Mechanics of Materials Philpot 3rd Edition Solutions Manual

8/20/2019 Mechanics of Materials Philpot 3rd Edition Solutions Manual

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P1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as

a compression member. If the axial normal stress in the member must be limited to 200 MPa,determine the maximum load P that the member can support.

Solution

The cross-sectional area of the stainless steel tube is

2 2 2 2 2

( ) [(60 mm) (50 mm) ] 863.938 mm4 4 A D d

The normal stress in the tube can be expressed as

P

A

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable

normal stress, rearrange this expression to solve for the maximum load P 2 2

max allow (200 N/mm )(863.938 mm ) 172,788 172.8 k N N P A Ans.


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P1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip

load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thicknessrequired for the tube.

Solution

From the definition of normal stress, solve for the minimum area required to support a 27-kip load

without exceeding a stress of 18 ksi

2

min

27 kips1.500 in.

18 ksi

P P A

A

The cross-sectional area of the aluminum tube is given by

2 2( )4

A D d

Set this expression equal to the minimum area and solve for the maximum inside diameter d

2 2 2

2 2 2

2 2 2

max

[(2.50 in.) ] 1.500 in.4

4(2.50 in.) (1.500 in. )

4(2.50 in.) (1.500 in. )

2.08330 in.

d

d

d

d

The outside diameter D, the inside diameter d , and the wall thickness t are related by

2 D d t Therefore, the minimum wall thickness required for the aluminum tube is

min

2.50 in. 2.08330 in.0.20835 in. 0.208 in.

2 2

D d t

Ans.


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P1.3 Two solid cylindrical rods (1) and (2)

are joined together at flange B and loaded, asshown in Figure P1.3/4. If the normal stress

in each rod must be limited to 40 ksi,

determine the minimum diameter required

for each rod.

FIGURE P1.3/4

Solution

Cut a FBD through rod (1). The FBD should include the free end of the rod at A.

As a matter of course, we will assume that the internal force in rod (1) is tension

(even though it obviously will be in compression). From equilibrium,

1

1

15 kips 0

15 kips 15 kips (C)

y F F

F

Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again,we will assume that the internal force in rod (2) is tension. Equilibrium of this

FBD reveals the internal force in rod (2):

2

2

30 kips 30 kips 15 kips 0

75 kips 75 kips (C)

y F F

F

Notice that rods (1) and (2) are in compression. In this situation, we areconcerned only with the stress magnitude; therefore, we will use the force

magnitudes to determine the minimum required cross-sectional areas. If

the normal stress in rod (1) must be limited to 40 ksi, then the minimum

cross-sectional area that can be used for rod (1) is

211,min

15 kips0.375 in.

40 ksi

F A

The minimum rod diameter is therefore

2 2

1,min 1 10.375 in. 0.6909 0.6919 i4

inn. . A d d

Ans.

Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of

222,min

75 kips 1.875 in.40 ksi

F A

The minimum diameter for rod (2) is therefore

2 2

2,min 2 21.875 in. 1.54509 1.545 in.7 in.

4 A d d

Ans.


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P1.4 Two solid cylindrical rods (1) and (2) are

oined together at flange B and loaded, as shown inFigure P1.3/4. The diameter of rod (1) is 1.75 in.

and the diameter of rod (2) is 2.50 in. Determine the

normal stresses in rods (1) and (2).

FIGURE P1.3/4

Solution

Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We

will assume that the internal force in rod (1) is tension (even though it obviously will

be in compression). From equilibrium,

1

1

15 kips 0

15 kips 15 kips (C)

y F F

F

Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we

will assume that the internal force in rod (2) is tension. Equilibrium of this FBD

reveals the internal force in rod (2):

2

2

30 kips 30 kips 15 kips 0

75 kips 75 kips (C)

y F F

F

From the given diameter of rod (1), the cross-sectional area of rod (1) is

2 2

1 (1.75 in.) 2.4053 in.4

A

and thus, the normal stress in rod (1) is

11 2

1

15 kips6.23627 ksi

2.4053 in6.24 ksi )

.(C

F

A

Ans.


2 2

2(2.50 in.) 4.9087 in.

4

A

Accordingly, the normal stress in rod (2) is

22 2

2

75 kips15.2789 ksi

2.4053 in.15.28 ksi (C)

F

A

Ans.


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P1.5 Axial loads are applied with rigid bearing plates to the

solid cylindrical rods shown in Figure P1.5/6. The diameterof aluminum rod (1) is 2.00 in., the diameter of brass rod (2)

is 1.50 in., and the diameter of steel rod (3) is 3.00 in.

Determine the axial normal stress in each of the three rods.

FIGURE P1.5/6

Solution

Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal

force in rod (1) is tension (even though it obviously will be in compression). From equilibrium,

1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C) y F F F

FBD through rod (1)

FBD through rod (2)

FBD through rod (3)

Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal

force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):

2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)

y F F F

Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in

rod (3) is:

3

3

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0

26 kips 26 kips (C)

y F F

F


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2 2

1 (2.00 in.) 3.1416 in.4

A

and thus, the normal stress in aluminum rod (1) is

11 2

1

16 kips5.0930 ksi

3.1416 in5.09 ksi (C)

.

F

A

Ans.


2 2

2(1.50 in.) 1.7671 in.

4 A

Accordingly, the normal stress in brass rod (2) is

22 2

2

14 kips7.9224 ksi

1.7671 in.7.92 ksi (T)

F

A Ans.

Finally, the cross-sectional area of rod (3) is

2 2

3 (3.00 in.) 7.0686 in.

4

A

and the normal stress in the steel rod is

33 2

3

26 kips3.6782 ksi

7.0686 in3.68 ksi (C)

.

F

A

Ans.


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P1.6 Axial loads are applied with rigid bearing plates to the

solid cylindrical rods shown in Figure P1.5/6. The normalstress in aluminum rod (1) must be limited to 18 ksi, the

normal stress in brass rod (2) must be limited to 25 ksi, and

the normal stress in steel rod (3) must be limited to 15 ksi.

Determine the minimum diameter required for each of thethree rods.

