MEEM 2150 MTU Vable Mechanics of Materials Solutions CH1

MEEM 2150 Solutions to HW#2 – Ch #1

Gershenson, Fall 2014 1.3� The cable shown in Fig. P1.3 has a diameter of 1/5 inch. If the maximum stress in the cable must be limited to 4 ksi (T), what is the maximum weight that can be lifted?

The free body diagram: By equilibrium of forces we have: N = W

The area of cross-section of the rope is:

The average axial stress is:

1.9� A 5 kg picture is hung using a wire as shown in Fig. P1.9. If the tensile stress in the wires cannot exceed 10MPa, determine the minimum diameter of the wire to the nearest millimeter.

M. Vable Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1 January 2013

This manual is for instructors only. If you are a student then you have an illegal copy. 1-3

1.3 The cable shown in Fig. P1.3 has a diameter of 1/5 inch. If the maximum stress in the cable must be limited to 4 ksi (T), what is the maximum weight that can be lifted?

Fig. P1.3 Solution d=1/5 in Wmax = ?

------------------------------------------------------------The free body diagram of the weight is shown below.

By equilibrium of forces we have:


The average axial stress is: or or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

Vmax 4ksi T� �d

W

N

N W=

A S4--- 1

5---© ¹§ · 2

31.41 10 3–� �in2= =

V NA---- W

31.41 10 3–� �----------------------------- 4 103� �d= = W 125.66lbd Wmax 125.6lb=









--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

Vmax 4ksi T� �d

W

N

N W=

A S4--- 1

5---© ¹§ · 2

31.41 10 3–� �in2= =

V NA---- W

31.41 10 3–� �----------------------------- 4 103� �d= = W 125.66lbd Wmax 125.6lb=









--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

Vmax 4ksi T� �d

W

N

N W=

A S4--- 1

5---© ¹§ · 2

31.41 10 3–� �in2= =

V NA---- W

31.41 10 3–� �----------------------------- 4 103� �d= = W 125.66lbd Wmax 125.6lb=









--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

Vmax 4ksi T� �d

W

N

N W=

A S4--- 1

5---© ¹§ · 2

31.41 10 3–� �in2= =

V NA---- W

31.41 10 3–� �----------------------------- 4 103� �d= = W 125.66lbd Wmax 125.6lb=



1.9 A 5 kg picture is hung using a wire as shown in Fig. P1.9. If the tensile stress in the wires cannot exceed 10MPa, determine the minimum diameter of the wire to the nearest millimeter.

Fig. P1.9 Solution m= 5kg dmin = ? nearest millimeter.

------------------------------------------------------------The free body diagram of the picture frame is shown in Fig. (a). By equilibrium of forces in the y-direction we have the following equation.

1

The average axial stress in the wire is: or or

The minimum diameter to the nearest miliimeter that satisfy the inequality is:

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

54o

Vmax 10MPad

2N 54sin W 5g= = or N N 0.618 5� � 9.81� � 30.315 N= = =

N N

W = 5g

(a)54o 54o

V NA---- 30.315

Sd2 4e� �--------------------- 10� � 106� �d= = d 1.96 10 3–� � mt

dmin 2 mm=

The free body diagram of the picture frame is shown in Fig. (a). By equilibrium of forces in the y-direction we have the following equation.

The average axial stress in the wire is:

The minimum diameter to the nearest millimeter that satisfy the inequality is:

1.16� A 70 kg. person is standing on a bathroom scale that has a dimensions 150 mm x 100 mm x 40 mm. Determine the bearing stress between the scales and the floor.

The weight of the person is: W = mg = (70) (9.81) = 686.7 N�

By making an imaginary cut between the scales and the floor we obtain the following free body diagram.

By equilibrium of forces we obtain: N = W = 686.7N






1



--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

54o

Vmax 10MPad

2N 54sin W 5g= = or N N 0.618 5� � 9.81� � 30.315 N= = =

N N

W = 5g

(a)54o 54o

V NA---- 30.315

Sd2 4e� �--------------------- 10� � 106� �d= = d 1.96 10 3–� � mt

dmin 2 mm=






1



--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

54o

Vmax 10MPad

2N 54sin W 5g= = or N N 0.618 5� � 9.81� � 30.315 N= = =

N N

W = 5g

(a)54o 54o

V NA---- 30.315

Sd2 4e� �--------------------- 10� � 106� �d= = d 1.96 10 3–� � mt

dmin 2 mm=






1



--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

54o

Vmax 10MPad

2N 54sin W 5g= = or N N 0.618 5� � 9.81� � 30.315 N= = =

N N

W = 5g

(a)54o 54o

V NA---- 30.315

Sd2 4e� �--------------------- 10� � 106� �d= = d 1.96 10 3–� � mt

dmin 2 mm=M. Vable Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1 January 2013


1.16 A 70 kg. person is standing on a bathroom scale that has a dimensions 150 mm x 100 mm x 40 mm. Determine the bearing stress between the scales and the floor.

Fig. P1.16 Solution m = 70 kg Scale: 150 x 100 �Vb= ?

------------------------------------------------------------The weight of the person is: W = mg = (70) (9.81) = 686.7 NBy making an imaginary cut between the scales and the floor we obtain the following free body diagram.

By equilibrium of forces we obtain: N = W = 686.7NThe area of cross-section is: A = (150) (100) = 15 (103) mm2 = 15.0 (10-3) m2

The average bearing stress is:

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

N

VbNA---- 686.7

15 10 3–� �----------------------= 45.78 10 3–� � N m2e= = Vb 45.8 kPa C� �=








--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

N

VbNA---- 686.7

15 10 3–� �----------------------= 45.78 10 3–� � N m2e= = Vb 45.8 kPa C� �=

The area of cross-section is: A = (150) (100) = 15 (103) mm2 = 15.0 (10-3) m2


1.18 Determine the average bearing stress at the bottom of a block shown in Fig. P1.18 in terms of the specific weight ữ, and the length dimensions a and h.

