Mechanics of Materials Solutions Chapter08 Probs52 64

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8.52 A steel pipe assembly supports aconcentrated load of 17 kN as shown in Fig.

P8.52. The outside diameter of the pipe is 142

mm and the wall thickness is 6.5 mm.Determine the normal stresses produced at

points H and K .

Fig. P8.52

Solution

Section properties

2 2 2 2 2

4 4 4 4 4

2 142 mm 2(6.5 mm) 129 mm

(142 mm) (129 mm) 2, 766.958 mm4 4

(142 mm) (129 mm) 6, 364, 867 mm64 64

z

d D t

A D d

I D d

π π

π π

= − = − =

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

Internal forces and moments17 kN 17,000 N

(17,000 N)(370 mm) 6,290,000 N-mm z

F

M

= =

= =

Stresses

axial 2

bending 4

17,000 N6.144 MPa (C)

2,766.958 mm

(6,290,000 N-mm)(142 mm/2)70.165 MPa

6,364,867 mm

z

z

F

A

M c

I

σ

σ

= = =

= = = ±

Normal stress at H

By inspection, the bending stress at H will be compression; therefore, the normal stress at H is:

6.144 MPa 70.165 MPa 76.309 MPa 76.3 MPa (C) H σ = − − = − = Ans.

Normal stress at K

By inspection, the bending stress at K will be tension; therefore, the normal stress at K is:

6.144 MPa 70.165 MPa 64.021 MPa 64.0 MPa (T) K σ = − + = + = Ans.





8.53 The screw of a clamp exerts a compressive

force of 300 lb on the wood blocks. Determine the

normal stresses produced at points H and K . The

clamp cross-sectional dimensions at the section ointerest are 1.25 in. by 0.375 in. thick.

Fig. P8.53

Solution

Section properties2

34

(0.375 in.)(1.250 in.) 0.468750 in.

(0.375 in.)(1.250 in.)0.061035 in.

12 z

A

I

= =

= =

Internal forces and moments300 lb

(300 lb)(3.75 in. 1.25 in./2) 1,312.50 lb-in. z

F

M

=

= + =

Stresses

axial 2

bending 4

300 lb640 psi (T)

0.468750 in.

(1,312.50 lb-in.)(1.250 in./2)13,440 psi

0.061035 in.

z

z

F

A

M c

I

σ

σ

= = =

= = = ±

Normal stress at H

By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:

640 psi 13,440 psi 14,080 psi 14,080 psi (T) H σ = + = = Ans.

Normal stress at K

By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:

640 psi 13,440 psi 12,800 psi 12,800 psi (C) K σ = − = − = Ans.





8.54 Determine the normal stresses produced at points H and K of the pier support shown in Fig. P8.54a.

Fig. P8.54a Fig. P8.54b Cross section a–a

Solution

Section properties2

39 4

(250 mm)(500 mm) 125,000 mm(250 mm)(500 mm)

2.60417 10 mm12

z

A

I

= =

= = ×

Internal forces and moments250 kN 400 kN 650 kN

(250 kN)(3.25 m) (400 kN)(2.25 m) 87.50 kN-m z

F

M

= + =

= − = −

Stresses

axial 2

2

bending 9 4

650,000 N5.20 MPa (C)

125,000 mm

(87.5 kN-m)(500 mm/2)(1,000)8.40 MPa

2.60417 10 mm

z

z

F

A

M c

I

σ

σ

= = =

= = = ±×

Normal stress at H

By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:

5.20 MPa 8.40 MPa 3.20 MPa 3.20 MPa (T) H σ = − + = = Ans.

Normal stress at K

By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:

5.20 MPa 8.40 MPa 13.60 MPa 13.60 MPa (C) K σ = − − = − = Ans.





8.55 A tubular steel column CD supportshorizontal cantilever arm ABC , as shown in Fig.

P8.55. Column CD has an outside diameter o

10.75 in. and a wall thickness of 0.365 in.Determine the maximum compression stress at

the base of column CD.

Fig. P8.55

Solution

Section properties

2 2 2 2 2

4 4 4 4 4

2 10.750 in. 2(0.365 in.) 10.020 in.

