Solution manual mechanics of materials 8th edition hibbeler chapter 11

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11–1. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 6.5 MPa and an
allowable shear stress of tallow = 500 kPa. Determine its
dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.

8 kN/m

2m

Ix =

1
(b)(1.25b)3 = 0.16276b4
12

Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3
Assume bending moment controls:
Mmax = 16 kN # m
sallow =

Mmax c
I

6.5(106) =

16(103)(0.625b)
0.16276b4

b = 0.21143 m = 211 mm

Ans.

h = 1.25b = 264 mm

Ans.

Check shear:
Qmax = 1.846159(10 - 3) m3

I = 0.325248(10 - 3) m4
tmax =

VQmax
16(103)(1.846159)(10 - 3)
= 429 kPa 6 500 kPa‚ OK
=
It
0.325248(10 - 3)(0.21143)

830

4m

2m


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11–2. The brick wall exerts a uniform distributed load
of 1.20 kip>ft on the beam. If the allowable bending stress
is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi, select the lightest wide-flange section with
the shortest depth from Appendix B that will safely support
the load.

1.20 kip/

4 ft

10 ft

ft

6 ft
b

Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming
bending controls the design and applying the flexure formula.
Sreq d =

=

44.55 (12)
= 24.3 in3
22

W12 * 22

A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B
V
for the W12 * 22 wide tw d
= 6.60 kip.

Shear Stress: Provide a shear stress check using t =
flange section. From the shear diagram, Vmax
tmax =

=

Vmax
tw d

6.60
0.260(12.31)

= 2.06 ksi 6 tallow = 12 ksi (O.K!)
Hence,

Use

9 in.
0.5 in.

Mmax
sallow

Two choices of wide flange section having the weight 22 lb>ft can be made. They
are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest.
Select

0.5 in.

0.5 in.

Ans.

W12 * 22

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11–3. The brick wall exerts a uniform distributed load
of 1.20 kip>ft on the beam. If the allowable bending stress
is sallow = 22 ksi, determine the required width b of the
flange to the nearest 14 in.

1.20 kip/

4 ft

10 ft

ft

6 ft
b
0.5 in.

0.5 in.
9 in.
0.5 in.

Section Property:
I =

1
1
(b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375
12
12

Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft.
sallow =
22 =

Mmax c
I
44.55(12)(5)
22.583b + 30.375

b = 4.04 in.
Use

b = 4.25 in.

Ans.

832


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*11–4. Draw the shear and moment diagrams for the
shaft, and determine its required diameter to the nearest
1
4 in. if sallow = 7 ksi and tallow = 3 ksi. The bearings at A
and D exert only vertical reactions on the shaft. The loading
is applied to the pulleys at B, C, and E.

14 in.

20 in.

15 in.

12 in.

E
A

C

B

D
35 lb

80 lb
110 lb

sallow =
7(103) =

Mmax c
I
1196 c
p 4 ;
4 c

c = 0.601 in.

d = 2c = 1.20 in.
Use d = 1.25 in.

Ans.

Check shear:
2

tmax =

0.625

108(4(0.625)
Vmax Q
3p )(p)( 2 )
= 117 psi 6 3 ksi OK
=
p
4
It
4 (0.625) (1.25)

•11–5.

Select the lightest-weight steel wide-flange beam
from Appendix B that will safely support the machine loading
shown. The allowable bending stress is sallow = 24 ksi and
the allowable shear stress is tallow = 14 ksi.

2 ft

Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft.
Assume bending controls the design. Applying the flexure formula.
Sreq¿d =

=
Select

W12 * 16

Mmax
sallow

30.0(12)
= 15.0 in3
24

A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B
V
for the W12 * 16 wide tw d
= 10.0 kip

Shear Stress: Provide a shear stress check using t =
flange section. From the shear diagram, Vmax
tmax =

=

Vmax
tw d
10.0
0.220(11.99)

= 3.79 ksi 6 tallow = 14 ksi (O.K!)
Hence,
Use

5 kip

5 kip

Ans.

