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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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There are still two unknowns in this equation; consequently, another equation is necessary

to obtain a solution. Let F 1 equal the force corresponding to the allowable stress s allow,1 in

member (1), and solve for the applied load P. (Note: The negative sign attached to F 1 can

be omitted here since we are interested only in the magnitude of load P.) First, we have

s ⎢ + ⎥

=

⎡⎛

⎟ ⎛ ⎝ ⎜

⎟ ⎛ ⎝ ⎜ ⎞ ⎤

A

L 1 A2

E2

2 2

1.8 2,000 70

allow,1 1

1 (160 N/mm )(3,600 mm ) ⎢

⎟ + 1

L A E

2 1 1

⎣ 1.4 3,600 200 ⎦

= (576,000 N)[1.25] = 720,000 N = 720 kN ≥ P

Next, repeat this process for member (2). Then solve Equation (e) for F 1 :

F

F L 2 A1

E1

=− (f)

L A E

1 2

Now substitute Equation (f) into Equation (a) to obtain

1

2

2

F

F L 2 A1

E1

+ = F2

1 +

L1

A2

E

2 ⎣

2 2

L

L

2

1

A

A

1

2

E1

E

2 ⎦

=

P

Finally, let F 2 equal the allowable force and solve for the corresponding applied force P:

⎡ L2

A1

E1

s ⎢ + ⎥

= 2 2

+ ⎛ ⎝ ⎜ 1.4⎞

⎟ ⎛ ⎝ ⎜

3,600 ⎞

A 1 (120 N/mm )(2,000 mm )1 ⎢

⎟ ⎛ allow,2 2

L A E

⎣ 1.8 2,000 ⎝ ⎜

1 2 2

= (240,000 N)[5.0] = 1,200,000 N = 1,200 kN ≥ P

Therefore, the maximum load P that can be applied to the flange at B is P = 720 kN.

Ans.

200

70

⎞ ⎤

⎟ ⎥

mecmovies

ExAmpLES

m5.5 A steel rod (1) is attached

to a steel post (2) at flange B. A

downward load of 110 kN is applied

to flange B. Rod (1) and post

(2) are attached to rigid supports

at A and C, respectively. Rod (1)

has a cross-sectional area of 800

mm 2 and an elastic modulus of

200 GPa. Post (2) has a crosssectional

area of 1,600 mm 2 and

an elastic modulus of 200 GPa.

(a) Compute the normal stress in

rod (1) and post (2).

(b) Compute the deflection of

flange B.

m5.6 An aluminum tube

(1) encases a brass core

(2). The two components

are bonded together to

form an axial member that

is subjected to a downward

force of 30 kN.

Tube (1) has an outer diameter

D = 30 mm and an

inner diameter d = 22 mm.

The elastic modulus of

the aluminum is 70 GPa.

The brass core (2) has a

diameter d = 22 mm and an

elastic modulus of 105 GPa. Compute the

normal stresses in tube (1) and core (2).

110

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