## Strength of materials |

### From inside the book

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Page 287

... we have, as before Vc = 193 N and MC = — 179 N-m Ans. One interesting

aspect of this double-integration

perfectly restrained at both ends, the redundants could be found by merely

changing the ...

... we have, as before Vc = 193 N and MC = — 179 N-m Ans. One interesting

aspect of this double-integration

**solution**is that, if the given beam had beenperfectly restrained at both ends, the redundants could be found by merely

changing the ...

Page 296

here by the area-moment method shows the close similarity of the two methods

and demonstrates the direct use of the moment diagram. We begin by drawing ...

**Solution**: This problem was solved on page 288 by double integration. Its**solution**here by the area-moment method shows the close similarity of the two methods

and demonstrates the direct use of the moment diagram. We begin by drawing ...

Page 355

This shortcut is applied in the second

reduce A and D to free ends by releasing them, applying the balancing moments

of - 8000 at A and +7200 at D, and carrying over half these amounts with the ...

This shortcut is applied in the second

**solution**(b) shown in Fig. 8-23. As a start,reduce A and D to free ends by releasing them, applying the balancing moments

of - 8000 at A and +7200 at D, and carrying over half these amounts with the ...

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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