W12 * 16

833

2 ft

5 kip

2 ft

5 kip

2 ft

2 ft


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11–6. The compound beam is made from two sections,
which are pinned together at B. Use Appendix B and select

the lightest-weight wide-flange beam that would be safe for
each section if the allowable bending stress is sallow = 24 ksi
and the allowable shear stress is tallow = 14 ksi. The beam
supports a pipe loading of 1200 lb and 1800 lb as shown.

C

A
B
6 ft

Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB.
Assuming bending controls the design, applying the flexure formula.
Sreq¿d =

=
Select

Mmax
sallow
19.2(12)
= 9.60 in3
24

A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B

W10 * 12

For member BC, Mmax = 8.00 kip # ft.
Sreq¿d =

=
Select

Mmax
sallow
8.00(12)
= 4.00 in3
24

A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in. B

W6 * 9

V
for the W10 * 12 widetw d
flange section for member AB. From the shear diagram, Vmax = 2.20 kip.
Shear Stress: Provide a shear stress check using t =

tmax =

=

Vmax
tw d
2.20
0.19(9.87)

= 1.17 ksi 6 tallow = 14 ksi (O.K!)
Use

Ans.

W10 * 12

For member BC (W6 * 9), Vmax = 1.00 kip.
tmax =

=

Vmax
tw d
1.00
0.17(5.90)

= 0.997 ksi 6 tallow = 14 ksi (O.K!)
Hence,
Use

1800 lb

1200 lb

W6 * 9

Ans.

834

6 ft

8 ft

10 ft


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11–7. If the bearing pads at A and B support only vertical
forces, determine the greatest magnitude of the uniform
distributed loading w that can be applied to the beam.
sallow = 15 MPa, tallow = 1.5 MPa.

w

A

B
1m

1m

150 mm
25 mm
150 mm
25 mm

The location of c, Fig. b, is
y =

0.1625(0.025)(0.15) + 0.075(0.15)(0.025)
©yA
=
©A
0.025(0.15) + 0.15(0.025)
= 0.11875 m

I =

+

1
(0.025)(0.153) + (0.025)(0.15)(0.04375)2
12
1
(0.15)(0.0253) + 0.15(0.025)(0.04375)2
12

= 21.58203125(10 - 6) m4
Referring to Fig. b,
Qmax = y¿A¿ = 0.059375 (0.11875)(0.025)
= 0.176295313(10 - 4) m3

Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure
formula with C = y = 0.11875 m,
sallow =

Mmax c
;
I

15(106) =

0.28125w(0.11875)
21.582(10 - 6)

W = 9.693(103) N>m
Referring to shear diagram, Fig. a, Vmax = 0.75 w.
tallow =

Vallow Qmax
;
It

1.5(106) =

0.75w C 0.17627(10 - 3) D
21.582(10 - 6)(0.025)

W = 6.122(103) N>m
= 6.12 kN>m (Control!)

Ans.

835


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*11–8. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 1.20 ksi and an
allowable shear stress of tallow = 100 psi. Determine its
smallest dimensions to the nearest 18 in. if it is rectangular
and has a height-to-width ratio of 1.5.

12 kip/ft

B

A
3 ft

3 ft

1.5 b
b

The moment of inertia of the beam’s cross-section about the neutral axis is
1
(b)(1.5b)3 = 0.28125b4. Referring to the moment diagram,
I =
12
Mmax = 45.375 kip # ft.
sallow =

Mmax c
;
I

1.2 =

45.375(12)(0.75b)
0.28125b4

b = 10.66 in
Referring to Fig. b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3. Referring to the
shear diagram, Fig. a, Vmax = 33 kip.
tmax =

Vmax Qmax
;
It

100 =

33(103)(0.28125b3)
0.28125b4(b)

b = 18.17 in (Control!)
Thus, use
b = 18

1
in
4

Ans.

836


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•11–9.

Select the lightest-weight W12 steel wide-flange
beam from Appendix B that will safely support the loading
shown, where P = 6 kip. The allowable bending stress
is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi.