The volume of material is:

The weight of the material is:

By making an imaginary cut at the base of the block, we obtain the following free body diagram

By force equilibrium we have: N = W = 505ữah2

The area of cross-section is: A = (100 a) (10 h) = 1000 ah









--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

W

N

VbNA---- 686.7

15 10 3–� �----------------------= 45.78 10 3–� � N m2e= = Vb 45.8 kPa C� �=M. Vable Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1 January 2013


1.18 Determine the average bearing stress at the bottom of a block shown in Fig. P1.18 in terms of the specific weight J, and the length dimensions a and h.

Fig. P1.18 Solution Vb = f(J,a,h) = ?

------------------------------------------------------------The volume of material is:

The weight of the material is: W = JV = 505Jah2

By making an imaginary cut at the base of the block, we obtain the following free body diagram.

By force equilibrium we have: N = W = 505Jah2



--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

h10 h

a

100 a

V 12--- 100a a+� �h 10h� � 505ah2= =

W

N

VbNA---- 505Jah2

1000ah--------------------= 0.505Jh= = V 0.505Jh C� �=











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

h10 h

a

100 a

V 12--- 100a a+� �h 10h� � 505ah2= =

W

N

VbNA---- 505Jah2

1000ah--------------------= 0.505Jh= = V 0.505Jh C� �=











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

h10 h

a

100 a

V 12--- 100a a+� �h 10h� � 505ah2= =

W

N

VbNA---- 505Jah2

1000ah--------------------= 0.505Jh= = V 0.505Jh C� �=











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

h10 h

a

100 a

V 12--- 100a a+� �h 10h� � 505ah2= =

W

N

VbNA---- 505Jah2

1000ah--------------------= 0.505Jh= = V 0.505Jh C� �=











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

h10 h

a

100 a

V 12--- 100a a+� �h 10h� � 505ah2= =

W

N

VbNA---- 505Jah2

1000ah--------------------= 0.505Jh= = V 0.505Jh C� �=

1.21 The Great pyramid of Giza shown in Figure 1.14d has a base of 757.71ft x 757.71ft and a height of 480.96 ft. Assume an average specific weight of ữ = 75 lb/ft3. Determine (a) the bearing stress at the base of the pyramid. (b) the average compressive stress at mid height.h.

a=757.7ft h=480.96ft ữ=75lb/ft3 σb =? σH =?

Fig. (a) shows the geometry of the pyramid. The volume of the pyramid can be found as shown below.

The weight of the pyramid is:

Fig. (b) shows the free body diagram of the entire pyramid. The total bearing

force is

The bearing stress at the base is:

The sides of square at half height will be have the base side. Thus b=a/2.

Thus the volume will be

The weight of top half of pyramid is:



1.21 The Great pyramid of Giza shown in Figure 1.14d has a base of 757.71ft x 757.71ft and a height of 480.96 ft. Assume an average specific weight of J = 75 lb/ft3. Determine (a) the bearing stress at the base of the pyramid. (b) the average compressive stress at mid height. Solution a= 757.7 ft h= 480.96 ftJ = 75 lb/ft3 = ? = ?

------------------------------------------------------------Fig. (a) shows the geometry of the pyramid. The volume of the pyramid can be found as shown below.

1


2

Fig. (b) shows the free body diagram of the entire pyramid. The total bearing force is

3The bearing stress at the base is:

4

The sides of square at half height will be have the base side. Thus b=a/2. Thus the volume will be


5

The compressive stress will be:

6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=


1.26 The schematic of a punch and die arrangement shown Fig. P1.26 is used to punch out thin plate objects of different shapes. The cross-section of the punch and the die is a circle of diameter 1 inch. A force of P = 6 kips is applied to the punch. If the plate thickness t = 1/8 inch, what would be the average shear stress in the plate along the path of the punch.

The free body diagram of the punch as it will go through the plate is shown below.

By equilibrium of forces we have the following equation.

1.33 A bolt passing through a piece of wood is shown in Fig. P1.33. Determine (a) the axial stress in the bolt. (b) the average shear stress in the bolt-head. (c) the average bearing stress between the bolt head and the wood. (d) the average shear stress in the wood.





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=





1


2



4



5


6

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vb VH

V 13---a2h 1

3--- 757.7� �2 480.96� � 92.040 106� � ft3= = =

W JV 75� � 92.040� � 106� � 6.903 109� �lb= = =

a

h

a

h/2

b

b

(a)

W

N

(b)

N W 6.903 109� �lb= =

VbNa2----- 6.903 109� �

757.7� �2--------------------------- 12024 psi= = =

Vb 12024 psi C� �=

VH13--- b� �2 h

2---© ¹§ · 1

6---b2h= =

WH JVHJ6---b2h= =

VHNH

b2--------

WH

b2--------- J

6---h 75� � 480.96� �

6-------------------------------- 6012psi= = = = =

VH 6012psi C� �=




Fig. P1.26 Solution P = 6 kips t = 1/8 inch d = 1 inch W�= ?

------------------------------------------------------------The free body diagram of the punch as it will go through the plate is shown below.