(10.750 in.) (10.020 in.) 11.908 in.4 4

(10.750 in.) (10.020 in.) 160.734 in.64 64

z

d D t

A D d

I D d

π π

π π

= − = − =

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

Internal forces and moments700 lb 900 lb 1,600 lb

(700 lb)(13 ft) (900 lb)(23 ft) 29,800 lb-ft 357,600 lb-in.

F

M

= + =

= + = =

Stresses

axial 2

bending 4

1,600 lb134.36 psi (C)

11.908 in.

(357,600 lb-in.)(10.75 in./2)11,958.27 psi

160.734 in.

F

A

M c

I

σ

σ

= = =

= = = ±

Maximum compression stress at base of column

compression 134.36 psi 11,958.27 psi 12,092.63 psi 12.09 ksi (C)σ = − − = − = Ans.





8.56 Determine the normal stresses acting at points H and K for the structure shown in Fig. P8.56a. Thecross-sectional dimensions of the vertical member are shown in Fig. P8.56b.

Fig. P8.56a

Fig. P8.56b Cross section

Solution

Section properties2

34

(4 in.)(8 in.) 32 in.

(4 in.)(8 in.)170.6667 in.

12 z

A

I

= =

= =

Internal forces and moments1,000 lb 2,200 lb 3,200 lb

(1,000 lb)(12 in. 8 in./2) 16,000 lb-in. z

F

M

= + =

= + =

Stresses

axial 2

bending 4

3,200 lb100 psi (C)

32 in.

(16,000 lb-in.)(8 in./2)375 psi

170.6667 in.

z

z

F

A

M c

I

σ

σ

= = =

= = = ±

Normal stress at H


100 psi 375 psi 475 psi 475 psi (C) H σ = − − = − = Ans.

Normal stress at K


100 psi 375 psi 275 psi 275 psi (T) K σ = − + = = Ans.





8.57 A W18 × 35 standard steel shape issubjected to a tension force P that is applied

15 in. above the bottom surface of the wide-

flange shape as shown in Fig. P8.57. If thetension normal stress of the upper surface o

the W-shape must be limited to 15 ksi,

determine the allowable force P that may be

applied to the member. Fig. P8.57

Solution

Section properties (from Appendix B)

2

4

Depth 17.7 in.

10.3 in.

510 in. z

d

A

I

=

=

=

Stresses

axial 2

2

bending 4 4 4

10.3 in.

(15 in. 17.7 in./2)(17.7 in./2) (6.15 in.)(8.85 in.) (54.4275 in. )

510 in. 510 in. 510 in.

z

z

F P

A

M c P P P

I

σ

σ

= =

−= = = =

Normal stress on the upper surface of the W-shape

The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses.

Since these stresses are expressed in terms of the unknown force P , the tension normal stress is given

by:2

upper surface 2 4

2 2

2

(54.4275 in. )

10.3 in. 510 in.

(0.097087 in. 0.106721 in. )(0.203808 in. )

P P

P P

σ

− −

−

= +

= +

=

The normal stress on the upper surface of the W-shape must be limited to 15 ksi; therefore,2

2

(0.203808 in. ) 15 ksi

15 ksi73.6 kips

0.203808 in.

P

P

−

−

≤

∴ ≤ = Ans.





8.58 A WT305 × 41 standard steel shape issubjected to a tension force P that is applied 250

mm above the bottom surface of the tee shape, as

shown in Fig. P8.58. If the tension normal stressof the upper surface of the WT-shape must be

limited to 120 MPa, determine the allowable

force P that may be applied to the member. Fig. P8.58

SolutionSection properties (from Appendix B)

2

6 4

Depth 300 mm

Centroid 88.9 mm (from flange to centroid)

5,230 mm

48.7 10 mm z

d

y

A

I

=

=

=

= ×

Stresses

4 2

axial 2(1.9120 10 mm )

5,230 mm

F P P

A

σ − −

= = = ×

bending 6 4

6 4

4 2

(250 mm 88.9 mm)(300 mm 88.9 mm)

48.7 10 mm

(161.1 mm)(211.1 mm)