P

P

9 ft

From the Moment Diagram, Fig. a, Mmax = 54 kip # ft.
Mmax
sallow

Sreq¿d =

54(12)
22

=

= 29.45 in3
Select W12 * 26

C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in. D

From the shear diagram, Fig. a, Vmax = 7.5 kip. Provide the shear-stress check
for W 12 * 26,

tmax =

=

Vmax
tw d
7.5
0.230(12.22)

= 2.67 ksi 6 tallow = 12 ksi (O.K!)
Hence
Use

Ans.

W12 * 26

837

6 ft

6 ft


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11–10. Select the lightest-weight W14 steel wide-flange
beam having the shortest height from Appendix B that
will safely support the loading shown, where P = 12 kip.
The allowable bending stress is sallow = 22 ksi and the
allowable shear stress is tallow = 12 ksi.

P

P

9 ft

From the moment diagram, Fig. a, Mmax = 108 kip # ft.
Mmax
sallow

Sreq¿d =

108(12)
22

=

= 58.91 in3
Select W14 * 43

C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in. D

From the shear diagram, Fig. a, Vmax = 15 kip . Provide the shear-stress check
for W14 * 43 ,
tmax =

=

Vmax
tw d
15
0.305(13.66)

= 3.60 ksi 6 tallow = 12 ksi‚ (O.K!)
Hence,
Use

Ans.

W14 * 43

838

6 ft

6 ft


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11–11. The timber beam is to be loaded as shown. If the ends
support only vertical forces, determine the greatest magnitude
of P that can be applied. sallow = 25 MPa, tallow = 700 kPa.

150 mm
30 mm
120 mm

40 mm
P
4m

A

y =

(0.015)(0.150)(0.03) + (0.09)(0.04)(0.120)
= 0.05371 m
(0.150)(0.03) + (0.04)(0.120)

I =

1
1
(0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 +
(0.04)(0.120)3 +
12
12

B

(0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4
Maximum moment at center of beam:
Mmax =

P
(4) = 2P
2

Mc
;
I

s =

25(106) =

(2P)(0.15 - 0.05371)
19.162(10 - 6)

P = 2.49 kN
Maximum shear at end of beam:
Vmax =

P
2

VQ
;
t =
It

700(103) =

P 1
C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D
2 2
19.162(10 - 6)(0.04)

P = 5.79 kN
Thus,
P = 2.49 kN

Ans.

839

4m

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*11–12. Determine the minimum width of the beam to
the nearest 14 in. that will safely support the loading of
P = 8 kip. The allowable bending stress is sallow = 24 ksi
and the allowable shear stress is tallow = 15 ksi.

P
6 ft

6 ft

6 in.

B
A

Beam design: Assume moment controls.
sallow =

Mc

;
I

24 =

48.0(12)(3)
1
3
12 (b)(6 )

b = 4 in.

Ans.

Check shear:
8(1.5)(3)(4)
VQ
= 0.5 ksi 6 15 ksi OK
= 1
3
It
12 (4)(6 )(4)

tmax =

•11–13.

Select the shortest and lightest-weight steel wideflange beam from Appendix B that will safely support the
loading shown.The allowable bending stress is sallow = 22 ksi
and the allowable shear stress is tallow = 12 ksi.

10 kip

6 kip

4 kip

A

B
4 ft

Beam design: Assume bending moment controls.
Sreq¿d =

60.0(12)
Mmax
=
= 32.73 in3
sallow
22

Select a W 12 * 26
Sx = 33.4 in3, d = 12.22 in., tw = 0.230 in.
Check shear:
tavg =

V
10.5
=

= 3.74 ksi 6 12 ksi
Aweb
(12.22)(0.230)

Use W 12 * 26

Ans.

