## Strength of Materials |

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Page 226

to the elastic curve at midspan is horizontal and ... In such beams, the deviation at

either support from the midspan tangent is equal to the

**Midspan Deflections**In a symmetrically loaded simple beam, the tangent drawnto the elastic curve at midspan is horizontal and ... In such beams, the deviation at

either support from the midspan tangent is equal to the

**midspan deflection**.Page 234

Deflections by the Method of Superposition In a supplementary method of

determining slopes and deflections, the results ... Solution: From case 7 of Table

6-2, the

^ (3L2 ...

Deflections by the Method of Superposition In a supplementary method of

determining slopes and deflections, the results ... Solution: From case 7 of Table

6-2, the

**midspan deflection**of an eccentrically placed concentrated load is EIS =^ (3L2 ...

Page 268

Compute the

in. 749. A timber beam 6 in. wide by 12 in. deep and 20 ft long is perfectly

restrained at both ends. It supports a uniformly distributed load of 240 lb/ft over its

entire ...

Compute the

**midspan deflection**of this beam if E = 30 X 106 psi. Ans. S = 0.187in. 749. A timber beam 6 in. wide by 12 in. deep and 20 ft long is perfectly

restrained at both ends. It supports a uniformly distributed load of 240 lb/ft over its

entire ...

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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