.5 in.
I =
1
12
(11)(0.53
) + 11(0.5)(2.8362 - 0.25)2
+ 2a
1
12
b(0.5)(5.53
) + 2(0.5)(5.5)(3.25 - 2.8362)2
Ans.
Ans.
Ans.qmax =
1
2
a
7(103
)(16.9531)
92.569
b = 641 lb>in.
qB =
1
2
a
7(103
)(11.9483)
92.569
b = 452 lb>in.
qA =
7(103
)(2.5862)
92.569
= 196 lb>in.
q =
VQ
I
Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3
QB = y2¿A2¿ = (3.4138)(7)(0.5) = 11.9483 in3
QA = y1¿A1¿ = (2.5862)(2)(0.5) = 2.5862 in3
+
1
12
(7)(0.53
) + (0.5)(7)(6.25 - 2.8362)2
= 92.569 in4

519
Here
Therefore
Here
Ans.
occurs at ; therefore
; therefore
QEDtmax =
2V
A
A = 2pRt
tmax =
V
pR t
y = 0tmax
t =
V
pR2
t
2R2
- y2
cos u =
2R2
- y2
R
t =
VQ
I t
=
V(2R2
t cos u)
pR3
t(2t)
=
V cosu
pR t
=
R3
t
2
[u -
sin 2u
2
] Η
2p
0
=
R3
t
2
[2p - 0] = pR3
t
I =
L
2p
0
R3
t sin2
u du = R3
t
L
2p
0
(1 - cos 2u)
2
du
dI = y2
dA = y2
R t du = R3
t sin2
u du
= R2
t [-cos (p - u) - (-cosu)] = 2R2
t cosu
Q =
L
p-u
u
R2
t sin u du = R2
t(-cosu) |
p-u
u
dQ = R2
t sin u du
y = R sin u
dQ = y dA = yR t du
dA = R t du
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 , where Hint: Choose a
differential area element . Using
formulate Q for a circular section from to and show
that where cos u = 2R2
- y2
>R.Q = 2R2
t cos u,
(p - u)u
dQ = ydA,dA = Rt du
A = 2prt.tmax = 2V>A
t
y
du
ds
R
u

520
Section Properties:
Shear Flow Resultant:
Shear Center: Summing moment about point A.
Ans.
Note that if , (I shape).e = 0b2 = b1
e =
3(b2
2
- b1
2
)
h + 6(b1 + b2)
Pe =
3Pb2
2
hCh + 6(b1 + b2)D
(h) -
3Pb1
2
hCh + 6(b1 + b2)D
(h)
Pe = AFfB2 h - AFfB1 h
=
3Pb2
2
hCh + 6(b1 + b2)D
(Ff)2 =
L
b2
0
q2 dx2 =
6P
hCh + 6(b1 + b2)D L
b2
0
x2 dx2
=
3Pb1
2
hCh + 6(b1 + b2)D
(Ff)1 =
L
b1
0
q1 dx1 =
6P
hCh + 6(b1 + b2)D L
b1
0
x1 dx1
q2 =
VQ2
I
=
PAht
2 x2B
t h2
12 Ch + 6(b1 + b2)D
=
6P
hCh + 6(b1 + b2)D
x2
q1 =
VQ1
I
=
PAht
2 x1B
t h2
12 Ch + 6(b1 + b2)D
=
6P
hCh + 6(b1 + b2)D
x1
Q2 = y¿A¿ =
h
2
(x2)t =
h t
2
x2
Q1 = y¿A¿ =
h
2
(x1)t =
h t
2
x1
I =
1
12
t h3
+ 2c(b1 + b2)ta
h
2
b
2
d =
t h2
12
Ch + 6(b1 + b2)D
7–63. Determine the location e of the shear center,
point O, for the thin-walled member having the cross
section shown where . The member segments have
the same thickness t.
b2 7 b1
Oh
b2 b1
t
e

