## Strength of Materials |

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Page 67

of the two

assemhlv. Solution: This problem is statically indeterminate in that we do not

know how the applied torque is apportioned to each

follow ...

of the two

**segments**. Compute the maximum shearing stress developed in tlieassemhlv. Solution: This problem is statically indeterminate in that we do not

know how the applied torque is apportioned to each

**segment**. The procedure wefollow ...

Page 91

Assume that a cutting plane a-a at a distance x from Ri divides the beam into two

the externally applied load is R i . To maintain equilibrium in this

Assume that a cutting plane a-a at a distance x from Ri divides the beam into two

**segments**. The free-body diagram of the left**segment**in Fig. 4— 3b shows thatthe externally applied load is R i . To maintain equilibrium in this

**segment**of the ...Page 104

Applying these forces to a free - body diagram of a beam

equilibrium of that

right of section b - b in Fig . 4 - 17 are held in equilibrium by the shear Pi P a - w

lb / ft Ty ...

Applying these forces to a free - body diagram of a beam

**segment**producesequilibrium of that

**segment**. Thus in Fig . 4 - 18 , the**segments**to the left andright of section b - b in Fig . 4 - 17 are held in equilibrium by the shear Pi P a - w

lb / ft Ty ...

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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