FIGURE P1.5/6

Solution

The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) thatincludes the free end A. We will assume that the internal force in rod (1) is tension (even though it

obviously will be in compression). From equilibrium,

1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)

y F F F

FBD through rod (1)

FBD through rod (2)

FBD through rod (3)

Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal

force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):

2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)

y F F F

Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in

rod (3) is:


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3

3

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0

26 kips 26 kips (C)

y F F

F

Notice that two of the three rods are in compression. In these situations, we are concerned only with the

stress magnitude; therefore, we will use the force magnitudes to determine the minimum required cross-sectional areas, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be

limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is

211,min

1

16 kips 0.8889 in.18 ksi

F A


2 2

1,min 1 10.8889 in. 1.0638 in 1.064 in..4

A d d

Ans.

The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of

222,min

2

14 kips0.5600 in.

25 ksi

F A

which requires a minimum diameter for rod (2) of

2 2

2,min 2 20.5600 in. 0.8444 in 0.844 in..4

A d d Ans.

The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required

for this rod is:

233,min

3

26 kips1.7333 in.

15 ksi

F A


2 2

3,min 3 31.7333 in. 1.4856 in 1.486 in..4

A d d

Ans.


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P1.7 Two solid cylindrical rods support a load o

P = 50 kN, as shown in Figure P1.7/8. If thenormal stress in each rod must be limited to 130

MPa, determine the minimum diameter required

for each rod.

FIGURE P1.7/8

Solution

Consider a FBD of joint B. Determine the angle between

rod (1) and the horizontal axis:

4.0 mtan 1.600 57.99462.5 m

and the angle between rod (2) and the horizontal axis:

2.3 mtan 0.7188 35.7067

3.2 m

Write equilibrium equations for the sum of forces in thehorizontal and vertical directions. Note: Rods (1) and (2)

are two-force members.

2 1cos(35.7067 ) cos(57.9946 ) 0 x F F F (a)

2 1sin(35.7067 ) sin(57.9946 ) 0 y F F F P (b)

Unknown forces F 1 and F 2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the

substitution method, Eq. (b) can be solved for F 2 in terms of F 1:

2 1

cos(57.9946 )

cos(35.7067 ) F F

(c)

Substituting Eq. (c) into Eq. (b) gives

1 1

1

1

cos(57.9946 )sin(35.7067 ) sin(57.9946 )

cos(35.6553 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289

F F P

F P

P P F

For the given load of P = 50 kN, the internal force in rod (1) is therefore:

1

50 kN40.6856 kN

1.2289 F


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Backsubstituting this result into Eq. (c) gives force F 2:

2 1

cos(57.9946 ) cos(57.9946 )(40.6856 kN) 26.5553 kN

cos(35.7067 ) cos(35.7067 ) F F

The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area

required for rod (1) is

211,min 2

1

(40.6856 kN)(1,000 N/kN)312.9664 mm

130 N/mm

F A


2 2

1,min 1 1312.9664 mm 19.9620 19.4

96 mmmm A d d

Ans.

The minimum area required for rod (2) is

222,min 2

2

(26.5553 kN)(1,000 N/kN)204.2718 mm

130 N/mm

F A


2 2

2,min 2 2

204.2718 mm 16.1272 16.4

13 mmmm A d d

Ans.


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P1.8 Two solid cylindrical rods support a load

of P = 27 kN, as shown in Figure P1.7/8. Rod(1) has a diameter of 16 mm and the diameter

of rod (2) is 12 mm. Determine the axial

normal stress in each rod.

FIGURE P1.7/8

Solution

Consider a FBD of joint B. Determine the angle between rod (1)and the horizontal axis:

4.0 mtan 1.600 57.99462.5 m

and the angle between rod (2) and the horizontal axis:

2.3 mtan 0.7188 35.7067

3.2 m

Write equilibrium equations for the sum of forces in the horizontaland vertical directions. Note: Rods (1) and (2) are two-force

members.

2 1cos(35.7067 ) cos(57.9946 ) 0 x F F F (a)

2 1

sin(35.7067 ) sin(57.9946 ) 0 y

F F F P (b)

Unknown forces F 1 and F 2 can be found from the simultaneous solution of Eqs. (a) and (b). Using thesubstitution method, Eq. (b) can be solved for F 2 in terms of F 1:

2 1

cos(57.9946 )

cos(35.7067 ) F F

(c)

Substituting Eq. (c) into Eq. (b) gives

1 1

1

1

cos(57.9946 )sin(35.7067 ) sin(57.9946 )

cos(35.6553 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289

F F P

F P

P P F

For the given load of P = 27 kN, the internal force in rod (1) is therefore:

1

27 kN21.9702 kN

1.2289 F


12/53



Backsubstituting this result into Eq. (c) gives force F 2:

2 1

cos(57.9946 ) cos(57.9946 )(21.9702 kN) 14.3399 kN

cos(35.7067 ) cos(35.7067 ) F F

The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is:

2 2

1 (16 mm) 201.0619 mm4

A

and the normal stress in rod (1) is:21

1 2

1

(21.9702 kN)(1,000 N/kN)109.2710 N/mm

201.0109.3 MPa (T)

619 mm

F

A Ans.

The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is:

2 2

2 (12 mm) 113.0973 mm4

A

and the normal stress in rod (2) is:

22

2 2

2

(14.3399 kN)(1,000 N/kN)126.7924 N/mm

113.0126.8 MPa (T)

973 mm

F

A Ans.


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P1.9 A simple pin-connected truss is loaded

and supported as shown in Figure P1.9. Allmembers of the truss are aluminum pipes that

have an outside diameter of 4.00 in. and a wall

thickness of 0.226 in. Determine the normal

stress in each truss member.

FIGURE P1.9

Solution

Overall equilibrium:Begin the solution by determining the

external reaction forces acting on the

truss at supports A and B. Writeequilibrium equations that include all

external forces. Note that only theexternal forces (i.e., loads and

reaction forces) are considered at this

time. The internal forces acting in the

truss members will be consideredafter the external reactions have been

computed. The free-body diagram

(FBD) of the entire truss is shown.