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Plate

Punch

Die Die

P

t

W

P

P W Sd� � t� �= or W PSdt-------- 6

S 1� � 1 8e� �-------------------------- 15.28ksi= = = W 15.3ksi=







or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Plate

Punch

Die Die

P

t

W

P

P W Sd� � t� �= or W PSdt-------- 6

S 1� � 1 8e� �-------------------------- 15.28ksi= = = W 15.3ksi=







or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Plate

Punch

Die Die

P

t

W

P

P W Sd� � t� �= or W PSdt-------- 6

S 1� � 1 8e� �-------------------------- 15.28ksi= = = W 15.3ksi=







or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Plate

Punch

Die Die

P

t

W

P

P W Sd� � t� �= or W PSdt-------- 6

S 1� � 1 8e� �-------------------------- 15.28ksi= = = W 15.3ksi=

The following free body diagrams can be drawn:

The area of cross-section of the bolt is: Abolt = π(1 ⁄ 4)2 ⁄ 4 = 49.09(10–3)in2

By equilibrium of forces in Fig. (a) we obtain: N = P = 1.5 kips

The average normal stress at section BB is:

By equilibrium of forces in Fig. (b) we obtain the following.

As per Fig(c), the area of cross-section over which the bearing stress acts is as given below.

By equilibrium of forces in Fig. (c) we obtain the following.

By equilibrium of forces in Fig. (d) we obtain the following.




Fig. P1.33

Solution Vbolt = ? Wbolt = ? Vbear = ? Wwood = ?------------------------------------------------------------

The following free body diagrams can be drawn.

The area of cross-section of the bolt is: By equilibrium of forces in Fig. (a) we obtain: N = P = 1.5 kipsThe average normal stress at section BB is:

or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =

VbearAbear P 1.5= = or Vbear1.5

0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=

Wwood S 3 4e� �> @ 1 2e� � P 1.5= = or Wwood1.5

3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=




Fig. P1.33




or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =


0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=


3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=




Fig. P1.33




or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =


0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=


3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=




Fig. P1.33




or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =


0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=


3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=




Fig. P1.33




or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =


0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=


3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=




Fig. P1.33




or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =


0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=


3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=




Fig. P1.33




or


or



or


or

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P= 1.5 kips

1/4 in

3/4 in

3/8 in

1/2 in

PW

3/8 in

1/4 in3/4 in

PVbear P

3/4 in

1/2 in

Wwood(a) (c) (d)

P

(b)

N

Abolt S 1 4e� �2 4e 49.09 10 3–� �in2= =

VboltN

Abolt------------- 1.5

49.09 10 3–� �----------------------------- 30.556ksi= = = Vbolt 30.56 ksi T� �=

W S 1 4e� �> @ 3 8e� � P 1.5= = or W 1.53S 32e� �

--------------------- 5.093ksi= = W 5.09 ksi=

Abear S 3 4e� �2 1 4e� �2–> @ 4e 0.3927in2= =


0.3927---------------- 3.8197ksi= = Vbear 3.82 ksi C� �=


3S 8e� �------------------ 1.273ksi= = Wwood 1.27ksi=

1.38 A metal plate welded to an I-beam is securely fastened to the foundation wall using four bolts of 1/ 2 in diameter as shown Figure P1.36. The allowable normal stress in the bolts is 15 ksi and the allowable shear stress is 12 ksi. Assume the load is equally distributed between the four bolts. Determine the maximum load P to the nearest pound the beam can support.

By equilibrium we have:

The stresses have to be less than the limiting values, hence

The maximum value of P is:

1.47 A butt joint is created by welding two plates to transmits a force of P = 250 kN as shown in Figure P1.47. Determine the average normal and shear stress on the plane AA of the weld.



1.38 A metal plate welded to an I-beam is sequrely fastened to the foundation wall using four bolts of 1/2 in diameter as shown Figure P1.36. The allowable normal stress in the bolts is 15 ksi and the allowable shear stress is 12 ksi. Assume the load is equally distibuted between the four bolts. Determine the maxi-mum load P to the nearest pound the beam can support. Solution

------------------------------------------------------------The FBD is:


1

2


3

4


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P60o

2N

2N

2V

2V

4N P 60sin= or N P 60sin4

-----------------© ¹§ · 0.2165P= =

4V P 60cos= or V P 60cos4

------------------© ¹§ · 0.125P= =

VavN

Abolt------------- 0.2165P

S 4e� � 1 2e� �2--------------------------------- 15 103� �psid= = or P 13603.9 lbsd

WavV

Abolt------------- 0.125P

S 4e� � 1 2e� �2--------------------------------- 12 103� �psid= = or P 18849.6 lbsd

Pmax 13603 lbs=




------------------------------------------------------------The FBD is:


1

2


3

4


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P60o

2N

2N

2V

2V


-----------------© ¹§ · 0.2165P= =


------------------© ¹§ · 0.125P= =

VavN

Abolt------------- 0.2165P

S 4e� � 1 2e� �2--------------------------------- 15 103� �psid= = or P 13603.9 lbsd

WavV

Abolt------------- 0.125P

S 4e� � 1 2e� �2--------------------------------- 12 103� �psid= = or P 18849.6 lbsd

Pmax 13603 lbs=




------------------------------------------------------------The FBD is:


1

2


3

4


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P60o

2N

2N

2V

2V


-----------------© ¹§ · 0.2165P= =


------------------© ¹§ · 0.125P= =

VavN

Abolt------------- 0.2165P

S 4e� � 1 2e� �2--------------------------------- 15 103� �psid= = or P 13603.9 lbsd

WavV

Abolt------------- 0.125P

S 4e� � 1 2e� �2--------------------------------- 12 103� �psid= = or P 18849.6 lbsd

Pmax 13603 lbs=




------------------------------------------------------------The FBD is:


1

2


3

4


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P60o

2N

2N

2V

2V


-----------------© ¹§ · 0.2165P= =


------------------© ¹§ · 0.125P= =

VavN

Abolt------------- 0.2165P

S 4e� � 1 2e� �2--------------------------------- 15 103� �psid= = or P 13603.9 lbsd

WavV

Abolt------------- 0.125P

S 4e� � 1 2e� �2--------------------------------- 12 103� �psid= = or P 18849.6 lbsd

Pmax 13603 lbs=

We draw the FBD below

By equilibrium

From geometry the cross-sectional area is:

The stresses are:

1.53 Fig. P1.53 shows a truss and the sequence of assembly of members at pin H, G, and C. All members of the truss have a cross-sectional area of 250 mm2 and all pins have a diameter of 15 mm. Determine (a) the axial stresses in members FG, FC, FD, and FE (b) the maximum shear stress in pin F



1.47 A butt joint is created by welding two plates to transmits a force of P = 250 kN as shown in Fig-ure P1.47. Determine the average normal and shear stress on the plane AA of the weld.