48.7 10 mm

(6.9832 10 mm )

z

z

M c P

I

P

P

σ

− −

− −= =

×

=×

= ×

Normal stress on the upper surface of the WT-shape

The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses.Since these stresses are expressed in terms of the unknown force P , the tension normal stress is given

by:4 2 4 2

upper surface

4 2

(1.9120 10 mm ) (6.9832 10 mm )

(8.8953 10 mm )

P P

P

σ − − − −

− −

= × + ×

= ×

The normal stress on the upper surface of the W-shape must be limited to 15 ksi; therefore,4 2

2

4 2

(8.8953 10 mm ) 120 MPa

120 N/mm134,903 N 134.9 kN

8.8953 10 mm

P

P

− −

− −

× ≤

∴ ≤ = =×

Ans.





8.59 A pin support consists of a vertical plate 60 mm wide by 10 mm thick. The pin

carries a load of 750 N. Determine the

normal stresses acting at points H and K forthe structure shown in Fig. P8.59.

Fig. P8.59

Solution

Section properties2

34

(60 mm)(10 mm) 600 mm

(60 mm)(10 mm)5,000 mm

12

A

I

= =

= =

Internal forces and moments750 N

(750 N)(30 mm 10 mm/2) 26,250 N-mm

F

M

=

= + =

Stresses

axial 2

bending 4

750 N1.25 MPa (T)

600 mm

(26,250 N-mm)(10 mm/2)26.25 MPa

5,000 mm

F

A

M c

I

σ

σ

= = =

= = = ±

Normal stress at H


1.25 MPa 26.25 MPa 25.00 MPa 25.0 MPa (C) H σ = − = − = Ans.

Normal stress at K


1.25 MPa 26.25 MPa 27.50 MPa 27.5 MPa (T) K σ = + = = Ans.





8.60 The tee shape shown in Fig. P8.60b is used as a short post to support a load of P = 2,000 lb. Theload P is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.60a. Determine

the normal stresses at points H and K , which are located on section a–a.

Fig. P8.60a

Fig. P8.60b Cross-sectional dimensions

Solution

Centroid location in x direction:

Shape width b height h Area Ai

xi

(from left) xi Ai

(in.) (in.) (in.2) (in.) (in.

3)

flange 12 2 24 1 24

stem 2 10 20 7 140

44 in.2 164 in.

3

3

2

164 in.3.7273 in. (from left side to centroid)

44 in.

8.2727 in. (from right side to centroid)

i i

i

x A x

A

Σ= = =

Σ

=

Moment of inertia about the z axis:

Shape I C d = xi – x d²A I C + d²A

(in.4) (in.) (in.

4) (in.

4)

flange 8 −2.7273 178.5160 186.5160

stem 166.6667 3.2727 214.2113 380.8790

Moment of inertia about the z axis (in.4) = 567.3940

Internal forces and moments

2,000 lb

(2,000 lb)(5 in. 3.7273 in.) 17,454.6 lb-in. z

F

M

=

= + =





Stresses

axial 2

,bending 4

,bending 4

2,000 lb45.4545 psi

44 in.

(17, 454.6 lb-in.)( 3.7273 in.)114.6620 psi

567.3940 in.

(17, 454.6 lb-in.)(8.2727 in.)254.4910 psi

567.3940 in.

z H

z

z K

z

F

A

M x

I

M x

I

σ

σ

σ

−= = = −

−= = = −

= = =

Normal stress at H

45.4545 psi 114.6620 psi 160.1165 psi 160.1 psi (C) H σ = − − = − = Ans.

Normal stress at K

45.4545 psi 254.4910 psi 209.0365 psi 209 psi (T) K σ = − + = = Ans.





8.61 The tee shape shown in Fig. P8.61b is used as a short post to support a load of P . The load P isapplied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.61a. The tension and

compression normal stresses in the post must be limited to 1,000 psi and 800 psi, respectively.

Determine the maximum magnitude of load P that satisfies both the tension and compression stresslimits.

Fig. P8.61a

Fig. P8.61b Cross-sectional dimensions

Solution



xi (from left) xi Ai

(in.) (in.) (in.2) (in.) (in.