840

4 ft

4 ft

4 ft


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11–14. The beam is used in a railroad yard for loading and
unloading cars. If the maximum anticipated hoist load is

12 kip, select the lightest-weight steel wide-flange section
from Appendix B that will safely support the loading. The
hoist travels along the bottom flange of the beam,
1 ft … x … 25 ft, and has negligible size. Assume the beam
is pinned to the column at B and roller supported at A.
The allowable bending stress is sallow = 24 ksi and
the allowable shear stress is tallow = 12 ksi.

x

27 ft

A

B

12 kip

15 ft

C

Maximum moment occurs when load is in the center of beam.
Mmax = (6 kip)(13.5 ft) = 81 lb # ft
sallow =

M
;
S

24 =

81(12)
Sreq¿d
Sreq¿d = 40.5 in3

Select a W 14 * 30, Sx = 42.0 in3, d = 13.84 in, tw = 0.270 in.
At x = 1 ft, V = 11.56 kip
t =

11.36
V
=
= 3.09 ksi 6 12 ksi
Aweb
(13.84)(0.270)

Use W14 * 30

Ans.

841


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11–15. The simply supported beam is made of timber that
has an allowable bending stress of sallow = 960 psi and an
allowable shear stress of tallow = 75 psi. Determine its
dimensions if it is to be rectangular and have a heightto-width ratio of 1.25.

5 kip/ft

6 ft

1
I =
(b)(1.25b)3 = 0.16276b4
12
Sreq¿d

b

Assume bending moment controls:
Mmax = 60 kip # ft

960 =

Mmax
Sreq¿d

60(103)(12)
0.26042 b3

b = 14.2 in.
Check shear:
tmax =

1.5(15)(103)
1.5V
=
= 88.9 psi 7 75 psi NO
A
(14.2)(1.25)(14.2)

Shear controls:
tallow =

6 ft

1.25 b

I
0.16276b4
=
=
= 0.26042b3
c
0.625b

sallow =

B

A

1.5(15)(103)
1.5V
=
A
(b)(1.25b)

b = 15.5 in.

Ans.

842


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*11–16. The simply supported beam is composed of two

W12 * 22 sections built up as shown. Determine the
maximum uniform loading w the beam will support if
the allowable bending stress is sallow = 22 ksi and the
allowable shear stress is tallow = 14 ksi.

w

Section properties:
24 ft

For W12 * 22 (d = 12.31 in. Ix = 156 in4 tw = 0.260 in. A = 6.48 in2)
I = 2c 156 + 6.48a

S =

12.31 2
b d = 802.98 in4
2

I
802.98
=
= 65.23 in3
c
12.31

Maximum Loading: Assume moment controls.
M = sallowS(72 w)(12) = 22(65.23)
w = 1.66 kip>ft
Check Shear:

tmax =

Ans.
(Neglect area of flanges.)

12(1.66)
Vmax
= 3.11 ksi 6 tallow = 14 ksi OK
=
Aw
2(12.31)(0.26)

•11–17.

The simply supported beam is composed of two
W12 * 22 sections built up as shown. Determine if the beam
will safely support a loading of w = 2 kip>ft. The allowable
bending stress is sallow = 22 ksi and the allowable shear
stress is tallow = 14 ksi.

w

24 ft

Section properties:
For W 12 * 22 (d = 12.31 in.

Ix = 156 in4

tw = 0.260 in.

A = 6.48 in2)

I = 2[156 + 6.48(6.1552)] = 802.98 in4
S =

802.98
I
=
= 65.23 in3
c
12.31

Bending stress:
smax =

144 (12)
Mallow
=
= 26.5 ksi 7 sallow = 22 ksi
S
65.23

No, the beam falls due to bending stress criteria.
Check shear:
tmax =

Ans.

(Neglect area of flanges.)

Vmax
24
=
= 3.75 ksi 6 tallow = 14 ksi OK
Aw
2(12.31)(0.26)

843


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11–18. Determine the smallest diameter rod that will
safely support the loading shown. The allowable bending
stress is sallow = 167 MPa and the allowable shear stress
is tallow = 97 MPa.

