## Strength of materials |

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Results 1-3 of 53

Page 101

zero at the wall, as shown in Fig. P-411. Ans. M = 6L 2 -to lb/ft 180 lb/ ft L 100 lb/ft

4' |J 10' 12001b Fig. P-413. 100 lb/ ft 6' B 4' | 4' 12001b Fig. P-414. 414.

the loads shown in Fig. P-415. Ans. Mac = -100x2 + 1400a: — 1600 f 200 lb/ ft 4'

B V 6001b ...

**Cantilever beam**carrying the uniformly varying load shown in Fig. P-410. 411.**Cantilever beam**carrying a distributed load varying from w lb/ft at the free end tozero at the wall, as shown in Fig. P-411. Ans. M = 6L 2 -to lb/ft 180 lb/ ft L 100 lb/ft

4' |J 10' 12001b Fig. P-413. 100 lb/ ft 6' B 4' | 4' 12001b Fig. P-414. 414.

**Cantilever beam**loaded as shown in Fig. P-414. 415.**Cantilever beam**carryingthe loads shown in Fig. P-415. Ans. Mac = -100x2 + 1400a: — 1600 f 200 lb/ ft 4'

B V 6001b ...

Page 208

Deflection of

any point is the distance from the point on the elastic curve to a tangent drawn to

the curve at some other point (Art. 6-3 and Fig. 6-8). As a consequence, the

tangential deviation is generally not equal to the deflection. In

however, the wall is usually assumed to be perfectly fixed, and hence the tangent

drawn to the elastic curve at the wall will be horizontal, as in Fig. 6-18. Therefore,

if the ...

Deflection of

**Cantilever Beams**It will be recalled that the tangential deviation atany point is the distance from the point on the elastic curve to a tangent drawn to

the curve at some other point (Art. 6-3 and Fig. 6-8). As a consequence, the

tangential deviation is generally not equal to the deflection. In

**cantilever beams**,however, the wall is usually assumed to be perfectly fixed, and hence the tangent

drawn to the elastic curve at the wall will be horizontal, as in Fig. 6-18. Therefore,

if the ...

Page 233

straints must be applied. Nevertheless, the conjugate-beam method offers an

occasional advantage in certain routine work, in that it permits direct application

of the definitions of shear and moment to the fictitious loading to find slope and

deflection without any need of an elastic curve. Now for a word about the need for

artificial constraints in certain cases. M For the

— diagram appears as in Fig. hi l 6-31b. This diagram cannot be applied directly

as a ...

straints must be applied. Nevertheless, the conjugate-beam method offers an

occasional advantage in certain routine work, in that it permits direct application

of the definitions of shear and moment to the fictitious loading to find slope and

deflection without any need of an elastic curve. Now for a word about the need for

artificial constraints in certain cases. M For the

**cantilever beam**in Fig. 6-31a, the— diagram appears as in Fig. hi l 6-31b. This diagram cannot be applied directly

as a ...

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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