## Strength of materials |

### From inside the book

Results 1-3 of 69

Page 133

With this change, the maximum flexure stress in any section is given by If - is

called the section modulus and denoted by Z, another common c variation of the

flexure formula is This variation is useful for beams of constant

...

With this change, the maximum flexure stress in any section is given by If - is

called the section modulus and denoted by Z, another common c variation of the

flexure formula is This variation is useful for beams of constant

**cross section**, as it...

Page 324

This shaded area is known as the kern of the

a circular section is a circle whose diameter is one- quarter the diameter of the

section. PROBLEMS 918. A compressive load P — 12,000 lb is applied, as in Fig.

This shaded area is known as the kern of the

**cross section**. Show that the kern ofa circular section is a circle whose diameter is one- quarter the diameter of the

section. PROBLEMS 918. A compressive load P — 12,000 lb is applied, as in Fig.

Page 499

Table 13-2 lists correction factors for various

to other than pure bending, as in Fig. 13-35, the system of coplanar forces acting

in the plane of curvature is reduced to a single force R acting at the cen- troid of ...

Table 13-2 lists correction factors for various

**cross sections**. For beams subjectedto other than pure bending, as in Fig. 13-35, the system of coplanar forces acting

in the plane of curvature is reduced to a single force R acting at the cen- troid of ...

### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Other editions - View all

### Common terms and phrases

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