Solution manual for mechanics of materials 3rd edition by philpot

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only determine the maximum load P that the membe

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determine the maximum load P that the member can support



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

max allow (200 N/mm )(863.938 mm ) 172,788 N 172.8 kN

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

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shown 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

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

Notice that rods (1) and (2) are in compression In this situation, we are

concerned 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

2 1

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

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

F A

F A

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of 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 8 kips 4 kips 4 kips 0 1 16 kips 16 kips (C)

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

F A

F A

F A

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stress 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 the

three rods

FIGURE P1.5/6

Solution

The internal forces in the three rods must be determined Begin with a FBD cut through rod (1) that

includes 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 8 kips 4 kips 4 kips 0 1 16 kips 16 kips (C)

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

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

2 1

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normal 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:

Write equilibrium equations for the sum of forces in the

horizontal and vertical directions Note: Rods (1) and (2)

are two-force members

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs (a) and (b) Using the

substitution method, Eq (b) can be solved for F2 in terms of F1:

2 1

cos(57.9946 )cos(35.7067 )

cos(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

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

2 1

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(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:

Write equilibrium equations for the sum of forces in the horizontal

and vertical directions Note: Rods (1) and (2) are two-force

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs (a) and (b) Using the

substitution method, Eq (b) can be solved for F2 in terms of F1:

2 1

cos(57.9946 )cos(35.7067 )

cos(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

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

1 (16 mm) 201.0619 mm4

and the normal stress in rod (1) is:

2 1

1

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

109.2710 N/mm201.0619 mm 109.3 MPa (T)

F A

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

2 (12 mm) 113.0973 mm4

and the normal stress in rod (2) is:

2 2

2

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

126.7924 N/mm113.0973 mm 126.8 MPa (T)

F A

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members 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 Write

equilibrium equations that include all

external 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 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:

Before beginning the process of determining the internal forces in the axial members, the geometry of

the 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:

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

y AC

Joint B:

Next, consider a FBD of joint B In this instance, the equilibrium

equations 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

y BC

AB AB AB

F A

2

20.125 kips

7.510 ksi2.67954 in 7.51 ksi (T)

AC AC AC

F A

2

21.260 kips

7.934 ksi2.67954 in 7.93 ksi (C)

BC BC BC

F A

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members of the truss are aluminum pipes that

have an outside diameter of 60 mm and a wall

thickness of 4 mm Determine the normal

stress in each truss member

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., loads

and reaction forces) are considered at this

time The internal forces acting in the truss

members will be considered after the external

reactions have been computed The

free-body diagram (FBD) of the entire truss is

shown The following equilibrium equations

Before beginning the process of determining the internal forces in the axial members, the geometry of

the 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

1.0 m1.5 m

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

y AB

Joint C:

Next, consider a FBD of joint C In this instance, the equilibrium

equations 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

F A

F A

F A

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members of the truss are aluminum pipes that

have an outside diameter of 42 mm and a wall

thickness of 3.5 mm Determine the normal

stress in each truss member

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 all

external 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 external

reactions have been computed The free-body

diagram (FBD) of the entire truss is shown The

following equilibrium equations can be written for

Before beginning the process of determining the internal forces in the axial members, the geometry of

the 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:

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

x AC

Joint B:

Next, consider a FBD of joint B In this instance, the equilibrium

equations 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

x BC

F A

F A

F A

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areas of 175 mm and 300 mm , respectively For a

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

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

2 2

maximum load P that may be supported by

Consider a FBD of ABC From the moment equilibrium

equation about joint A, the relationship between the force in

bar (1) and the load P is:

1 1

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axial 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

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)

In the interval between B and C where a  x a b, and write

the following equilibrium equation:

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

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 at

the following locations:

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

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

Equilibrium: Draw a FBD for the interval between B and C

where a  x a b, and write the following equilibrium

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fully 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 the

splice plates for an applied load of P =

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

5,000 lb2

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

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 a 0.5-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

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:

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 to

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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

625 lb

5,658.8427 psi0.110447 in 5,660 psi

V A

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

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

available to transmit shear stress in each bolt

2

2 bolt