Solution------------------------------------------------------------

We draw the FBD below:

By equilbrium 1


2

The stresses are

3

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P P

900 mm

50 mm 200 mm60o 60o

Fig. P1.47

A

A

60o

N

V

P60o

h

N P 60sin 216.5 kN= = V P 60cos 125 kN= =

AAA h 0.2� � 50 10 3–� �60sin

----------------------© ¹§ · 0.2� � 57.735 10 3–� �> @ 0.2� � 11.547 10 3–� �m2= = = =

VAAN

AAA----------- 216.5 103� �

11.547 10 3–� �-------------------------------- 18.75 106� �N m2e= = = WAA

VAAA----------- 125

11.547 10 3–� �-------------------------------- 10.83 106� �N m2e= = =

VAA 18.75 MPa T� �= WAA 10.83 MPa=




Solution------------------------------------------------------------


By equilbrium 1


2

The stresses are

3

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P P

900 mm

50 mm 200 mm60o 60o

Fig. P1.47

A

A

60o

N

V

P60o

h

N P 60sin 216.5 kN= = V P 60cos 125 kN= =

AAA h 0.2� � 50 10 3–� �60sin

----------------------© ¹§ · 0.2� � 57.735 10 3–� �> @ 0.2� � 11.547 10 3–� �m2= = = =

VAAN

AAA----------- 216.5 103� �

11.547 10 3–� �-------------------------------- 18.75 106� �N m2e= = = WAA

VAAA----------- 125

11.547 10 3–� �-------------------------------- 10.83 106� �N m2e= = =

VAA 18.75 MPa T� �= WAA 10.83 MPa=




Solution------------------------------------------------------------


By equilbrium 1


2

The stresses are

3

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P P

900 mm

50 mm 200 mm60o 60o

Fig. P1.47

A

A

60o

N

V

P60o

h

N P 60sin 216.5 kN= = V P 60cos 125 kN= =

AAA h 0.2� � 50 10 3–� �60sin

----------------------© ¹§ · 0.2� � 57.735 10 3–� �> @ 0.2� � 11.547 10 3–� �m2= = = =

VAAN

AAA----------- 216.5 103� �

11.547 10 3–� �-------------------------------- 18.75 106� �N m2e= = = WAA

VAAA----------- 125

11.547 10 3–� �-------------------------------- 10.83 106� �N m2e= = =

VAA 18.75 MPa T� �= WAA 10.83 MPa=




Solution------------------------------------------------------------


By equilbrium 1


2

The stresses are

3

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P P

900 mm

50 mm 200 mm60o 60o

Fig. P1.47

A

A

60o

N

V

P60o

h

N P 60sin 216.5 kN= = V P 60cos 125 kN= =

AAA h 0.2� � 50 10 3–� �60sin

----------------------© ¹§ · 0.2� � 57.735 10 3–� �> @ 0.2� � 11.547 10 3–� �m2= = = =

VAAN

AAA----------- 216.5 103� �

11.547 10 3–� �-------------------------------- 18.75 106� �N m2e= = = WAA

VAAA----------- 125

11.547 10 3–� �-------------------------------- 10.83 106� �N m2e= = =

VAA 18.75 MPa T� �= WAA 10.83 MPa=




Solution------------------------------------------------------------


By equilbrium 1


2

The stresses are

3

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P P

900 mm

50 mm 200 mm60o 60o

Fig. P1.47

A

A

60o

N

V

P60o

h

N P 60sin 216.5 kN= = V P 60cos 125 kN= =

AAA h 0.2� � 50 10 3–� �60sin

----------------------© ¹§ · 0.2� � 57.735 10 3–� �> @ 0.2� � 11.547 10 3–� �m2= = = =

VAAN

AAA----------- 216.5 103� �

11.547 10 3–� �-------------------------------- 18.75 106� �N m2e= = = WAA

VAAA----------- 125

11.547 10 3–� �-------------------------------- 10.83 106� �N m2e= = =

VAA 18.75 MPa T� �= WAA 10.83 MPa=




Solution------------------------------------------------------------


By equilbrium 1


2

The stresses are

3

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

P P

900 mm

50 mm 200 mm60o 60o

Fig. P1.47

A

A

60o

N

V

P60o

h

N P 60sin 216.5 kN= = V P 60cos 125 kN= =

AAA h 0.2� � 50 10 3–� �60sin

----------------------© ¹§ · 0.2� � 57.735 10 3–� �> @ 0.2� � 11.547 10 3–� �m2= = = =

VAAN

AAA----------- 216.5 103� �

11.547 10 3–� �-------------------------------- 18.75 106� �N m2e= = = WAA

VAAA----------- 125

11.547 10 3–� �-------------------------------- 10.83 106� �N m2e= = =

VAA 18.75 MPa T� �= WAA 10.83 MPa=

The free body diagram of the entire truss can be drawn as shown below in Fig. (a). By equilibrium of moment about point E, we obtain: Ay(12) - 4(9) - 2(6) - 3(3) = 0 or Ay = 4.75 kN

By equilibrium of forces in the y-direction, we obtain:

Ay + Ey - 4 - 2 - 3 = 0 or Ey = 4.25 kN

By equilibrium of forces in the x-direction, we obtain:

Ex = 0

We can make an imaginary cut through members FG, FC, and DC to obtain the free body diagram shown in Fig. (b).