25 N/m
15 N/m

15 N/m

1.5 m

Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Assume bending
controls the design. Applying the flexure formula.
sallow =
167 A 10

6

B =

Mmax c
I
24.375
p
4

A d2 B

A d2 B 4

d = 0.01141 m = 11.4 mm

Ans.

Shear Stress: Provide a shear stress check using the shear formula with
I =

p
A 0.0057074 B = 0.8329 A 10 - 9 B m4
4

Qmax =

4(0.005707) 1
c (p) A 0.0057062 B d = 0.1239 A 10 - 6 B m3
3p
2

From the shear diagram, Vmax = 30.0 N.
tmax =

=

Vmax Qmax
It
30.0 C 0.1239(10 - 6) D

0.8329 (10 - 9)(0.01141)

= 0.391 MPa 6 tallow = 97 MPa (O.K!)

844

1.5 m


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11–19. The pipe has an outer diameter of 15 mm.
Determine the smallest inner diameter so that it will safely
support the loading shown. The allowable bending stress
is sallow = 167 MPa and the allowable shear stress is
tallow = 97 MPa.

25 N/m
15 N/m

15 N/m

1.5 m

Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Q. Assume
bending controls the design. Applying the flexure formula.
sallow =
167 A 106 B =

Mmax c
I

24.375(0.0075)
p
4

C 0.00754 - A 2i B 4 D
d

di = 0.01297 m = 13.0 mm

Ans.

Shear Stress: Provide a shear stress check using the shear formula with
I =

p
A 0.00754 - 0.0064864 B = 1.0947 A 10 - 9 B m4
4

Qmax =

4(0.0075) 1
4(0.006486) 1
c (p) A 0.00752 B d c (p) A 0.0064862 B d
3p
2
3p
2

= 99.306 A 10 - 9 B m3
From the shear diagram, Vmax = 30.0 N. Q

tmax =

=

Vmax Qmax
It
30.0 C 99.306(10 - 9) D

1.0947(10 - 9)(0.015 - 0.01297)

= 1.34 MPa 6 tallow = 97 MPa (O.K!)

845

1.5 m


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*11–20. Determine the maximum uniform loading w
the W12 * 14 beam will support if the allowable bending

stress is sallow = 22 ksi and the allowable shear stress is
tallow = 12 ksi.

w
10 ft
10 ft

From the moment diagram, Fig. a, Mmax = 28.125 w. For W12 * 14, Sx = 14.9 in3,
d = 11.91 in and tw = 0.200 in.
sallow =
22 =

Mmax
S
28.125 w (12)
14.9
Ans.

w = 0.9712 kip>ft = 971 lb>ft

From the shear diagram, Fig. a, Vmax = 7.5(0.9712) = 7.284 kip. Provide a shear
stress check on W12 * 14,
tmax =

=

Vmax
tw d
7.284
0.200(11.91)

= 3.06 ksi 6 tallow = 12 ksi (O.K)

846


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•11–21.

Determine if the W14 * 22 beam will safely
support a loading of w = 1.5 kip>ft. The allowable bending
stress is sallow = 22 ksi and the allowable shear stress
is tallow = 12 ksi.

w
10 ft
10 ft

For W14 * 22, Sx = 29.0 in3, d = 13.74 in and tw = 0.23 in. From the moment
diagram, Fig. a, Mmax = 42.1875 kip # ft.

smax =

=

Mmax
S
42.1875(12)
29.0

= 17.46 ksi 6 sallow = 22 ksi (O.K!)
From the shear diagram, Fig. a, Vmax = 11.25 kip.
tmax =

=

Vmax
tw d

11.25
0.23(13.74)

= 3.56 ksi 6 tallow = 12 ksi (O.K!)
Based on the investigated results, we conclude that W14 * 22 can safely support
the loading.

847


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11–22. Determine the minimum depth h of the beam to
the nearest 18 in. that will safely support the loading shown.
The allowable bending stress is sallow = 21 ksi and the
allowable shear stress is tallow = 10 ksi. The beam has a
uniform thickness of 3 in.

