## Strength of materials |

### From inside the book

Results 1-3 of 52

Page 68

Since the function of this

torque T, the load must be directed perpendicular to the radius p in order to

produce the maximum effect. It is true, but difficult to prove here, that in circular ...

Since the function of this

**resisting**load dP is to produce resistance to the appliedtorque T, the load must be directed perpendicular to the radius p in order to

produce the maximum effect. It is true, but difficult to prove here, that in circular ...

Page 369

+ -^=,350 X 10W [1 = 1- Ad2] 7NA = (1350 X 106)-(60 X 1 03X HI)2 = 611 X 106

mm4 The

equivalent of the steel, the maximum stress is Therefore the

...

+ -^=,350 X 10W [1 = 1- Ad2] 7NA = (1350 X 106)-(60 X 1 03X HI)2 = 611 X 106

mm4 The

**resisting**moment in terms of the maximum wood stress is In the woodequivalent of the steel, the maximum stress is Therefore the

**resisting**moment that...

Page 498

Applying the flexure formula, we find that the

bh2 Myp = ffyp "T (O) 0 « At section b-b, the section is elastic over the depth 2yh

but plastic outside this depth, as shown by the stress distribution in part (c).

Applying the flexure formula, we find that the

**resisting**moment at this section isbh2 Myp = ffyp "T (O) 0 « At section b-b, the section is elastic over the depth 2yh

but plastic outside this depth, as shown by the stress distribution in part (c).

### What people are saying - Write a review

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

### Common terms and phrases

allowable stresses aluminum angle area-moment assumed axes axial load beam in Fig beam loaded beam shown bending bolts cantilever beam caused centroid column components compressive stress Compute the maximum concentrated load connector cross section deformations Determine the maximum diameter elastic curve element end moments equal equivalent Euler's formula factor of safety fibers Figure flange flexure formula free-body diagram Hence Hooke's law horizontal Illustrative Problem kips lb/ft length loaded as shown main plate maximum shearing stress maximum stress method midspan deflection Mohr's circle moments of inertia neutral axis obtain plane plastic positive product of inertia proportional limit radius reaction rectangular resisting resultant rivet rotation segment shaft shear center shear diagram shearing force shown in Fig slope Solution span static steel strain tensile stress thickness three-moment equation torque torsional U.S. Customary Units uniformly distributed load vertical shear weld zero