By equilibrium of moment about point C in Fig. (b), we obtain:

NFG sin30)(6) + (4.25)(6) - 3(3) = 0 or NFG = -5.5 kN

By equilibrium of moment about point E, we obtain:



1.53 Fig. P1.53 shows a truss and the sequence of assembly of members at pin H, G, and C. All mem-bers of the truss have a cross-sectional area of 250 mm2 and all pins have a diameter of 15 mm. Determine (a) the axial stresses in members FG, FC, FD, and FE (b) the maximum shear stress in pin F.

Solution Am = 250 mm2 dp = 15 mm VFG = ? VFC = ? VFD = ? VFE = ? (WF)max = ?------------------------------------------------------------

The free body diagram of the entire truss can be drawn as shown below in Fig. (a) .By equilibrium of moment about point E, we obtain: Ay(12) - 4(9) - 2(6) - 3(3) = 0 or

1

By equilibrium of forces in the y-direction, we obtain: Ay + Ey - 4 - 2 - 3 = 0 or

2By equilbrium of forces in the x-direction, we obtain:

3We can make an imaginary cut through members FG, FC, and DC to obtain the free body diagram shown in Fig. (b).

By equilibrium of moment about point C in Fig. (b), we obtain: NFG sin30)(6) + (4.25)(6) - 3(3) = 0 or

4By equilibrium of moment about point E, we obtain:NFC sin30)(6) + 3(3) = 0

5Fig. (c) is the free body diagram of joint D. By equilibrium of forces in the y-direction we obtain:

6Fig. (d) represents the free body diagram of joint F. By equilibrium of forces in the x-direction we obtain:

4 kN 2 kN 3 kN

300 300

3 mA B C D E

F

G

H

3 m 3 m 3 m

HAHB

HGHC

Pin H

GHGC

GF

Pin G

FGFC

FDFE

Pin F

Fig. P1.53

Ay = 4.75 kN

4 kN 2 kN 3 kN

300 300

3 mB C D

Ey

F

G

H

3 m 3 m 3 m

Ex

Ay

(a)

Ey = 4.25 kN

Ex = 0

60o60o30o

30o

NFGNFDNFC

NFGNDFNDE

3 kN

NDC

3 kN

300

D

Ey

F

3 m

Ex

C

3 m

NFG

NFCNDC

(b) (c) (d)

NFG = -5.5 kN

NFC = -3 kN

NDF = 3 kN






1








4 kN 2 kN 3 kN

300 300

3 mA B C D E

F

G

H

3 m 3 m 3 m

HAHB

HGHC

Pin H

GHGC

GF

Pin G

FGFC

FDFE

Pin F

Fig. P1.53

Ay = 4.75 kN

4 kN 2 kN 3 kN

300 300

3 mB C D

Ey

F

G

H

3 m 3 m 3 m

Ex

Ay

(a)

Ey = 4.25 kN

Ex = 0

60o60o30o

30o

NFGNFDNFC

NFGNDFNDE

3 kN

NDC

3 kN

300

D

Ey

F

3 m

Ex

C

3 m

NFG

NFCNDC

(b) (c) (d)

NFG = -5.5 kN

NFC = -3 kN

NDF = 3 kN






1








4 kN 2 kN 3 kN

300 300

3 mA B C D E

F

G

H

3 m 3 m 3 m

HAHB

HGHC

Pin H

GHGC

GF

Pin G

FGFC

FDFE

Pin F

Fig. P1.53

Ay = 4.75 kN

4 kN 2 kN 3 kN

300 300

3 mB C D

Ey

F

G

H

3 m 3 m 3 m

Ex

Ay

(a)

Ey = 4.25 kN

Ex = 0

60o60o30o

30o

NFGNFDNFC

NFGNDFNDE

3 kN

NDC

3 kN

300

D

Ey

F

3 m

Ex

C

3 m

NFG

NFCNDC

(b) (c) (d)

NFG = -5.5 kN

NFC = -3 kN

NDF = 3 kN

NFC sin30)(6) + 3(3) = 0 NFC = -3 kN

Fig. (c) is the free body diagram of joint D. By equilibrium of forces in the y-direction we obtain:

NDF =3kN

Fig. (d) represents the free body diagram of joint F. By equilibrium of forces in the x-direction we obtain:

NFE sin60 - NFG cos30 - NFC cos30 = 0 or� NFE = -8.5 kN

The axial stresses in each member can be found as shown below.

σFG = NFG / A = -5.5(103) / 250(10-6) = -22(106) N/m2

σFC = NFC / A = -3(103) / 250(10-6) = -12(106) N/m2

σFD = NFD / A = 3(103) / 250(10-6) = 12(106) N/m2

σFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2

The answers are:

The free body diagram of pin F is shown in Fig. (e). By making imaginary cuts at different sections of the pin the free body diagrams shown in Figs. (f), (g), and (h) are obtained.

From Fig. (f)

From Fig. (g)



NFE sin60 - NFG cos30 - NFC cos30 = 0 or

7The axial stresses in each member can be found as shown below.VFG = NFG / A = -5.5(103) / 250(10-6) = -22(106) N/m2

VFC = NFC / A = -3(103) / 250(10-6) = -12(106) N/m2 �VFD = NFD / A = 3(103) / 250(10-6) = 12(106) N/m2

VFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2

The answers are: ; ; ; The free body diagram of pin F is shown in Fig. (e). By making imaginary cuts at different sections of the pin the free body diagrams shown in Figs. (f), (g), and (h) are obtained.