The following equilibrium equations

can be written for this structure:2 kips

2 ki

0

ps

x

x

x F A

A

(6 ft) (5 kips)(14 ft) (2 kips)(7 ft)

14 kips

0

y

A y B

B

M

5 kips 0

9 kips

y y y

y

F A B

A

Method of joints:

Before beginning the process of determining the internal forces in the axial members, the geometry ofthe truss will be used to determine the magnitude of the inclination angles of members AC and BC . Use

the definition of the tangent function to determine AC and BC :

7 fttan 0.50 26.565

14 ft

7 fttan 0.875 41.186

8 ft

AC AC

BC BC


14/53



Joint A:

Begin the solution process by considering a FBD of joint A. Consideronly those forces acting directly on joint A. In this instance, two axial

members, AB and AC , are connected at joint A. Additionally, two

reaction forces, A x and A y, act at joint A. Tension forces will be

assumed in each truss member.

cos(26.565 ) 0 x AC AB x F F F A (a)

sin(26.565 ) 0 y AC y

F F A (b)

Solve Eq. (b) for F AC :

9 kips

sin(26.565 ) sin(26.520.125 kip

65 )s

y

AC

A F

and then compute F AB using Eq. (a):

cos(26.565 )

(20.125 kips)cos(26.5 16.000 kips65 ) ( 2 kips)

AB AC x F F A

Joint B: Next, consider a FBD of joint B. In this instance, the equilibriumequations associated with joint B seem easier to solve than those that

would pertain to joint C . As before, tension forces will be assumed in

each truss member.

cos(41.186 ) 0 x AB BC F F F (c)

sin(41.186 ) 0 y BC y F F B (d)

Solve Eq. (d) for F BC :

14 kips

sin(41.186 ) sin(41.18

21.260 kip

6

s

)

y

BC

B F

Eq. (c) can be used as a check on our calculations:

cos(41.186 )

( 16.000 kips) ( 21.260 kips)cos(41.186 ) 0

x AB BC F F F

Checks!

Section properties:For each of the three truss members:

2 2 24.00 in. 2(0.226 in.) 3.548 in. (4.00 in.) (3.548 in.) 2.67954 in.4

d A

Normal stress in each truss member:

2

16.000 kips5.971 ksi

2.679545.97 ksi (C)

in.

AB AB

AB

F

A

Ans.

2

20.125 kips7.510 ksi

2.679547.51 ksi (T)

in.

AC

AC

AC

F

A Ans.

2

21.260 kips7.934 ksi

2.679547.93 ksi (C)

in.

BC

BC

BC

F

A

Ans.


15/53





have an outside diameter of 60 mm and a wall

thickness of 4 mm. Determine the normal


FIGURE P1.10

Solution

Overall equilibrium:Begin the solution by determining the

external reaction forces acting on the truss at

supports A and B. Write equilibrium

equations that include all external forces.

Note that only the external forces (i.e., loadsand reaction forces) are considered at this

time. The internal forces acting in the trussmembers will be considered after the external

reactions have been computed. The free-

body diagram (FBD) of the entire truss isshown. The following equilibrium equations

can be written for this structure:

12 k

12

N 0

kN x

x x F A

A

(1 m) (15 kN)(4.3 m) 064.5 kN y

A y B B

M

15 kN

49.5 kN

0

y

y y y F

A

A B

Method of joints:

Before beginning the process of determining the internal forces in the axial members, the geometry ofthe truss will be used to determine the magnitude of the inclination angles of members AB and BC . Use

the definition of the tangent function to determine AB and BC :

1.5 m

tan 1.50 56.3101.0 m

1.5 mtan 0.454545 24.444

3.3 m

AB AB

BC BC


16/53



Joint A:



reaction forces, A x and A y, act at joint A. Tension forces will be assumed

in each truss member.

cos(56.310 ) 0 x AC AB x F F F A (a)

sin(56.310 ) 0 y y AB

F A F (b)

Solve Eq. (b) for F AB:

49.5 kN

sin(56.310 ) sin(56.310 )59.492 kN

y

AB

A F

and then compute F AC using Eq. (a):

cos(56.310 )

( 59.492 kN)cos(56.3 45.00010 ) ( 12 kN) kN

AC AB x F F A

Joint C: Next, consider a FBD of joint C . In this instance, the equilibriumequations associated with joint C seem easier to solve than those that

would pertain to joint B. As before, tension forces will be assumed in

each truss member.

cos(24.444 ) 12 kN 0 x AC BC F F F (c)

sin(24.444 ) 15 kN 0 y BC F F (d)

Solve Eq. (d) for F BC :

15 kN

sin(24.444 )

36.249 kN BC

F

Eq. (c) can be used as a check on our calculations:

cos(24.444 ) 12 kN 0

(45.000 kN) ( 36.249 kN)cos(24.444 ) 12 kN 0

x AC BC F F F

Checks!


2 2 260 mm 2(4 mm) 52 mm (60 mm) (52 mm) 703.7168 mm4

d A


2

( 59.492 kN)(1,000 N/kN)84.539 MPa

7084.5 MPa (C)

3.7168 mm

AB AB

AB

F

A

Ans.

2

(45.000 kN)(1,000 N/kN)63.946 MPa

7063.9 MPa

3.7168)

mm(T AC

AC

AC

F

A Ans.

2

( 36.249 kN)(1,000 N/kN)51.511 MPa

7051.5 MPa (C)

3.7168 mm

BC BC

BC

F

A

Ans.


17/53





have an outside diameter of 42 mm and a wall

thickness of 3.5 mm. Determine the normal


FIGURE P1.11

Solution

Overall equilibrium:Begin the solution by determining the external

reaction forces acting on the truss at supports A

and B. Write equilibrium equations that include allexternal forces. Note that only the external forces

(i.e., loads and reaction forces) are considered at

this time. The internal forces acting in the truss

members will be considered after the externalreactions have been computed. The free-body

diagram (FBD) of the entire truss is shown. The

following equilibrium equations can be written forthis structure:

30 kN

30 kN

0 y

y

y F A

A

(30 kN)(4.5 m) (15 kN)(1.6 m) (5.6 m)

19.821 kN

0

x

A x

B

M B

15 kN 0

15 kN 15 kN ( 19.821 kN 34.821) kN

x x

x x

x

x

F A B

A A B

Method of joints:Before beginning the process of determining the internal forces in the axial members, the geometry ofthe truss will be used to determine the magnitude of the inclination angles of members AC and BC . Use

the definition of the tangent function to determine AC and BC :

1.6 mtan 0.355556 19.573

4.5 m

4 mtan 0.888889 41.634

4.5 m

AC AC

BC BC


18/53



Joint A:



reaction forces, A x and A y, act at joint A. Tension forces will be

assumed in each truss member.

