## 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

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

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 inFig. 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. Inthe 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

resultant of any single load. The difference between the

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 theresultant of any single load. The difference between the

**end moments**Mb andMa is ...

Page 303

The

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