From Fig. (f) V1 = 5.5 kN 8From Fig. (g) V2x - 8.5 sin60 = 0 V2x = 7.36 kN

-V2y - 3 + 8.5 cos60 = 0 V2y = 1.25 kN

9

From Fig (h) V3 = 8.5 kN 10From equations (8), (9) and (10), we see that the maximum shear force is 8.5kN, thus the maximum sheer stress will be between members FD and FE.

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

NFE = -8.5 kN

VFG 22 MPa (C)= VFC 12 MPa (C)= VFD 12 MPa (T)= VFE 34 MPa (C)=

60o

FGFC

FD

FE

Pin F

60o5.5 kN

3kN

8.5 kN60o

(e) (f) (g) (h)3kN

60o

FG

5.5 kN

V1

FD

FE

3kN

8.5 kN60o

V2x

V2y

FE

8.5 kN60o

V3

V2 V2x2 V2y

2+ 8.50kN= =

WF� �max

V3Apin----------- 8.5 103� �

S4--- 0.015·� �

2------------------------ 48.1 106� �= = = WF� �max 48.1 MPa=






VFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2



-V2y - 3 + 8.5 cos60 = 0 V2y = 1.25 kN

9


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

NFE = -8.5 kN


60o

FGFC

FD

FE

Pin F

60o5.5 kN

3kN

8.5 kN60o

(e) (f) (g) (h)3kN

60o

FG

5.5 kN

V1

FD

FE

3kN

8.5 kN60o

V2x

V2y

FE

8.5 kN60o

V3

V2 V2x2 V2y

2+ 8.50kN= =

WF� �max

V3Apin----------- 8.5 103� �

S4--- 0.015·� �

2------------------------ 48.1 106� �= = = WF� �max 48.1 MPa=






VFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2



-V2y - 3 + 8.5 cos60 = 0 V2y = 1.25 kN

9


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

NFE = -8.5 kN


60o

FGFC

FD

FE

Pin F

60o5.5 kN

3kN

8.5 kN60o

(e) (f) (g) (h)3kN

60o

FG

5.5 kN

V1

FD

FE

3kN

8.5 kN60o

V2x

V2y

FE

8.5 kN60o

V3

V2 V2x2 V2y

2+ 8.50kN= =

WF� �max

V3Apin----------- 8.5 103� �

S4--- 0.015·� �

2------------------------ 48.1 106� �= = = WF� �max 48.1 MPa=






VFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2



-V2y - 3 + 8.5 cos60 = 0 V2y = 1.25 kN

9


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

NFE = -8.5 kN


60o

FGFC

FD

FE

Pin F

60o5.5 kN

3kN

8.5 kN60o

(e) (f) (g) (h)3kN

60o

FG

5.5 kN

V1

FD

FE

3kN

8.5 kN60o

V2x

V2y

FE

8.5 kN60o

V3

V2 V2x2 V2y

2+ 8.50kN= =

WF� �max

V3Apin----------- 8.5 103� �

S4--- 0.015·� �

2------------------------ 48.1 106� �= = = WF� �max 48.1 MPa=

From Fig. (h)

From equations (8), (9) and (10), we see that the maximum shear force is 8.5kN, thus the maximum sheer tress will be between members FD and FE.

1.56 All pins shown are in single shear and have a diameter of 40 mm. All members have square cross- sections. Determine the maximum shear stress in the pins and the axial stress in member BD.

The free body diagram of the entire structure can be drawn as shown in Fig. (a). By equilibrium of moment about point A we obtain:

Ex(3) - 250(2.5) = 0 or Ex = 208.33 kN

By equilibrium of force in the x-direction we obtain:

Ax = Ex = 208.33 kN or Ax = 208.33 kN

By equilibrium of force in the y-direction we obtain:






VFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2



-V2y - 3 + 8.5 cos60 = 0 V2y = 1.25 kN

9


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

NFE = -8.5 kN


60o

FGFC

FD

FE

Pin F

60o5.5 kN

3kN

8.5 kN60o

(e) (f) (g) (h)3kN

60o

FG

5.5 kN

V1

FD

FE

3kN

8.5 kN60o

V2x

V2y

FE

8.5 kN60o

V3

V2 V2x2 V2y

2+ 8.50kN= =

WF� �max

V3Apin----------- 8.5 103� �

S4--- 0.015·� �

2------------------------ 48.1 106� �= = = WF� �max 48.1 MPa=






VFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2



-V2y - 3 + 8.5 cos60 = 0 V2y = 1.25 kN

9


--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

NFE = -8.5 kN


60o

FGFC

FD

FE

Pin F

60o5.5 kN

3kN

8.5 kN60o

(e) (f) (g) (h)3kN

60o

FG

5.5 kN

V1

FD

FE

3kN

8.5 kN60o

V2x

V2y

FE

8.5 kN60o

V3

V2 V2x2 V2y

2+ 8.50kN= =

WF� �max

V3Apin----------- 8.5 103� �

S4--- 0.015·� �

2------------------------ 48.1 106� �= = = WF� �max 48.1 MPa=



1.56 All pins shown are in single shear and have a diameter of 40 mm. All members have square cross-sections. Determine the maximum shear stress in the pins and the axial stress in member BD.

Fig. P1.56 Solution d = 40mm VBD = ? Wmax = ?