3)

flange 12 2 24 1 24

stem 2 10 20 7 140

44 in.2 164 in.

3

3

2

164 in. 3.7273 in. (from left side to centroid)44 in.

8.2727 in. (from right side to centroid)

i i

i

x A x A

Σ= = =Σ

=



(in.4) (in.) (in.

4) (in.

4)

flange 8 −2.7273 178.5160 186.5160

stem 166.6667 3.2727 214.2113 380.8790

Moment of inertia about the z axis (in.4) = 567.3940


(5 in. 3.7273 in.) (8.7273 in.) z

F P

P P

=

= + =





Stresses

2

axial 2

2

,bending 4

2

,bending 4

(0.022727 in. )44 in.

(8.7273 in.) ( 3.7273 in.)(0.057331 in. )

567.3940 in.

(8.7273 in.) (8.2727 in.)(0.127246 in. )

567.3940 in.

z H

z

z K

z

F P P

A

M x P P

I

M x P P

I

σ

σ

σ

−

−

−

= = − = −

−= = = −

= = =

Compression stress limit (at H )2 2 2

2

(0.022727 in. ) (0.057331 in. ) (0.080058 in. )

(0.080058 in. ) 800 psi

9,992.76 lb

H P P P

P

P

σ − − −

−

= − − = −

≤

∴ ≤

Tension stress limit (at K )2 2 2

2

(0.022727 in. ) (0.127246 in. ) (0.104519 in. )

(0.104519 in. ) 1,000 psi

9,567.64 lb

K P P P

P

P

σ − − −

−

= − + =

≤

∴ ≤

Maximum magnitude of load P

max 9,570 lb P = Ans.





8.62 The tee shape shown in Fig. P8.62b is used as a post that supports a load of P = 25 kN. Note thatthe load P is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.62a. Determine the

normal stresses at points H and K .

Fig. P8.62a Fig. P8.62b Cross-sectional dimensions

Solution



xi (from left) xi Ai

(mm) (mm) (mm2) (mm) (mm

3)

stem 20 130 2,600 65 169,000

flange 120 20 2,400 140 336,000

5,000 505,000

3

2

505,000 mm 101.0 mm (from left side to centroid)5,000 mm

49.0 mm (from right side to centroid)

i i

i

x A x A

Σ= = =

Σ

=



(mm4) (mm) (mm

4) (mm

4)

stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67

flange 80,000.00 39.0 3,650,400.00 3,730,400.00

Moment of inertia about the z axis (mm4) = 10,761,666.67

Internal forces and moments25 kN 25,000 N

(25,000 N)(400 mm 49.0 mm) 11,225,000 N-mm z

F

M

= =

= − + = −





Stresses

axial 2

,bending 4

,bending 4

25,000 N5 MPa

5,000 mm

( 11, 225,000 N-mm)( 101.0 mm)105.35 MPa

10,761,666.67 mm

( 11, 225,000 N-mm)(49.0 mm)51.11 MPa

10,761,666.67 mm

z H

z

z K

z

F

A

M x

I

M x

I

σ

σ

σ

−= = = −

− −= = =

−= = = −

Normal stress at H

5 MPa 105.35 MPa 100.4 MPa (T) H σ = − + = Ans.

Normal stress at K

5 MPa 51.11 MPa 56.1 MPa (C) K σ = − − = Ans.





8.63 The tee shape shown in Fig. P8.63b is used as a post that supports a load of P , which is applied 400mm from the flange of the tee shape, as shown in Fig. P8.63a. The tension and compression normal

stresses in the post must be limited to 165 MPa and 80 MPa, respectively. Determine the maximum

magnitude of load P that satisfies both the tension and compression stress limits.