4 kip/ft

h

A
B
12 ft

The section modulus of the rectangular cross-section is
S =

I
=
C

1
12

(3)(h3)
h>2

= 0.5 h2

From the moment diagram, Mmax = 72 kip # ft.
Sreq¿d =

Mmax
sallow

0.5h2 =

72(12)
21

h = 9.07 in
Use

h = 9 18 in

Ans.

From the shear diagram, Fig. a, Vmax = 24 kip . Referring to Fig. b,
9.125 9.125
ba
b (3) = 31.22 in3 and

Qmax = y¿A¿ = a
4
2
1
I =
(3) A 9.1253 B = 189.95 in4 . Provide the shear stress check by applying
12
shear formula,
tmax =

=

Vmax Qmax
It
24(31.22)
189.95(3)

= 1.315 ksi 6 tallow = 10 ksi (O.K!)

848

6 ft


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11–23. The box beam has an allowable bending stress
of sallow = 10 MPa and an allowable shear stress of
tallow = 775 kPa. Determine the maximum intensity w of the
distributed loading that it can safely support. Also, determine
the maximum safe nail spacing for each third of the length of
the beam. Each nail can resist a shear force of 200 N.

w
30 mm
250 mm
30 mm
150 mm
30 mm

Section Properties:
I =

1
1
(0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4
12
12

QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3
Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06)

= 0.96375 A 10 - 3 B m3
Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending
controls the design. Applying the flexure formula.
sallow =
10 A 106 B =

Mmax c
I
4.50w (0.125)
0.1877 (10 - 3)

w = 3336.9 N>m
Shear Stress: Provide a shear stress check using the shear formula. From the shear
diagram, Vmax = 3.00w = 10.01 kN.
tmax =

=

Vmax Qmax
It
10.01(103) C 0.96375(10 - 3) D
0.1877(10 - 3)(0.06)

= 857 kPa 7 tallow = 775 kPa (No Good!)
Hence, shear stress controls.
tallow =
775 A 103 B =

Vmax Qmax
It

3.00w C 0.96375(10 - 3) D
0.1877(10 - 3)(0.06)

w = 3018.8 N>m = 3.02 kN>m

Ans.

Shear Flow: Since there are two rows of nails, the allowable shear flow is
2(200)
400
=
q =
.
s
s

849

6m


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11–23.

Continued

For 0 … x 6 2 m and 4 m 6 x … 6 m, the design shear force is
V = 3.00w = 9056.3 N.
q =

VQA
I

9056.3 C 0.495(10 - 3) D
400
=
s
0.1877(10 - 3)
s = 0.01675 m = 16.7 mm

Ans.

For 2 m 6 x 6 4 m, the design shear force is V = w = 3018.8 N.
q =

VQA
I

3018.8 C 0.495(10 - 3) D
400

=
s
0.1877(10 - 3)
s = 0.05024 m = 50.2 mm

Ans.

850


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*11–24. The simply supported joist is used in the
construction of a floor for a building. In order to keep the
floor low with respect to the sill beams C and D, the ends of
the joists are notched as shown. If the allowable shear stress
for the wood is tallow = 350 psi and the allowable bending
stress is sallow = 1500 psi, determine the height h that will
cause the beam to reach both allowable stresses at the same
time. Also, what load P causes this to happen? Neglect the
stress concentration at the notch.

P

2 in.

15 ft
B

h

15 ft
D

A
10 in.
C

Bending Stress: From the moment diagram, Mmax = 7.50P. Applying the flexure
formula.
Mmax c
I

salllow =

7.50P(12)(5)

1500 =

1
12

(2)(103)

P = 555.56 lb = 556 lb

Ans.

Shear Stress: From the shear diagram, Vmax = 0.500P = 277.78 lb. The notch is the
critical section. Using the shear formula for a rectangular section.
tallow =
350 =

3Vmax
2A
3(277.78)
2(2) h

h = 0.595 in.

Ans.