------------------------------------------------------------The free body diagram of the entire structure can be drawn as shown in Fig. (a). By equilibrium of moment about point A we obtain: Ex(3) - 250(2.5) = 0 or Ex = 208.33 kN

By equilibrium of force in the x-direction we obtain:Ax = Ex = 208.33 kN or Ax = 208.33 kNBy equilibrium of force in the y-direction we obtain:Ay = 250 kN The free body diagram of member CE can be drawn as shown in Fig. (b)By equilibrium of moment about point C we obtain: NBD(2.5) - 208.33(3) = 0 or NBD = 250 kN

The axial stress in member BD is: or

By equilibrium of force in the x-direction in Fig. (b) we obtain: Cx = Ex = 208.33 kN or Cx = 208.33 kNBy equilibrium of force in the x-direction in Fig. (b) we obtain��Cy = NBD = 250 kN or Cy = 250 kN

The resultant force at A:

The resultant force at The force on the pin B and D will equal NBD. Thus, pins A & C will have maximum shear.

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

50 kN/m

AB C

D

E

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

B C

D

E

250 kNAy

Ax

Ex

2.5 m 2.5 m

D

EEx

Cy

CxNBD(a) (b)

VBD250 103� �

0.05� � 0.05� �------------------------------- 100 106� � N m2e= = VBD 100MPa T� �=

FA Ax2 Ay

2+ 325kN= =

FC Cx2 Cy

2+ 325kN= =

Wmax325 103� �

S 4e� � 0.04� �2--------------------------------- 258.6 106� � N m2e= = Wmax 259MPa=











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

50 kN/m

AB C

D

E

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

B C

D

E

250 kNAy

Ax

Ex

2.5 m 2.5 m

D

EEx

Cy

CxNBD(a) (b)

VBD250 103� �

0.05� � 0.05� �------------------------------- 100 106� � N m2e= = VBD 100MPa T� �=

FA Ax2 Ay

2+ 325kN= =

FC Cx2 Cy

2+ 325kN= =

Wmax325 103� �

S 4e� � 0.04� �2--------------------------------- 258.6 106� � N m2e= = Wmax 259MPa=

Ay = 250 kN

The free body diagram of member CE can be drawn as shown in Fig. (b)�

By equilibrium of moment about point C we obtain:

NBD(2.5) - 208.33(3) = 0 or NBD = 250 kN

The axial stress in member BD is:

By equilibrium of force in the x-direction in Fig. (b) we obtain:

Cx = Ex = 208.33 kN or Cx = 208.33 kN

By equilibrium of force in the x-direction in Fig. (b) we obtain:

Cy = NBD = 250 kN or Cy = 250 kN


FA = sqrt(A2x + Ay2) = 325kN

The resultant force at FC = sqrt(C2x + Cy2) = 325kN�

The force on the pin B and D will equal NBD.

Thus, pins A & C will have maximum shear.

1.78 Two cast iron pipes are adhesively bonded together over a length of 200 mm. The outer diameters of the two pipes are 50 mm and 70 mm and wall thickness of each pipe is 10 mm. The two pipes separated while transmitting a torque of 2 kN-m. What was the shear stress in the adhesive just before the two pipes separated?











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

50 kN/m

AB C

D

E

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

B C

D

E

250 kNAy

Ax

Ex

2.5 m 2.5 m

D

EEx

Cy

CxNBD(a) (b)

VBD250 103� �

0.05� � 0.05� �------------------------------- 100 106� � N m2e= = VBD 100MPa T� �=

FA Ax2 Ay

2+ 325kN= =

FC Cx2 Cy

2+ 325kN= =

Wmax325 103� �

S 4e� � 0.04� �2--------------------------------- 258.6 106� � N m2e= = Wmax 259MPa=











--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

50 kN/m

AB C

D

E

3 m

2.5 m 2.5 m

150 mm50 mm

200 mm

B C

D

E

250 kNAy

Ax

Ex

2.5 m 2.5 m

D

EEx

Cy

CxNBD(a) (b)

VBD250 103� �

0.05� � 0.05� �------------------------------- 100 106� � N m2e= = VBD 100MPa T� �=

FA Ax2 Ay

2+ 325kN= =

FC Cx2 Cy

2+ 325kN= =

Wmax325 103� �

S 4e� � 0.04� �2--------------------------------- 258.6 106� � N m2e= = Wmax 259MPa=

The free body diagram of the smaller pipe and the differential circular surface area over is shown below.

By moment equilibrium about the axis of the pipe we have the following.

1.81 It is proposed to use 1/2 inch bolts in a 10 inch diameter coupling for transferring a torque of�100 in-kips from a 4 inch diameter shaft onto another. The maximum average shear stress in the bolts is to be limited to 20 ksi. How many bolts and at what radius should the bolts be placed on the coupling?

The free body diagram of the coupling can be drawn after making an imaginary cut through the bolts as shown below. By equilibrium of moment about the shaft axis we obtain:




Fig. P1.78 Solution (do)small = 50 mm (di)small = 30 mm (do)big = 70 mm (di)big = 50 mm

T = 2 kN-m L = 200 mm W�= ?------------------------------------------------------------



1

2

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

T

T

W

L

(do)small = 2Ro

T

dx

dA = (RodT)dx dT

dV = W dA

T Ro VdA³ RoW Ad

A³ RoW Ro Td� � xd

0

2S

³0

L

³ WRo2 Td

0

2S

³ xd

0

L

³ WRo2 2S> @ L� �= = = = =

W T

2SRo2L

----------------- 2 103� �

2S 0.025� �2 0.2� �---------------------------------------- 2.546 106� �N m2e= = =

W 2.55MPa=





T = 2 kN-m L = 200 mm W�= ?------------------------------------------------------------



1

2

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

T

T

W

L

(do)small = 2Ro

T

dx

dA = (RodT)dx dT

dV = W dA

T Ro VdA³ RoW Ad


0

2S

³0

L

³ WRo2 Td

0

2S

³ xd

0

L

³ WRo2 2S> @ L� �= = = = =

W T

2SRo2L

----------------- 2 103� �

2S 0.025� �2 0.2� �---------------------------------------- 2.546 106� �N m2e= = =