cos(19.573 ) 0 x x AC F A F (a)

sin(19.573 ) 0 y y AC AB

F A F F (b)

Solve Eq. (a) for F AC :

34.821 kN

cos(19.573 ) cos(19.573 )36.957 kN x

AC

A F

and then compute F AB using Eq. (b):

sin(19.573 )

(30.000 kN) (36.957 kN)sin(19. 17.619573 ) kN

AB y AC F A F

Joint B:

Next, consider a FBD of joint B. In this instance, the equilibriumequations associated with joint B seem easier to solve than those that

would pertain to joint C . As before, tension forces will be assumed in

each truss member.

cos(41.634 ) 0 x x BC F B F (c)

sin(41.634 ) 0 y BC AB F F F (d)

Solve Eq. (c) for F BC :

( 19.821 kN)

cos(41.634 ) cos(41.634 )26.520 kN x

BC

B F

Eq. (d) can be used as a check on our calculations:

sin(41.634 )( 26.520 kN)sin(41.634 ) (17.619 kN) 0

y BC AB F F F Checks!


2 2 242 mm 2(3.5 mm) 35 mm (42 mm) (35 mm) 423.3296 mm4

d A


2

(17.619 kN)(1,000 N/kN)41.620 MPa

42

41.6 MPa

3.3296

)

mm

(T AB AB

AB

F

A

Ans.

2

(36.957 kN)(1,000 N/kN)87.301 MPa

4287.3 MPa

3.3296)

mm(T AC

AC

AC

F

A Ans.

2

( 26.520 kN)(1,000 N/kN)62.647 MPa

4262.6 MPa (C)

3.3296 mm

BC BC

BC

F

A

Ans.


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P1.12 The rigid beam BC shown in Figure P1.12 is

supported by rods (1) and (2) that have cross-sectionalareas of 175 mm

2 and 300 mm

2, respectively. For a

uniformly distributed load of w = 15 kN/m, determine

the normal stress in each rod. Assume L = 3 m and a =

1.8 m.

FIGURE P1.12

Solution

Equilibrium: Calculate the internal forces in members (1) and (2).

1

2

1

2

1.8 m(3 m) (15 kN/m)(1.8 m) 0

2

1.8 m(3 m) (15 kN/m)(1.8 m) 3 m 0

8.100 kN

18.900

2

kN

C

B

M F

M

F

F

F

Stresses:

211 2

1

(8.100 kN)(1,000 N/kN)46.286 N/mm

175 m46.3 MPa

m

F

A Ans.

22

2 22

(18.900 kN)(1,000 N/kN)

63.000 N/mm300 m 63.0 MPam

F

A

Ans.


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P1.13 Bar (1) in Figure P1.15 has a cross-

sectional area of 0.75 in.2. If the stress in bar

(1) must be limited to 30 ksi, determine the

maximum load P that may be supported by

the structure.

FIGURE P1.13

Solution

Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi,

the maximum force that may be carried by bar (1) is2

1,max 1 1 (30 ksi)(0.75 in. ) 22.5 kips F A

Consider a FBD of ABC . From the moment equilibriumequation about joint A, the relationship between the force in

bar (1) and the load P is:

1

1

(6 ft) (10 ft) 0

6 ft

10 ft

A M F P

P F

Substitute the maximum force F 1,max = 22.5 kips into this relationship to obtain the maximum load that

may be applied to the structure:

1

6 ft 6 ft(22.5 kips)

10 ft 10 ft

13.50 kips P F Ans.


21/53



P1.14 The rectangular bar shown in Figure

P1.14 is subjected to a uniformly distributedaxial loading of w = 13 kN/m and a

concentrated force of P = 9 kN at B.

Determine the magnitude of the maximum

normal stress in the bar and its location x.Assume a = 0.5 m, b = 0.7 m, c = 15 mm, and

d = 40 mm.

FIGURE P1.14

Solution

Equilibrium: Draw a FBD for the interval between A and B where

0 x a . Write the following equilibrium equation:

(13 kN/m)(1.2 m ) (9 kN) 0

(13 kN/m)(1.2 m ) (9 kN)

x F x F

F x

The largest force in this interval occurs at x = 0 where F = 6.6

kN.

In the interval between B and C where a x a b , and writethe following equilibrium equation:

(13 kN/m)(1.2 m ) 0

(13 kN/m)(1.2 m )

x F x F

F x

The largest force in this interval occurs at x = a where F = 9.1kN.

Maximum Normal Stress:

max

(9.1 kN)(1,000 N/kN)

(15 mm 15.17 MPa)(40 mm) at 0.5 m x Ans.


22/53



P1.15 The solid 1.25-in.-diameter rod shown in

Figure P1.15 is subjected to a uniform axialdistributed loading along its length of w = 750 lb/ft.

Two concentrated loads also act on the rod: P =

2,000 lb and Q = 1,000 lb. Assume a = 16 in. and b

= 32 in. Determine the normal stress in the rod atthe following locations:

(a) x = 10 in.

(b) x = 30 in.FIGURE P1.15

Solution

(a) x = 10 in.Equilibrium: Draw a FBD for the interval between A and B

where 0 x a , and write the following equilibrium equation:

(750 lb/ft)(1 ft/12 in.)(48 in. )

(2,000 lb) (1,000 lb) 0

(62.5 lb/in.)(48 in. ) 3,000 lb

x F x

F

F x

At x = 10 in., F = 5,375 lb.

Stress: The normal stress at this location can be calculated as follows.

2 2

2

(1.25 in.) 1.227185 in.4

5,375 lb4,379 4,380 p.944 psi

1si

.227185 in.

A

Ans.

(b) x = 30 in.

Equilibrium: Draw a FBD for the interval between B and C where a x a b , and write the following equilibriumequation:

(750 lb/ft)(1 ft/12 in.)(48 in. )

(1,000 lb) 0

(62.5 lb/in.)(48 in. ) 1,000 lb

x F x

F

F x

At x = 30 in., F = 2,125 lb.

Stress: The normal stress at this location can be calculated as follows.

2 1,730 ps2,125 lb

1,731.606 psi1.227185 i

in.

Ans.


