## Strength of materials |

### From inside the book

Results 1-3 of 38

Page 218

This difference is so negligible in comparison with possible variations in the

given data that for all practical purposes we may assume the

to be equivalent to the actual maximum deflection. Indeed, it may be shown that,

for a ...

This difference is so negligible in comparison with possible variations in the

given data that for all practical purposes we may assume the

**midspan deflection**to be equivalent to the actual maximum deflection. Indeed, it may be shown that,

for a ...

Page 224

Ans. EIS = 428 N • m3 down 900 N/m Ri R2 Figure P-670 6-7

elastic curve at midspan is horizontal and parallel to the unloaded beam. In such

beams ...

Ans. EIS = 428 N • m3 down 900 N/m Ri R2 Figure P-670 6-7

**MIDSPAN****DEFLECTIONS**In a symmetrically loaded simple beam, the tangent drawn to theelastic curve at midspan is horizontal and parallel to the unloaded beam. In such

beams ...

Page 265

gives the

SSB! PM - 4,3), + Sfflfi - 4,4), + SSI £7>> = 27 000 + 60 000 + 918 = 87 920 lb- ft3

Substituting numerical values for E and /, we then convert the right-hand side to ...

gives the

**midspan deflection**,* including the dead weight of the beam, as E,y -SSB! PM - 4,3), + Sfflfi - 4,4), + SSI £7>> = 27 000 + 60 000 + 918 = 87 920 lb- ft3

Substituting numerical values for E and /, we then convert the right-hand side to ...

### 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