W 2.55MPa=





T = 2 kN-m L = 200 mm W�= ?------------------------------------------------------------



1

2

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

T

T

W

L

(do)small = 2Ro

T

dx

dA = (RodT)dx dT

dV = W dA

T Ro VdA³ RoW Ad


0

2S

³0

L

³ WRo2 Td

0

2S

³ xd

0

L

³ WRo2 2S> @ L� �= = = = =

W T

2SRo2L

----------------- 2 103� �

2S 0.025� �2 0.2� �---------------------------------------- 2.546 106� �N m2e= = =

W 2.55MPa=



1.81 It is proposed to use 1/2 inch bolts in a 10 inch diameter coupling for transferring a torque of 100 in-kips from a 4 inch diameter shaft onto another. The maximum average shear stress in the bolts is to be limited to 20 ksi. How many bolts and at what radius should the bolts be placed on the coupling?

Fig. P1.81 Solution dbolt = 1/2 inch T = 100 in-kips n = number of bolts = ? r = ?

------------------------------------------------------------The free body diagram of the coupling can be drawn after making an imaginary cut through the bolts as shown below. By equilibrium of moment about the shaft axis we obtain:T = nVr = 100(103) or V = 100(103)/(nr)The shear stress in the bolt can be written as given below.

1

1As per Eq. 1 there are several for n and r. The minimum value of r is rmin = 2+0.25=2.25 corresponding to n = 11.3 i.e., n = 12.The maximum value of r is rmax = 5-0.25=4.75 corresponding to n = 5.36 i.e., n = 6.the table below gives all possible answers between n = 6 to n = 12.

For minimizing machining and assembly cost 6 bolts should be used.--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

n (number of bolts) 6 7 8 9 10 11r in inches 4.25 3.64 3.18 2.83 2.55 2.31

TT

Wbolt 20ksid

WboltVA---- 100 103� �

nr� � Sd2 4e� �------------------------------- 20 103� �d= = or nr 5

S0.52 4e� �-------------------------t or nr 25.465t

T

V

V

VV

V

V r

T = nVr = 100(103) or V = 100(103)/(nr)

�The shear stress in the bolt can be written as given below.

As per Eq. 1 there are several for n and r.� The minimum value of r is rmin = 2+0.25=2.25 corresponding to n = 11.3 i.e., n = 12.

The maximum value of r is rmax = 5-0.25=4.75 corresponding to n = 5.36 i.e., n = 6. The table below gives all possible answers between n = 6 to n = 12.

For minimizing machining and assembly cost 6 bolts should be used.

1.86 Show the stress components of a point in plane stress on the square






1




TT

Wbolt 20ksid

WboltVA---- 100 103� �

nr� � Sd2 4e� �------------------------------- 20 103� �d= = or nr 5

S0.52 4e� �-------------------------t or nr 25.465t

T

V

V

VV

V

V r






1




TT

Wbolt 20ksid

WboltVA---- 100 103� �

nr� � Sd2 4e� �------------------------------- 20 103� �d= = or nr 5

S0.52 4e� �-------------------------t or nr 25.465t

T

V

V

VV

V

V r






1




TT

Wbolt 20ksid

WboltVA---- 100 103� �

nr� � Sd2 4e� �------------------------------- 20 103� �d= = or nr 5

S0.52 4e� �-------------------------t or nr 25.465t

T

V

V

VV

V

V r



1.86 Show the stress components of a point in plane stress on.the square

Fig. P1.86

Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vxx 27ksi C� �= Wxy 18ksi=

Wyx 18ksi= Vyy 85ksi T� �=

x

y

x

y85

27

1818

2718

18

85

1.97 Show the non-zero stress components on the A, B, and C faces of the cube.

1.99 Show the non-zero stress components in the r, Ư̆, and x cylindrical coordinate system on the A, B, and C faces of the stress element shown.



1.86 Show the stress components of a point in plane stress on.the square

Fig. P1.86

Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vxx 27ksi C� �= Wxy 18ksi=

Wyx 18ksi= Vyy 85ksi T� �=

x

y

x

y85

27

1818

2718

18

85M. Vable Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1 January 2013


1.97 Show the non-zero stress components on the A,B, and C faces of the cube.

Fig. P1.97

Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vxx 0= Wxy 15– ksi= Wxz 0=

Wyx 15– ksi= Vyy 10ksi C� �= Wyz 25ksi=

Wzx 0= Wzy 25ksi= Vzz 20ksi T� �=

xz

y.A

.B

.C

xz

y10

25 15

20

25

15



1.97 Show the non-zero stress components on the A,B, and C faces of the cube.

Fig. P1.97

Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vxx 0= Wxy 15– ksi= Wxz 0=

Wyx 15– ksi= Vyy 10ksi C� �= Wyz 25ksi=

Wzx 0= Wzy 25ksi= Vzz 20ksi T� �=

xz

y.A

.B

.C

xz

y10

25 15

20

25

15



1.99 Show the non-zero stress components in the r, T��and x�cylindrical coordinate system on the A,B, and C faces of the stress element shown

Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Fig. P1.99

Vrr 10ksi C� �= WrT 22ksi= Wrx 32ksi=

WTr 22ksi= VTT 0= WTx 25ksi=

Wxr 32ksi= WxT 25ksi= Vxx 20ksi T� �= xT

r

BA

C

r

x

T

20

3225

32

10

22

22

25




Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Fig. P1.99




r

BA

C

r

x

T

20

3225

32

10

22

22

25




Solution------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Fig. P1.99




r

BA

C

r

x

T

20

3225

32

10

22

22

25

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MEEM 2150 MTU Vable Mechanics of Materials Solutions CH1