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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.

472

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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 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.

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

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.

478

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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 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.

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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