11–25. The simply supported joist is used in the
construction of a floor for a building. In order to keep the
floor low with respect to the sill beams C and D, the ends of
the joists are notched as shown. If the allowable shear stress
for the wood is tallow = 350 psi and the allowable bending
stress is sallow = 1700 psi, determine the smallest height h
so that the beam will support a load of P = 600 lb. Also,
will the entire joist safely support the load? Neglect the
stress concentration at the notch.

P

B

tallow =

1.5V
;
A

350 =

D

A
10 in.

600
= 300 lb
2

1.5(300)
(2)(h)

h = 0.643 in.
smax =

Ans.

4500(12)(5)
Mmax c
= 1620 psi 6 1700 psi OK
= 1
3
I
12 (2)(10)

Yes, the joist will safely support the load.

Ans.

851

h

15 ft

C

The reaction at the support is

2 in.

15 ft


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11–26. Select the lightest-weight steel wide-flange beam
from Appendix B that will safely support the loading
shown. The allowable bending stress is sallow = 22 ksi and
the allowable shear stress is tallow = 12 ksi.

5 kip
18 kip ft
B
A
6 ft

From the moment diagram, Fig. a, Mmax = 48 kip # ft.
Sreq¿d =

=

Mmax
sallow
48(12)
22

= 26.18 in3

Select W 14 * 22 C Sx = 29.0 in3, d = 13.74 in. and tw = 0.230 in. D
From the shear diagram, Fig. a, Vmax = 5 kip. Provide the shear stress check for
W 14 * 22,
tmax =

=

Vmax
twd

5
0.230(13.74)

= 1.58 ksi 6 tallow = 12 ksi‚ (O.K!)
Use

Ans.

W14 * 22

W12 * 22 would work also.

852

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11–27. The T-beam is made from two plates welded
together as shown. Determine the maximum uniform
distributed load w that can be safely supported on the beam
if the allowable bending stress is sallow = 150 MPa and the
allowable shear stress is tallow = 70 MPa.

w

A
1.5 m

1.5 m
200 mm
20 mm
200 mm
20 mm

The neutral axis passes through centroid c of the beam’s cross-section. The location
of c, Fig. b, is
y =

0.21(0.02)(0.2) + 0.1(0.2)(0.02)

©yA
=
©A
0.02(0.2) + 0.2(0.02)
= 0.155 m

I =

1
(0.2)(0.023) + 0.2(0.02)(0.055)2
12

+

1
(0.02)(0.23) + 0.02(0.2)(0.055)2
12

= 37.667 (10 - 6) m4
Referring to Fig. b,
Qmax = y¿A¿ = 0.0775(0.155)(0.02)
= 0.24025(10 - 3) m3
Referring to the moment diagram, Mmax = -3.375 w. Applying the flexure
formula with C = y = 0.155 m,
sallow =

Mmax c
;
I

150(106) =

3.375 w (0.155)
37.667(10 - 6)

w = 10.80(103) N>m
= 10.8 kN>m (Control!)

Ans.

Referring to the shear diagram, Vmax = 1.5w.
tallow =

Vmax Qmax
;
It

70(106) =

1.5 w C 0.24025(10 - 3) D
37.667(10 - 6)(0.02)

w = 146.33(103) N>m
= 146 kN>m

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*11–28. The beam is made of a ceramic material having
an allowable bending stress of sallow = 735 psi and an
allowable shear stress of tallow = 400 psi. Determine the
width b of the beam if the height h = 2b.

15 lb
10 lb

6 lb/in.

2 in.

6 in.

2 in.

h
b

Bending Stress: From the moment diagram, Mmax = 30.0 lb # in. Assume bending
controls the design. Applying the flexure formula.

sallow =

Mmax c
I
30.0

735 =

1
12

A 2b2 B

(b) (2b)3

b = 0.3941 in. = 0.394 in.

Ans.

Shear Stress: Provide a shear stress check using the shear formula for a rectangular
section. From the shear diagram, Vmax = 19.67 lb.
tmax =

=

3Vmax
2A
3(19.67)
2(0.3941)(2)(0.3941)

= 94.95 psi 6 tallow = 400 psi (O.K!)

854