521
Section Properties:
Shear Center: Summing moments about point A,
Ans.e =
3b2
2(d + 3b)
Pe = c
3b2
sin 45°
2d(d + 3b)
Pd(2d sin 45°)
Pe = Ff(2d sin 45°)
Ff =
L
b
0
qfdx =
3P sin 45°
d(d + 3b) L
b
0
xdx =
3b
2
sin 45°
2d(d + 3b)
P
qf =
VQ
I
=
P(td sin 45°)x
td2
3 (d + 3b)
=
3P sin 45°
d(d + 3b)
x
Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x
=
td2
3
(d + 3b)
I =
1
12
a
t
sin 45°
b(2d sin 45°)3
+ 2Cbt(d sin 45°)2
D
*7–64. Determine the location e of the shear center,
section shown. The member segments have the same
thickness t.
45Њ
45Њ
d
O
b
e

522
Section Properties:
Shear Center: Summing moments about point A.
Ans.e =
7
10
a
Pe = 2a
P
20
ba + a
3
10
Pb2a
Pe = 2(Fw)1 (a) + Ff(2a)
Ff =
L
a
0
q2 dx =
3P
20a2
L
a
0
(a + 2x)dx =
3
10
P
(Fw)1 =
L
a
0
q1 dy =
3P
20a3
L
a
0
y2
dy =
P
20
q2 =
VQ2
I
=
PCat
2 (a + 2x)D
10
3 a3
t
=
3P
20a2
(a + 2x)
q1 =
VQ1
I
=
PA1
2 y2
B
10
3 a3
t
=
3P
20a3
y2
Q2 = ©y¿A¿ =
a
2
(at) + a(xt) =
at
2
(a + 2x)
Q1 = y1
œ
A¿ =
y
2
(yt) =
t
2
y2
I =
1
12
(2t)(2a)3
+ 2CatAa2
B D =
10
3
a3
t
•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.
t
a
e
a
a
O

523
Summing moments about A.
Ans.e =
223
3
a
F2 =
V
a
(a) +
2
3
a
V
2a
b(a) =
4V
3
q2 = q1 +
V(a>2)(t)(a>4)
1
4 t a3
= q1 +
V
2a
q1 =
V(a)(t)(a>4)
1
4 t a3
=
V
a
I =
1
12
(t)(a)3
+
1
12
a
t
sin 30°
b(a)3
=
1
4
t a3
Pe = F2 a
13
2
ab
section shown. a
a
a
60Њ
60Њ
O
e

524
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 . In
order to satisfy this equilibrium requirement. F1 and F2 must be equal to zero.
Shear Center: Summing moments about point A.
Ans.
Also,
The shear flows through the section as indicated by F1, F2, F3.
However,
To satisfy this equation, the section must tip so that the resultant of
Also, due to the geometry, for calculating F1 and F3, we require .
Hence, Ans.e = 0
F1 = F3
= P
:
+ F3
:
+ F2
:
F1
:
+
:©Fx Z 0
Pe = F2(0) e = 0
(©Fx Z 0)
thickness t.
b
b
O
t
e
h
2
h
2

525
Thus,
Summing moments about A:
Ans.e = 1.26 r
e =
r (1.9634 + 2)
3.15413
Pe =
Pr
3.15413 L
p>2
-p>2
(0.625 + cos u)du
Pe =
L
p>2
-p>2
(q r du)r
q =
VQ
I
=
P(0.625 + cos u)t r2
3.15413 t r3
Q = 0.625 t r2
+ t r2
cos u
Q = a
r
2
bt a
r
4
+ rb +
L
p>2
u
r sin u(t r du)
I = 1.583333t r3
+ t r3
a
p
2
b = 3.15413t r3
Isemi-circle = t r3
a
p
2
b
Isemi-circle =
L
p>2
-p>2
(r sin u)
2
t r du = t r3
L
p>2
-p>2
sin2
u du
= 1.583333t r3
+ Isemi-circle
I = (2)c
1
12
(t)(r>2)3
+ (r>2)(t)ar +
r
4
b
2
d + Isemi-circle
*7–68. Determine the location e of the shear center,
point O, for the beam having the cross section shown. The
thickness is t.
e
O
r
1
—
2
r
1
—
2
r

