Strength of materials |
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Page 262
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, support B has 6EIA settled a
distance A ...
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, support B has 6EIA settled a
distance A ...
Page 263
u; lb/ft v w lb/ft (b) Load simply supported (a) Restrained beam (c) End couple
loading Fig. 7-1 1. — Restrained beam resolved into simple beam loadings.
these end slopes by adding the loading in Fig. 7-1 lc, we must have 6i = 6[ and
02 = 02, which requires that Mb be greater than Ma in order for 62 to be greater
than 6{. In other words, the larger end moment acts at the wall that is closer to the
resultant of any single load. The difference between the end moments Mb and
Ma is ...
u; lb/ft v w lb/ft (b) Load simply supported (a) Restrained beam (c) End couple
loading Fig. 7-1 1. — Restrained beam resolved into simple beam loadings.
these end slopes by adding the loading in Fig. 7-1 lc, we must have 6i = 6[ and
02 = 02, which requires that Mb be greater than Ma in order for 62 to be greater
than 6{. In other words, the larger end moment acts at the wall that is closer to the
resultant of any single load. The difference between the end moments Mb and
Ma is ...
Page 303
The final moments in Fig. 8-19d are obtained by superposition of the FEMs in Fig.
8-19b and of the distributed moments and carry-over moments in Fig. 8-1 9e.
Sign Convention. In the preceding discussion, signs were based on conventional
bending moments. This convention requires the carry-over moment to be of
opposite sign and often leads to confusion regarding the magnitude and sign of
the unbalanced moment to be distributed, especially when more than two
members frame ...
The final moments in Fig. 8-19d are obtained by superposition of the FEMs in Fig.
8-19b and of the distributed moments and carry-over moments in Fig. 8-1 9e.
Sign Convention. In the preceding discussion, signs were based on conventional
bending moments. This convention requires the carry-over moment to be of
opposite sign and often leads to confusion regarding the magnitude and sign of
the unbalanced moment to be distributed, especially when more than two
members frame ...
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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