## Strength of materials |

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Page 46

A timber block 10 in. square is reinforced on each side by a

wide and t in. thick. Determine the thickness t of the

will support an axial load of 300,000 lb without exceeding a maximum timber

stress of 1200 psi or a maximum

106 psi and for

by three rods as shown in Fig. P-235. Each copper rod has an area of 1.5 sq in.,

E = 17 X ...

A timber block 10 in. square is reinforced on each side by a

**steel**plate 10 in.wide and t in. thick. Determine the thickness t of the

**steel**plates so the assemblywill support an axial load of 300,000 lb without exceeding a maximum timber

stress of 1200 psi or a maximum

**steel**stress of 20,000 psi. E for timber is 1.5 X106 psi and for

**steel**it is 30 X 106 psi. 235. A rigid block of weight W is supportedby three rods as shown in Fig. P-235. Each copper rod has an area of 1.5 sq in.,

E = 17 X ...

Page 366

neutral axis as the

deformations in Eq. (a). In other words, the equivalent wood area is n times as

wide as the

desired, an equivalent

by

directly to either the equivalent wood section or the equivalent

the ...

neutral axis as the

**steel**fibers they replace in order to satisfy the criterion of equaldeformations in Eq. (a). In other words, the equivalent wood area is n times as

wide as the

**steel**it replaces; the equivalent wood section is shown in Fig. 10-lb. Ifdesired, an equivalent

**steel**section can be set up by replacing the original woodby

**steel**- as wide, as in Fig. 10-1c. n The flexure formula can now be applieddirectly to either the equivalent wood section or the equivalent

**steel**section. Withthe ...

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

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 because

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 because

it is cheap ...

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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