526
(1)
From Eq, (1).
Ans.e =
t
12I
(6h1 h2
b + 3h2
b2
- 8h1
3
b) =
b(6h1 h2
+ 3h2
b - 8h1
3
)
2h3
+ 6bh2
- (h - 2h1)3
I =
t
12
(2h3
+ 6bh2
- (h - 2h1)3
)
Pe =
Pt
2I
[h1 h2
b - h1
2
hb +
h2
b2
2
+ hh1
2
b -
4
3
h1
3
b]
F =
L
q2 dx =
Pt
2I L
b
0
[h1 (h - h1) + hx]dx =
Pt
2I
ah1 hb - h1
2
b +
hb2
2
b
q2 =
VQ2
I
=
Pt
2I
(h1 (h - h1) + hx)
Q2 = ©y¿A¿ =
1
2
(h - h1)h1 t +
h
2
(x)(t) =
1
2
t[h1 (h - h1) + hx]
V =
L
q1 dy =
Pt
2I L
h1
0
(hy - 2h1 y + y2
)dy =
Pt
2I
c
hh1
2
2
-
2
3
h1
3
d
q1 =
VQ
I
=
Pt(hy - 2h1 y + y2
)
2I
Q1 = y¿A¿ =
1
2
(h - 2h1 + y)yt =
t(hy - 2h1 y + y2
)
2
=
th3
6
+
bth2
2
-
t(h - 2h1)3
12
I =
1
12
(t)(h3
) + 2b(t)a
h
2
b
2
+
1
12
(t)[h3
- (h - 2h1)3
]
Pe = F(h) + 2V(b)
•7–69. Determine the location e of the shear center,
thickness t.
h
b
e
O
h1
h1

527
(1)
dQ = y dA = r sinu(t r du) = r2
t sin u du
=
r3
t
2
2 (2a - 2 sina cosa) =
r3
t
2
(2a - sin 2a)
=
r3
t
2
c ap + a -
sin 2(p + a)
2
b - ap - a -
sin 2(p - a)
2
b d
=
r3
t
2
(u -
sin 2u
2
) Η
p+a
p-a
I = r3
t
L
sin2
u du = r3
t
L
p+a
p-a
1 - cos 2u
2
du
dI = y2
dA = r2
sin2
u(t r du) = r3
t sin2
udu
y = r sin u
dA = t ds = t r du
P e = r
L
dF
7–70. Determine the location e of the shear center, point O,
for the thin-walled member having the cross section shown.
e
r O
a
a
t
Q = r2
t
L
u
p-a
sinu du = r2
t (-cosu)|
u
p-a
= r2
t(-cosu - cosa) = -r2
t(cosu + cosa)
L
dF =
L
q ds =
L
q r du
q =
VQ
I
=
P(-r2
t)(cosu + cosa)
r3
t
2 (2a - sin 2a)
=
-2P(cosu + cosa)
r(2a - sin 2a)
L
dF =
2P r
r(2a - sin 2a) L
p+p
p-a
(cosu + cosa) du =
-2P
2a - sin 2a
(2a cosa - 2 sina)
From Eq. (1);
Ans.e =
4r (sina - a cosa)
2a - sin 2a
P e = r c
4P
2a - sin 2a
(sina - a cosa)d
=
4P
2a - sin 2a
(sina - a cosa)

528
Section Properties:
(Q.E.D)
Shear Stress: Applying the shear formula ,
At
At
At
Resultant Shear Force: For segment AB.
Ans.= 9957 lb = 9.96 kip
=
L
0.8947 in
2.8947 in
A3173.76 - 40.12y2
2
B dy
=
L
0.8947 in
2.8947 in
A1586.88 - 20.06y2
2
B (2dy)
VAB =
L
tAB dA
y2 = 2.8947 in. tAB = 1419 psi
= {1586.88 - 20.06y2
2
} psi
tAB =
VQ2
It
=
35(103
)(79.12 - y2
2
)
872.49(2)
y1 = -2.8947 in. tCB = 355 psi
y1 = 0, tCB = 523 psi
= {522.77 - 20.06y1
2
} psi
tCB =
VQ1
It
=
35(103
)(104.25 - 4y1
2
)
872.49(8)
t =
VQ
It
= 79.12 - y2
2
Q2 = y2
œ
A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2)
= 104.25 - 4y1
2
Q1 = y1
œ
A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8)
= 872.49 in4
+
1
12
(2)A63
B + 2(6)(11 - 5.1053)2
INA =
1
12
(8)A83
B + 8(8)(5.1053 - 4)2
y =
©yA
©A
=
4(8)(8) + 11(6)(2)
8(8) + 6(2)
= 5.1053 in.
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 . Show that
INA = 872.49 in4
.
V = 35 kip
2 in.
3 in.
3 in.
6 in.
8 in.
A
B
C
V