23/53



P1.16 Two 6 in. wide wooden boards are

to be joined by splice plates that will befully glued on the contact surfaces. The

glue to be used can safely provide a shear

strength of 120 psi. Determine the smallest

allowable length L that can be used for thesplice plates for an applied load of P =

10,000 lb. Note that a gap of 0.5 in. is

required between boards (1) and (2). FIGURE P1.16

Solution

Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb appliedload on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as

V , equilibrium in the horizontal direction requires

0

10,000 lb5,000 lb

2

x F P V V

V

In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on

board (2) must be at least

2

min

5,000 lb41.6667 in.

120 psi A

The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least2

glue joint

41.6667 in.6.9444 in.

6 in. L

Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1)

and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the

boards in order to resist the 10,000 lb applied load.

The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore,

the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer.

Altogether, the length of the splice plates must be at least

min 6.9444 in. 6.9444 in. 0.5 in 14.39 in.. L Ans.


24/53



P1.17 For the clevis connection shown in Figure

P1.17, determine the maximum applied load P thatcan be supported by the 10-mm-diameter pin if the

average shear stress in the pin must not exceed 95

MPa.

FIGURE P1.17

Solution

Consider a FBD of the bar that is connected by the clevis,including a portion of the pin. If the shear force acting on each

exposed surface of the pin is denoted by V , then the shear force

on each pin surface is related to the load P by:

0 2 x F P V V P V

The area of the pin surface exposed by the FBD is simply the cross-sectional area of the pin:

2 2 2

pin pin (10 mm) 78.539816 mm4 4 A d

If the average shear stress in the pin must be limited to 95 MPa, the maximum shear force V on a single

cross-sectional surface must be limited to2 2

bolt (95 N/mm )(78.539816 mm ) 7,461.283 NV A

Therefore, the maximum load P that may be applied to the connection is

2 2(7,461.283 N) 14,922.565 14.92 kN N P V Ans.


25/53



P1.18 For the connection shown in Figure P1.18,

determine the average shear stress produced in the 3/8-in. diameter bolts if the applied load is P = 2,500 lb.

FIGURE P1.18

Solution

There are four bolts, and it is assumed that each bolt supports an equal portion of the external load P .

Therefore, the shear force carried by each bolt is

2,500 lb625 lb

4 boltsV

The bolts in this connection act in single shear. The cross-sectional area of a single bolt is

2 2 2 2

bolt bolt (3 / 8 in.) (0.375 in.) 0.110447 in.4 4 4

A d

Therefore, the average shear stress in each bolt is

2

bolt

625 lb5,658.8427 psi

0.110447 in.5,660 psi

V

A Ans.


26/53



P1.19 The five-bolt connection shown in Figure P1.19 must

support an applied load of P = 265 kN. If the average shear stressin the bolts must be limited to 120 MPa, determine the minimum

bolt diameter that may be used for this connection.

FIGURE P1.19

Solution

There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P .Therefore, the shear force carried by each bolt is

265 kN53 kN 53,000 N

5 boltsV

Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at

least:

2

2

53,000 N441.6667 mm

120 N/mmV A

Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are

available to transmit shear stress in each bolt.

22

bolt

441.6667 mm220.8333 mm per surface

2 surfaces per bolt 2 surfaces

V A A

The minimum bolt diameter must be

2 2

bolt bolt16.77 m220.8333 mm 16.7682 mm m

4d d

Ans.


27/53



P1.20 A coupling is used to connect a 2 in. diameter

plastic pipe (1) to a 1.5 in. diameter pipe (2), asshown in Figure P1.20. If the average shear stress in

the adhesive must be limited to 400 psi, determine

the minimum lengths L1 and L2 required for the joint

if the applied load P is 5,000 lb.

FIGURE P1.24

Solution

To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is

2

adhesive

5,000 lb12.5 in.

400 psiV

V A

Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and thecircumference C 1 of pipe (1) is

1 1(2.0 in.) 6.2832 in.C D

The minimum length L1 is therefore2

1

1

12.5 in.1.9894 in.

6.2832 i1.989 i

n n.

.

V A

LC

Ans.

Consider the coupling on pipe (2). The circumference C 2 of pipe (2) is

2 2 (1.5 in.) 4.7124 in.C D

The minimum length L2 is therefore2

2

2

12.5 in.2.6526 in.

4.71242.65 in.

in.

V A

LC

Ans.


28/53



1.21 A hydraulic punch press is used to

punch a slot in a 0.50-in. thick plate, asillustrated in Fig. P1.21. If the plate shears

at a stress of 30 ksi, determine the

minimum force P required to punch the

slot.

FIGURE P1.21

Solution

The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug isgiven by

perimeter 2(3.00 in.) + (0.75 in.) 8.35619 in.

Thus, the area subjected to shear stress is2 perimeter plate thickness (8.35619 in.)(0.50 in.) 4.17810 in.

V A

Given that the plate shears at = 30 ksi, the force required to remove the slug is therefore2

min(30 ksi)(4.17810 in. ) 125.343 kips 125.3 kips

V P A Ans.


29/53



P1.22 The handle shown in Figure P1.22 is

attached to a 40-mm-diameter shaft with asquare shear key. The forces applied to the

lever are P = 1,300 N. If the average shear

stress in the key must not exceed 150 MPa,

determine the minimum dimension a that must be used if the key is 25 mm long. The overall

length of the handle is L = 0.70 m.

FIGURE P1.22

Solution

To determine the shear force V that must be resisted by the shear key, sum moments about the center ofthe shaft (which will be denoted O):

700 mm 700 mm 40 mm(1,300 N) (1,300 N) 0

2 2 2

45,500 N

O M V

V

Since the average shear stress in the key must not exceed 150 MPa, the shear area required is

2

2

45,500 N

303.3333 mm150 N/mmV

V

A

The shear area in the key is given by the product of its length L (i.e., 25 mm) and its width a. Therefore,

the minimum key width a is2303.3333 mm

12.1333 mm25 m

12.1m

3 mmV A

a L

Ans.


30/53



P1.23 An axial load P is supported by the short steel

column shown in Figure P1.23. The column has a cross-sectional area of 14,500 mm

2. If the average normal stress

in the steel column must not exceed 75 MPa, determine the

minimum required dimension a so that the bearing stress

between the base plate and the concrete slab does notexceed 8 MPa. Assume b = 420 mm.

