## Strength of materials |

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Page 91

Equivalence of longitudinal and

obtain [ZMM = 0] (r dr r d0) dx — (r' dr dx)r d0 = 0 Canceling out the common

product r d0 dr dx, we see that I 1- = -r A pictorial view illustrating this equivalence

...

Equivalence of longitudinal and

**torsional**shearing stress. over which they act, weobtain [ZMM = 0] (r dr r d0) dx — (r' dr dx)r d0 = 0 Canceling out the common

product r d0 dr dx, we see that I 1- = -r A pictorial view illustrating this equivalence

...

Page 96

This resisting couple is created by a

cross section of the spring; it is represented by T = PR. The magnified view of the

cross section in Fig. 3-13b shows the stress distribution that created the ...

This resisting couple is created by a

**torsional**shearing stress distributed over thecross section of the spring; it is represented by T = PR. The magnified view of the

cross section in Fig. 3-13b shows the stress distribution that created the ...

Page 490

However, by adding a pair of equal, oppositely directed, and collinear forces of

magnitude P (shown dashed) at the centroid of the rivet group, the applied

eccentric load P is replaced by a central load P and the

However, by adding a pair of equal, oppositely directed, and collinear forces of

magnitude P (shown dashed) at the centroid of the rivet group, the applied

eccentric load P is replaced by a central load P and the

**torsional**couple T = Pe, ...### What people are saying - Write a review

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### Common terms and phrases

allowable stresses aluminum angle applied load area-moment method assumed axial load beam in Fig beam loaded beam shown bolt bronze cantilever beam caused centroid column compressive stress concentrated load continuous beam cross section deﬂection deformation Determine the maximum diameter elastic curve element end moments equal equilibrium equivalent exploratory section factor of safety fibers ﬂexural flexural stress Flgure free-body diagram Hence horizontal ILLUSTRATIVE PROBLEMS kN-m kN/m length loaded as shown maximum shearing stress maximum stress midspan midspan deﬂection Mohr’s circle neutral axis normal stress obtain plane positive principal stresses proportional limit radius reaction resisting restrained beam resultant rivet segment shaft shear center shear diagram shearing force shearing strain shown in Fig simply supported beam slope Solution span static equilibrium statically indeterminate steel strain tangent drawn tensile stress three-moment equation torque torsional uniformly distributed load value of E18 vertical shear weld