529
Section Properties:
Shear Flow:
Hence, the shear force resisted by each nail is
Ans.
Ans.FD = qDs = (460.41 lb>in.)(3 in.) = 1.38 kip
FC = qCs = (65.773 lb>in.)(3 in.) = 197 lb
qD =
VQD
I
=
4.5(103
)(42.0)
410.5
= 460.41 lb>in.
qC =
VQC
I
=
4.5(103
)(6.00)
410.5
= 65.773 lb>in.
QD = yœ
2A¿ = 3.50(12)(1) = 42.0 in2
QC = yœ
1A¿ = 1.5(4)(1) = 6.00 in2
= 410.5 in4
+
1
12
(1)A123
B + 1(12)(7 - 3.50)2
+
1
12
(2)A43
B + 2(4)(3.50 - 2)2
INA =
1
12
(10)A13
B + (10)(1)(3.50 - 0.5)2
y =
© yA
©A
=
0.5(10)(1) + 2(4)(2) + 7(12)(1)
10(1) + 4(2) + 12(1)
= 3.50 in.
*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 The beam is subjected to a
shear of V = 4.5 kip.
s = 3 in.
1 in.
12 in.
3 in.
B
V
1 in.
10 in.
A
1 in.
1 in.

530
Section Properties:
Ans.
Shear Flow:
Ans.
Ans.
Ans.qC =
VQC
I
=
2(103
)(0.16424)(10-3
)
86.93913(10-6
)
= 3.78 kN>m
qB =
VQB
I
=
2(103
)(52.57705)(10-6
)
86.93913(10-6
)
= 1.21 kN>m
qA =
VQA
I
= 0
= 0.16424A10-3
B m3
= 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015)
QC = ©y¿A¿
QB =
'
y œ
1A¿ = 0.03048(0.115)(0.015) = 52.57705A10-6
B m3
QA = 0
= 86.93913A10-6
B m4
+
1
12
(0.015)A0.33
B + 0.015(0.3)(0.165 - 0.08798)2
+
1
12
(0.03)A0.1153
B + 0.03(0.115)(0.08798 - 0.0575)2
INA =
1
12
(0.2)A0.0153
B + 0.2(0.015)(0.08798 - 0.0075)2
= 0.08798 m
=
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)
y =
© yA
©A
•7–73. The member is subjected to a shear force of
. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
V = 2 kN
300 mm
A
B
C
100 mm
200 mm
V ϭ 2 kN

531
Section Properties:
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
.
Ans.V = 4100 lb = 4.10 kip
150 =
V(3.50)
95.667
q =
VQ
I
2(75) = 150 lb>in
Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3
= 95.667 in4
INA =
1
12
(1)A103
B + 2c
1
12
(4)A0.53
B + 4(0.5)A1.752
B d
7–74. The beam is constructed from four boards glued
together at their seams. If the glue can withstand
what is the maximum vertical shear V that the
beam can support?
75 lb>in.,
3 in.
4 in.
3 in.
3 in.
0.5 in.
0.5 in.
0.5 in.0.5 in.
V
Section Properties:
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is
.
Ans.V = 749 lb
150 =
V(11.25)
56.167
q =
VQ
I
2(75) = 150 lb>in
Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3
INA =
1
12
(10)A53
B -
1
12
(9)A43
B = 56.167 in4
7–75. Solve Prob. 7–74 if the beam is rotated 90° from the
position shown.
3 in.
4 in.
3 in.
3 in.
0.5 in.
0.5 in.
0.5 in.0.5 in.
V

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