## Strength of materials |

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Page 261

729. For the

maximum EIS. Ans. M = — 1840 ft-lb; EIS = 9120 lb-ft3 730. Determine the end

moment and maximum deflection for the perfectly

shown in Fig. P-730. 2 3 Ans. M = (3L - 2a); EIy = - ^ (L - a) Fig. P-730. Fig. P-731.

732. Determine the end moments for the

Ans. Ma : wlb/ft 2 ~ 2 Fig. P-732. Fig. P-733. 733. For the

in Fig.

729. For the

**restrained beam**shown in Fig. P-729, compute the end moment andmaximum EIS. Ans. M = — 1840 ft-lb; EIS = 9120 lb-ft3 730. Determine the end

moment and maximum deflection for the perfectly

**restrained beam**loaded asshown in Fig. P-730. 2 3 Ans. M = (3L - 2a); EIy = - ^ (L - a) Fig. P-730. Fig. P-731.

732. Determine the end moments for the

**restrained beam**shown in Fig. P-732.Ans. Ma : wlb/ft 2 ~ 2 Fig. P-732. Fig. P-733. 733. For the

**restrained beam**shownin Fig.

Page 262

The beam shown in Fig. P-735 is only partially restrained at its ends. wL3 The

slope at the left end is directed up to the left, whereas at the right 72EI end the

slope is wL3 72EI directed up to the right. Determine the end moments. 736. The

beam shown in Fig. P-735 is perfectly restrained at A but only 7 3 partially

restrained at B, where the slope is — — directed up to the right. Solve iSEI for the

end moments. 737. In the perfectly

B has 6EIA ...

The beam shown in Fig. P-735 is only partially restrained at its ends. wL3 The

slope at the left end is directed up to the left, whereas at the right 72EI end the

slope is wL3 72EI directed up to the right. Determine the end moments. 736. The

beam shown in Fig. P-735 is perfectly restrained at A but only 7 3 partially

restrained at B, where the slope is — — directed up to the right. Solve iSEI for the

end moments. 737. In the perfectly

**restrained beam**shown in Fig. P-737, supportB has 6EIA ...

Page 264

Determine the wall moment in the propped beam described in Prob. 705 (page

252). 740. Solve for the wall moment in the propped beam shown in Fig. P-740. ,

4 ns. M = —1120 ft-lb 60 lb /ft 90 lb/ ft 8' Fig. P-740. Fig. P-741. 741. Compute the

moment at the restrained end of the propped beam shown in Fig. P-741. 742.

Determine the end moments in the

253). 743. Compute the end moments in the

Determine the wall moment in the propped beam described in Prob. 705 (page

252). 740. Solve for the wall moment in the propped beam shown in Fig. P-740. ,

4 ns. M = —1120 ft-lb 60 lb /ft 90 lb/ ft 8' Fig. P-740. Fig. P-741. 741. Compute the

moment at the restrained end of the propped beam shown in Fig. P-741. 742.

Determine the end moments in the

**restrained beam**described in Prob. 710 (page253). 743. Compute the end moments in the

**restrained beam**described in Prob.### What people are saying - Write a review

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allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bolt cantilever beam caused centroid CN CN column compressive stress Compute the maximum concentrated load concrete cover plate cross section deformation Determine the maximum diameter elastic curve end moments equal equivalent Euler's formula factor of safety fibers flange flexure formula free-body diagram ft long ft-lb Hence hinged Hooke's law horizontal ILLUSTRATIVE PROBLEMS lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan midspan deflection modulus Mohr's circle moments of inertia neutral axis obtain plane plastic positive product of inertia proportional limit radius ratio reaction Repeat Prob resisting restrained beam resultant segment shaft shear center shear diagram shearing force shown in Fig Solution Solve Prob span static steel strain tensile stress thickness torque torsional uniformly distributed load vertical shear weld zero