FIGURE P1.23

Solution

Since the normal stress in the steel column must not exceed 75 MPa, the maximum column load is2 2

max(75 N/mm )(14,500 mm ) 1,087,500 N P A

The maximum column load must be distributed over a large enough area so that the bearing stress

between the base plate and the concrete slab does not exceed 8 MPa; therefore, the minimum plate areais

2

min 2

1,087,500 N135,937.5 mm

8 N/mmb

P A

The area of the plate is a ×b. Since b = 420, the minimum length of a must be2

min

2

135,937.5 mm

135,937.5 mm

420 mm324 mm

A a b

a

Ans.


31/53



P1.24 The two wooden boards shown in Figure P1.24 are

connected by a 0.5-in.-diameter bolt. Washers are installedunder the head of the bolt and under the nut. The washer

dimensions are D = 2 in. and d = 5/8 in. The nut is tightened

to cause a tensile stress of 9,000 psi in the bolt. Determine

the bearing stress between the washer and the wood.

FIGURE P1.24

Solution

The tensile stress in the bolt is 9,000 psi; therefore, the tension force that acts in the bolt is

2 2

bolt bolt bolt (9,000 psi) (0.5 in.) (9,000 psi)(0.196350 in. ) 1,767.146 lb4

F A

The contact area between the washer and the wood is

2 2 2

washer (2 in.) (0.625 in.) 2.834796 in.4

A

Thus, the bearing stress between the washer and the wood is

2

1,767.146 lb

2.834796 in.623 psi

b Ans.


32/53



P1.25 For the beam shown in Figure P1.25, the

allowable bearing stress for the material under

the supports at A and B is b = 800 psi.Assume w = 2,100 lb/ft, P = 4,600 lb, a = 20 ft,

and b = 8 ft. Determine the size of square

bearing plates required to support the loadingshown. Dimension the plates to the nearest ½

in.

FIGURE P1.25

Solution

Equilibrium: Using the FBD shown,

calculate the beam reaction forces.

20 ft(20 ft) (2,100 lb/ft)(20 ft) (4,600 lb)(28 ft) 0

2

27,440 lb

A y

y

M B

B

20 ft(20 ft) (2,100 lb/ft)(20 ft) (4,600 lb)(8 ft) 0

2

19,160 lb

B y

y

M A

A

Bearing plate at A: The area of the bearing plate required for support A is

219,160 lb 23.950 in.800 psi

A A

Since the plate is to be square, its dimensions must be2 use 5 in.23.95 5 in. bearing plate at0 in. 4.894 in . Awidth Ans.

Bearing plate at B : The area of the bearing plate required for support B is

227,440 lb 34.300 in.800 psi

B A

Since the plate is to be square, its dimensions must be2 use 6 in.34.30 6 in. bearing plate at0 in. 5.857 in . Bwidth Ans.


33/53



P1.26 The d = 15-mm-diameter solid rod shown in

Figure P1.26 passes through a D = 20-mm-diameterhole in the support plate. When a load P is applied

to the rod, the rod head rests on the support plate.

The support plate has a thickness of b = 12 mm.

The rod head has a diameter of a = 30 mm and thehead has a thickness of t = 10 mm. If the normal

stress produced in the rod by load P is 225 MPa,

determine:(a) the bearing stress acting between the support

plate and the rod head.

(b) the average shear stress produced in the rodhead.

(c) the punching shear stress produced in the

support plate by the rod head.

FIGURE P1.26

Solution

The cross-sectional area of the rod is:

2 2

rod (15 mm) 176.715 mm4

A

The tensile stress in the rod is 225 MPa; therefore, the tension force in the rod is2 2

rod rod rod (225 N/mm )(176.715 mm ) 39,760.782 N F A

(a) The contact area between the support plate and the rod head is

2 2 2

contact (30 mm) (20 mm) 392.699 mm

4

A

Thus, the bearing stress between the support plate and the rod head is

2

39,760.782 N

392.699 mm101.3 MPa

b Ans.

(b) In the rod head, the area subjected to shear stress is equal to the perimeter of the rod times the

thickness of the head.2(15 mm)(10 mm) 471.239 mm

V A

and therefore, the average shear stress in the rod head is

2

39,760.782 N

471.

84.4 MP

m

a

239 m

Ans.

(c) In the support plate, the area subjected to shear stress is equal to the product of the rod head

perimeter and the thickness of the plate.2(30 mm)(12 mm) 1,130.973 mmV A

and therefore, the average punching shear stress in the support plate is

2

39,760.782 N

1,13035.2 MPa

.973 mm Ans.


34/53



P1.27 The rectangular bar is connected to the support bracket with

a circular pin, as shown in Figure P1.27. The bar width is w = 1.75in. and the bar thickness is 0.375 in. For an applied load of P =

5,600 lb, determine the average bearing stress produced in the bar

by the 0.625-in.-diameter pin.

FIGURE P1.27

SolutionThe average bearing stress produced in the bar by the pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the bar thickness. Therefore, the average bearing stress

in the bar is

5,600 lb23,893.33 psi

(0.625 in.)(0.375 i23,

n.900

) psi

b Ans.


35/53



P1.28 The steel pipe column shown in Figure P1.28 has an

outside diameter of 8.625 in. and a wall thickness of 0.25 in.The timber beam is 10.75 in wide, and the upper plate has the

same width. The load imposed on the column by the timber

beam is 80 kips. Determine

(a) The average bearing stress at the surfaces between the pipecolumn and the upper and lower steel bearing plates.

(b) The length L of the rectangular upper bearing plate if its

width is 10.75 in. and the average bearing stress between thesteel plate and the wood beam is not to exceed 500 psi.

(c) The dimension “a” of the square lower bearing plate if the

average bearing stress between the lower bearing plate andthe concrete slab is not to exceed 900 psi.

Figure P1.28

Solution

(a) The area of contact between the pipe column and one of the bearing plates is simply the cross-sectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d :

2 2 8.625 in. 2(0.25 in.) 8.125 in. D d t d D t

The pipe cross-sectional area is

2 2 2 2 2

pipe (8.625 in.) (8.125 in.) 6.5777 in.4 4

A D d

Therefore, the bearing stress between the pipe and one of the bearing plates is

2

80 kips12.1623 ksi

6.5777 in.

12.16 ksib

b

P

A

Ans.

