## Strength of Materials |

### From inside the book

Results 1-3 of 48

Page 152

However, for materials relatively weak in tension and strong in compression,

such as cast iron, it is desirable to use beams that are unsymmetrical with respect

to the

a ...

However, for materials relatively weak in tension and strong in compression,

such as cast iron, it is desirable to use beams that are unsymmetrical with respect

to the

**neutral axis**. With such a cross section, the stronger fibers can be located ata ...

Page 488

13-28, in which a symmetrical section is subjected to loads inclined to the axes of

symmetry. Resolving the loading into ... In (b), the X axis is the

whereas in (c) the Y axis becomes the

13-28, in which a symmetrical section is subjected to loads inclined to the axes of

symmetry. Resolving the loading into ... In (b), the X axis is the

**neutral axis**,whereas in (c) the Y axis becomes the

**neutral axis**. Each of these conditions ...Page 489

From this, we see that unless Ix = Iy or tan 6 = 0 or oo, the

perpendicular to the load. Also observe, as shown in Fig. 13-28a, that the

is ...

From this, we see that unless Ix = Iy or tan 6 = 0 or oo, the

**neutral axis**is notperpendicular to the load. Also observe, as shown in Fig. 13-28a, that the

**neutral****axis**is inclined from the X axis in the same angular sense as the plane of loadingis ...

### 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

acting actual allowable angle applied assumed axes axis beam shown bending bending moment cantilever carries caused centroid circle CN CN column compressive compressive stress compute concentrated consider constant couple cross section deflection deformation Determine developed diameter direction distance distributed distributed load effect elastic curve element equal equation equivalent expressed flange flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia lb/ft length limit load loaded as shown material maximum method midspan moments negative neutral axis obtain occurs plane plate positive Prob PROBLEMS produce reaction reference relation resisting respect restrained resultant rivet segment shear diagram shearing stress shown in Fig shows simply supported slope Solution Solve span steel strain strength supported Table tangent tensile thickness varies vertical wall weight weld yield zero