Solution



xi

(from left) xi Ai


3)

stem 20 130 2,600 65 169,000

flange 120 20 2,400 140 336,000

5,000 505,000

3

2

505,000 mm101.0 mm (from left side to centroid)

5,000 mm


i i

i

x A x

A

Σ= = =

Σ

=



(mm4) (mm) (mm

4) (mm

4)

stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67

flange 80,000.00 39.0 3,650,400.00 3,730,400.00



(400 mm 49.0 mm) (449.0 mm) z

F P

P P

=

= − + = −





Stresses

4 2

axial 2

3 2

,bending 4

3 2

,bending 4

(2 10 mm )5,000 mm

( 449 mm) ( 101.0 mm)(4.21394 10 mm )

10,761,666.67 mm

( 449 mm) (49.0 mm)(2.04439 10 mm )

10,761,666.67 mm

z H

z

z K

z

F P P

A

M x P P

I

M x P P

I

σ

σ

σ

− −

− −

− −

= = − = − ×

− −= = = ×

−= = = − ×

Tension stress limit (at H )4 2 3 2

3 2

3 2 2

(2 10 mm ) (4.21394 10 mm )

(4.01394 10 mm )

(4.01394 10 mm ) 165 N/mm

41,106.7 N

H P P

P

P

P

σ − − − −

− −

− −

= − × + ×

= ×

× ≤

∴ ≤

Compression stress limit (at K )4 2 3 2 3 2

3 2 2

(2 10 mm ) (2.04439 10 mm ) (2.24439 10 mm )

(2.24439 10 mm ) 80 N/mm

35,644.43 N

K P P P

P

P

σ − − − − − −

− −

= − × − × = − ×

× ≤

∴ ≤

Maximum magnitude of load P

max 35.6 kN P = Ans.





8.64 The tee shape shown in Fig. P8.64b is used as a post that supports a load of P = 25 kN, which isapplied 400 mm from the flange of the tee shape as, shown in Fig. P8.64a. Determine the magnitudes

and locations of the maximum tension and compression normal stresses within the vertical portion BC o

the post.


Solution



xi

(from left) xi Ai


3)

stem 20 130 2,600 65 169,000

flange 120 20 2,400 140 336,000

5,000 505,000

3

2

505,000 mm101.0 mm (from left side to centroid)

5,000 mm


i i

i

x A x

A

Σ= = =

Σ

=



(mm4) (mm) (mm

4) (mm

4)

stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67

flange 80,000.00 39.0 3,650,400.00 3,730,400.00


Internal forces and moments(25 kN)cos35 20.4788 kN 20,478.8 N (vertical component)

(25 kN)sin35 14.3394 kN 14,339.4 N (horizontal component)

at B (20,478.8 N)(400 mm 49.0 mm) 9,194,981.2 N-mm

at C (20,478.8 N)(400 m

z

z

F

V

M

M

= ° = =

= ° = =

= − + = −

= − m 49.0 mm) (14,339.4 N)(1,200 mm) 8,012, 298.8 N-mm+ + =




Normal stress at H at location B

axial 2

,bending 4

20,478.8 N4.0958 MPa

5,000 mm

( 9,194,981.2 N-mm)( 101.0 mm)86.2964 MPa

10,761,666.67 mm

4.0958 MPa 86.2964 MPa 82.2 MPa

z H

z

H

F

A

M x

I

σ

σ

σ

−= = = −

− −= = =

= − + =

Normal stress at H at location C

,bending 4

(8,012,298.8 N-mm)( 101.0 mm)75.1967 MPa

10,761,666.67 mm

4.0958 MPa 75.1967 MPa 79.3 MPa

z H

z

H

M x

I σ

σ

−= = = −

= − − = −

Normal stress at K at location B

,bending 4

( 9,194,981.2 N-mm)(49.0 mm)41.8666 MPa

10,761,666.67 mm

4.0958 MPa 41.8666 MPa 46.0 MPa

z K

z

K

M x

I σ

σ

−= = = −

= − − = −

Normal stress at K at location C

,bending 4

(8,012,298.8 N-mm)(49.0 mm)36.4816 MPa

10,761,666.67 mm

4.0958 MPa 36.4816 MPa 32.4 MPa

z K

z

K

M x

I σ

σ

= = =

= − + =

Maximum tension stress

max tension 82.2 MPa (T) at location Bσ = Ans.

Maximum compression stress

max compression 79.3 MPa (C) at location C σ = Ans.

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Mechanics of Materials Solutions Chapter08 Probs52 64