(b) The bearing stress between the timber beam and the upper bearing plate must not exceed 500 psi

(i.e., 0.5 ksi). To support a load of 80 kips, the contact area must be at least

280 kips 160 in.0.5 ksi

b

b

P A

If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be2160 in.

14.8837 in. beam width 10.75

14.88 in

.in.

b A L Ans.

(c) The bearing stress between the concrete slab and the lower bearing plate must not exceed 900 psi(i.e., 0.9 ksi). To support the 80-kip pipe load, the contact area must be at least

280 kips 88.8889 in.0.9 ksi

b

b

P A

Since the lower bearing plate is square, its dimension a must be288.8889 in. 9.4 9.43 in281 n. .ib A a a a Ans.


36/53



P1.29 A clevis-type pipe hanger supports an 8-in-

diameter pipe as shown in Figure P1.29. The hangerrod has a diameter of 1/2 in. The bolt connecting the

top yoke and the bottom strap has a diameter of 5/8 in.

The bottom strap is 3/16-in.-thick by 1.75-in.-wide by

36-in.-long. The weight of the pipe is 2,000 lb.Determine the following:

(a) the normal stress in the hanger rod

(b) the shear stress in the bolt(c) the bearing stress in the bottom strap

FIGURE P1.29

Solution

(a) The normal stress in the hanger rod is

2 2

rod

rod 2

(0.5 in.) 0.196350 in.4

2,000 lb10,185.917 psi

0.196350 in.10,190 psi

A

Ans.

(b) The cross-sectional area of the bolt is:

2 2

bolt (0.625 in.) 0.306796 in.4

A

The bolt acts in double shear; therefore, its average shear stress is

bolt 2

2,000 lb3,259.493 psi

2(0.306796 in.3,260 i

) ps Ans.

(c) The bearing stress in the bottom strap is based on the projected area of the bolt in contact with thestrap. Also, keep in mind that there are two ends of the strap that contact the bolt. The bearing stress isthus

2,000 lb8,533.334 psi

2(0.625 in.)(3/168,530 psi

in.)b

Ans.


37/53



P1.30 Rigid bar ABC shown in Figure P1.30 is

supported by a pin at bracket A and by tie rod (1).Tie rod (1) has a diameter of 5 mm, and it is

supported by double-shear pin connections at B and

D. The pin at bracket A is a single-shear

connection. All pins are 7 mm in diameter.Assume a = 600 mm, b = 300 mm, h = 450 mm, P

= 900 N, and = 55°. Determine the following:

(a) the normal stress in rod (1)(b) the shear stress in pin B (c) the shear stress in pin A

FIGURE P1.30

Solution

Equilibrium: Using the FBD shown, calculatethe reaction forces that act on rigid bar ABC .

1

1

sin(36.87 )(600 mm)

(900 N)sin(55 )(900 mm) 0

1,843.092 N

A M F

F

(1,843.092 N)cos(36.87 ) (900 N)cos(55 ) 0

958.255 N

x x

x

F A

A

(1,843.092 N)sin(36.87 ) (900 N)sin(55 ) 0

368.618 N

y y

y

F A

A

The resultant force at A is

2 2(958.255 N) ( 368.618 N) 1,026.709 N A

(a) Normal stress in rod (1).

2 2

rod

rod 2

(5 mm) 19.635 mm4

1,843.092 N

19.635 mm93.9 MPa

A

Ans.

(b) Shear stress in pin B . The cross-sectional area of a 7-mm-diameter pin is:

2 2

pin (7 mm) 38.485 mm4

A

Pin B is a double shear connection; therefore, its average shear stress is

pin 2

1,843.092 N

2(38.485 m23.9 M a

)P

m B Ans.

(c) Shear stress in pin A.

Pin A is a single shear connection; therefore, its average shear stress is

pin 2

1,026.709 N

38.485 m26.7 MPa

m A Ans.


38/53



P1.31 The bell crank shown in Figure P1.31 is in

equilibrium for the forces acting in rods (1) and (2).The bell crank is supported by a 10-mm-diameter

pin at B that acts in single shear. The thickness of

the bell crank is 5 mm. Assume a = 65 mm, b =

150 mm, F 1 = 1,100 N, and = 50°. Determine thefollowing:

(a) the shear stress in pin B

(b) the bearing stress in the bell crank at B

FIGURE P1.31

Solution

Equilibrium: Using the FBD shown, calculate the

reaction forces that act on the bell crank.

2

2

(1,100 N)sin(50 )(65 mm)

(150 mm) 0

365.148 N

B M

F

F

(1,100 N)cos(50 )

365.148 N 0

341.919 N

x x

x

F B

B

(1,100 N)sin(50 ) 0

842.649 N

y y

y

F B

B

The resultant force at B is2 2(341.919 N) ( 842.649 N) 909.376 N B

(a) Shear stress in pin B . The cross-sectional area of the 10-mm-diameter pin is:

2 2

pin (10 mm) 78.540 mm4

A

Pin B is a single shear connection; therefore, its average shear stress is

pin 2

909.376 N

78.540 mm11.58 MPa B Ans.

(b) Bearing stress in the bell crank at B . The average bearing stress produced in the bell crank by the

pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the bell crank thickness. Therefore, the average bearing stress in the bell crank is

909.376 N

(10 mm)(5 mm)18.19 MPa

b Ans.


39/53



P1.32 The beam shown in Figure P1.32 is

supported by a pin at C and by a short link AB. If w = 30 kN/m, determine the average

shear stress in the pins at A and C . Each

pin has a diameter of 25 mm. Assume L =

1.8 m and = 35°.

FIGURE P1.32

Solution

Equilibrium: Using the FBD shown,

calculate the reaction forces that act on the

beam.

1

1

1.8 msin(35 )(1.8 m) (30 kN/m)(1.8 m) 0

247.0731 kN

C M F

F

(47.0731 kN)cos(35 ) 0

38.5600 kN

x x

x

F C

C

1.8 m(1.8 m) (30 kN/m)(1.8 m) 0

2

27.0000 kN

B y

y

M C

C

The resultant force at C is2 2(38.5600 kN) (27.0000 kN) 47.0731 kNC

Shear stress in pin A. The cross-sectional area of a 25-mm-diameter pin is:

2 2

pin (25 mm) 490.8739 mm4

A

Pin A is a single shear connection; therefore, its average shear stress is

pin 2

47,073.1 N

490.8739 m95.9 MPa

m A Ans.

