## Strength of materials |

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Results 1-3 of 49

Page 143

Referring to the table of properties of WF beams (page 571) and starting at the

bottom, we find that the first beam whose section

8 WF 35, with Z = 31.1 in.3. In the 10-in. group we find a 10 WF 29 beam with Z ...

Referring to the table of properties of WF beams (page 571) and starting at the

bottom, we find that the first beam whose section

**modulus**is more than 30 in.3 is8 WF 35, with Z = 31.1 in.3. In the 10-in. group we find a 10 WF 29 beam with Z ...

Page 146

This table enables you to select rapidly the lightest shape of beam to support any

required section

lists the weight of a concrete encasement in lb/ft, and the product — for checking

...

This table enables you to select rapidly the lightest shape of beam to support any

required section

**modulus**. In addition to section**modulus**and shape, the tablelists the weight of a concrete encasement in lb/ft, and the product — for checking

...

Page 398

In one proposed method — that of the double-

formula is extended to intermediate columns stressed above the proportional limit

by replacing the constant

The ...

In one proposed method — that of the double-

**modulus**theory* — the Eulerformula is extended to intermediate columns stressed above the proportional limit

by replacing the constant

**modulus**£ by a reduced**modulus**Tj, viz., P Ev2 (11-6)The ...

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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