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Solution manual mechanics of materials 8th edition hibbeler chapter 07


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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.

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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 ϭ 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.

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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 ϭ 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.

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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

200 mm


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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.

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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

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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.

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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.

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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

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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

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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

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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 Η
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.

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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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*7–28. Determine the 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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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–30. 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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