Shear stress in pin C .Pin C is a double shear connection; therefore, its average shear stress is

pin 2

47,073.1 N

2(490.8739 m47.9 M a

)P

mC

Ans.


40/53



P1.33 The bell-crank mechanism shown

in Figure P1.33 is in equilibrium for anapplied load of P = 7 kN applied at A.

Assume a = 200 mm, b = 150 mm, and =65°. Determine the minimum diameter d

required for pin B for each of the followingconditions:

(a) The average shear stress in the pin may

not exceed 40 MPa.(b) The bearing stress in the bell crank

may not exceed 100 MPa.

(c) The bearing stress in the support

bracket may not exceed 165 MPa.FIGURE P1.33

Solution

Equilibrium: Using the FBD shown, calculate

the reaction forces that act on the bell crank.

2

2

(7,000 N)sin(65 )(200 mm)

(150 mm) 0

8,458.873 N

B M

F

F

(7,000 N)cos(65 )

8,458.873 N 0

11,417.201 N

x x

x

F B

B

(7,000 N)sin(65 ) 0

6,344.155 N

y y

y

F B

B

The resultant force at B is2 2( 11,417.201 N) (6,344.155 N) 13,061.423 N B

(a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pinat B is

2

2

13,061.423 N326.536 mm

40 N/mmV A

Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin

is2

pin163.268 mm

2

V A A

and therefore, the pin must have a diameter of

24 (163.268 m 14.m 4) 2 mmd

Ans.


41/53



(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the

bell crank must equal or exceed

2

2

13,061.423 N130.614 mm

100 N/mmb A

The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d . Calculate the

required pin diameter d :2130.614 mm

8

16.

3

mm

3 mmd Ans.

(c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6-

mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit

on the support bracket is

2

2

13,061.423 N79.160 mm

165 N/mmb

A

279.160 mm

2(6 mm)6.60 mmd Ans.


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P1.34 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial

load of 150 kN. Determine the maximum normal and shear stresses in the bar.

Solution

The maximum normal stress in the steel bar is

max

(150 kN)(1,000 N/kN)

(25 mm)(75 mm)80 MPa

F

A Ans.

The maximum shear stress is one-half of the maximum normal stressmax

max2

40 MPa

Ans.


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P1.35 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The

maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine therequired diameter for the rod.

Solution

Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is

2

min

max

92 kips3.0667 in.

30 ksi

F A

For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be

2

min

max

92 kips3.8333 in.

2 2(12 ksi)

F A

Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the

normal and shear stress limits.

The minimum rod diameter D is therefore

2 2

min min3.8333 in. 2.2092 in. 2.21 in.4

d d

Ans.


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P1.36 An axial load P is applied to the

rectangular bar shown in Figure P1.36. Thecross-sectional area of the bar is 400 mm

2.

Determine the normal stress perpendicular to

plane AB and the shear stress parallel to

plane AB if the bar is subjected to an axialload of P = 70 kN.

FIGURE P1.36

Solution

The angle for the inclined plane is 35°. The

normal force N perpendicular to plane AB isfound from

cos (40 kN) cos 35 57.3406 kN N P

and the shear force V parallel to plane AB is

sin (70 kN)sin 35 40.1504 kNV P

The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is

22400 mm

488.3098 mmcos cos35

n

A A

The normal stress n perpendicular to plane AB is

2

(57.3406 kN)(1,000 N/kN)117.4268 MPa

488117.4 MP

.3098 mma

n

n

N

A Ans.

The shear stress nt parallel to plane AB is

2

(40.1504 kN)(1,000 N/kN)82.2231 MPa

482.2 MP

88.309a

8 mmnt

n

V

A Ans.


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P1.37 An axial load P is applied to the 1.75 in.

by 0.75 in. rectangular bar shown in FigureP1.37. Determine the normal stress

perpendicular to plane AB and the shear stress

parallel to plane AB if the bar is subjected to

an axial load of P = 18 kips.

FIGURE P1.37

Solution

The angle for the inclined plane is 60°. Thenormal force N perpendicular to plane AB is

found from

cos (18 kips) cos 60 9.0 kips N P


sin (18 kips)sin60 15.5885 kipsV P

The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane

AB is2

21.3125 in./ cos 2.6250 in.cos60

n A A


2

9.0 kips3.4286 ksi

2.6250 in3.43 ks

.i

n

n

N

A Ans.


2

15.5885 kips5.9385 ksi

2.62505.94 ks

ii

n.nt

n

V

A Ans.


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P1.38 A compression load of P = 80 kips is applied to a 4 in. by 4

in. square post, as shown in Figure P1.38/39. Determine the normalstress perpendicular to plane AB and the shear stress parallel to

plane AB.

FIGURE P1.38/39

Solution

The angle for the inclined plane is 55°. The normal force N

perpendicular to plane AB is found fromcos (80 kips)cos55 45.8861 kips N P


sin (80 kips)sin55 65.5322 kipsV P

The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area

along inclined plane AB is2

216 in./ cos 27.8951 in.

cos55n A A


2

45.8861 kips1.6449 ksi

27.89511.645 ksi

in.n

n

N

A Ans.


2

65.5322 kips2.3492 ksi

27.89512.35 ks

ii

n.nt

n

V

A Ans.


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P1.39 Specifications for the 50 mm × 50 mm square bar

shown in Figure P1.38/39 require that the normal and shearstresses on plane AB not exceed 120 MPa and 90 MPa,

respectively. Determine the maximum load P that can be

applied without exceeding the specifications.

FIGURE P1.38/39

Solution

The general equations for normal and shear stresses on an inclined plane in terms of the angle are

(1 cos2 )2

n P A

(a)

and

sin22

nt

P

A (b)

The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm

2, and the angle for plane AB is

55°.

The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can besupported by the square bar is found from Eq. (a):

2 2

2 2(2,500 mm )(120 N/mm ) 911,882 N1 cos2 1 cos2(55 )

n A P

The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear

stress limit is2 22 2(2,500 mm )(90 N/mm )

478,880 Nsin 2 sin 2(55 )

nt A P

Thus, the maximum load that can be supported by the bar is

max 479 kN P Ans.


48/53

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