## Strength of materials |

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Page 371

1002 carries a uniformly distributed load of 2000 lb/ft on a simply supported span

16 ft long. If E. = 30 X 106 psi and E„ = 1.5 X 106 psi, compute the midspan

deflection. Ans. S = 0.394 in. 1021. In Prob. 1016, determine the shear flow

developed between the steel and wood, and between the wood and aluminum.

Express the results as a function of the vertical shear V. Ans. 0.1297; 0.124V 10-4

. Reinforced

it is cheap ...

1002 carries a uniformly distributed load of 2000 lb/ft on a simply supported span

16 ft long. If E. = 30 X 106 psi and E„ = 1.5 X 106 psi, compute the midspan

deflection. Ans. S = 0.394 in. 1021. In Prob. 1016, determine the shear flow

developed between the steel and wood, and between the wood and aluminum.

Express the results as a function of the vertical shear V. Ans. 0.1297; 0.124V 10-4

. Reinforced

**Concrete**Beams**Concrete**is an excellent building material becauseit is cheap ...

Page 375

To stress the

iUbkd)(jd)] Me = i(600)(10)(6.27)(12.91) = 243,000 in.-lb To stress the steel to its

limit, the required bending moment is [M, = f.A.jd] M, = 16,000(1.5)(12.9l) =

310,000 in.-lb The safe bending moment is therefore 243,000 in.-lb. Since the

reinforced. PROBLEMS 1024. In a reinforced

A, = 1 .5 sq ...

To stress the

**concrete**to its maximum limit will require a bending moment [M, =iUbkd)(jd)] Me = i(600)(10)(6.27)(12.91) = 243,000 in.-lb To stress the steel to its

limit, the required bending moment is [M, = f.A.jd] M, = 16,000(1.5)(12.9l) =

310,000 in.-lb The safe bending moment is therefore 243,000 in.-lb. Since the

**concrete**governs, we conclude there is too much steel; hence the beam is over-reinforced. PROBLEMS 1024. In a reinforced

**concrete**beam, 6 = 10in.,d = 16 in.,A, = 1 .5 sq ...

Page 376

The dimensions of a reinforced

2sqin., and n = 12. If the allowable stresses are fc ^ 800 psi and /, ^ 18,000 psi,

determine the maximum bending moment that may be applied. In what state of

reinforcement is the beam? Ans. M = 507,000 in.-lb; over-reinforced 1030. In a

reinforced beam, 6 = 10 in., d = 18 in., A, = 2 sq in., and n = 12. Determine the

safe uniformly distributed load that can be carried on a simply supported span 12

ft long if fc ...

The dimensions of a reinforced

**concrete**beam are b = 12 in., d = 18 in., A, =2sqin., and n = 12. If the allowable stresses are fc ^ 800 psi and /, ^ 18,000 psi,

determine the maximum bending moment that may be applied. In what state of

reinforcement is the beam? Ans. M = 507,000 in.-lb; over-reinforced 1030. In a

reinforced beam, 6 = 10 in., d = 18 in., A, = 2 sq in., and n = 12. Determine the

safe uniformly distributed load that can be carried on a simply supported span 12

ft long if fc ...

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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