## Strength of materials |

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Page 40

Ferdinand Leon Singer. 221. A solid

concentrically in a hollow steel tube. Compute the internal diameter of the steel

tube so that no contact pressure exists when the

compressive load of 84,800 lb. Assume p. = \ and Ea = 10 X 106 psi. Solution:

The axial compressive stress in the

Ferdinand Leon Singer. 221. A solid

**aluminum**shaft of 3-in. diameter fitsconcentrically in a hollow steel tube. Compute the internal diameter of the steel

tube so that no contact pressure exists when the

**aluminum**shaft carries an axialcompressive load of 84,800 lb. Assume p. = \ and Ea = 10 X 106 psi. Solution:

The axial compressive stress in the

**aluminum**is ...Page 48

-10"- 15"

Fig. P-244 is firmly attached to unyielding supports. Find the stress caused in

each material by applying an axial load P = 44,000 lb. Ans. Sa = 8000 psi; S. =

16,000 psi A - 2 in.' Fig. P-244 and P-245. 245. Refer to Prob. 244. What

maximum load P can be applied without exceeding an allowable stress of 10,000

psi for

length of the ...

-10"- 15"

**Aluminum**.E-l0xl06 ^ =1.5 in.2 Steel £-30xl06 244. The bar shown inFig. P-244 is firmly attached to unyielding supports. Find the stress caused in

each material by applying an axial load P = 44,000 lb. Ans. Sa = 8000 psi; S. =

16,000 psi A - 2 in.' Fig. P-244 and P-245. 245. Refer to Prob. 244. What

maximum load P can be applied without exceeding an allowable stress of 10,000

psi for

**aluminum**or 18,000 psi for steel ? Can a larger load P be carried if thelength of the ...

Page 56

Find the stress in each material at 200°F. For

106 psi, and a = 12.8 X 10-6. For bronze, A = 3 sq in., E = 12 X 106 psi, and a =

10.5 X 10-6. For each steel bolt, A = 0.75 sq in., E = 30 X 106 psi, and a = 6.5 X l0

-6. Ans. S, = 4760 psi 272. At what temperature will the steel bolts in Prob. 271 be

stressed to 18,000 psi? l"k- 3" i''h- Fig. P-271 and P-272. 15"

A - 1.5 in.2 Steel £ = 30xl06 A =2 in.2 Fig. P-273 and P-274. 273. The composite

bar ...

Find the stress in each material at 200°F. For

**aluminum**, A = 2 sq in., E = 10 X106 psi, and a = 12.8 X 10-6. For bronze, A = 3 sq in., E = 12 X 106 psi, and a =

10.5 X 10-6. For each steel bolt, A = 0.75 sq in., E = 30 X 106 psi, and a = 6.5 X l0

-6. Ans. S, = 4760 psi 272. At what temperature will the steel bolts in Prob. 271 be

stressed to 18,000 psi? l"k- 3" i''h- Fig. P-271 and P-272. 15"

**Aluminum**£-10xl06A - 1.5 in.2 Steel £ = 30xl06 A =2 in.2 Fig. P-273 and P-274. 273. The composite

bar ...

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allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bolt cantilever beam caused centroid CN CN column compressive stress Compute the maximum concentrated load concrete cover plate cross section deformation Determine the maximum diameter elastic curve end moments equal equivalent Euler's formula factor of safety fibers flange flexure formula free-body diagram ft long ft-lb Hence hinged Hooke's law horizontal ILLUSTRATIVE PROBLEMS lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan midspan deflection modulus Mohr's circle moments of inertia neutral axis obtain plane plastic positive product of inertia proportional limit radius ratio reaction Repeat Prob resisting restrained beam resultant segment shaft shear center shear diagram shearing force shown in Fig Solution Solve Prob span static steel strain tensile stress thickness torque torsional uniformly distributed load vertical shear weld zero