## Strength of materials |

### From inside the book

Results 1-3 of 60

Page 123

midway between that load and the

this rule we locate the position of each load when the moment at that load is a

maximum, and compute the value of each such maximum moment. The

maximum shearing force occurs at, and is equal to, the maximum reaction. The

maximum reaction for a group of moving loads on a span occurs either at the left

reaction when the leftmost load is over that reaction, or at the right reaction when

the rightmost ...

midway between that load and the

**resultant**of all loads then on the span. Withthis rule we locate the position of each load when the moment at that load is a

maximum, and compute the value of each such maximum moment. The

maximum shearing force occurs at, and is equal to, the maximum reaction. The

maximum reaction for a group of moving loads on a span occurs either at the left

reaction when the leftmost load is over that reaction, or at the right reaction when

the rightmost ...

Page 314

9-lc), the

effects. Thus, the

dA and S/dA. Dividing this by the area dA gives the

directed normal to the cross section. Similarly, at a point B in the same section,

also at a distance y from the neutral axis but above it, the

difference between the axial and flexural stresses. If tensile stress is denoted by a

positive ...

9-lc), the

**resultant**stress at A is equal to the superposition of the two separateeffects. Thus, the

**resultant**force at A is the vector sum of the collinear forces SadA and S/dA. Dividing this by the area dA gives the

**resultant**stress <S = Sa + S/directed normal to the cross section. Similarly, at a point B in the same section,

also at a distance y from the neutral axis but above it, the

**resultant**stress is thedifference between the axial and flexural stresses. If tensile stress is denoted by a

positive ...

Page 341

Be careful not to confuse stress trajectories with lines of constant stress. Stress

trajectories are lines of principal stress direction but of variable stress intensity.

ILLUSTRATIVE PROBLEMS 943. A shaft 4 in. in diameter that rotates at 1800

rpm is subjected to bending loads that produce a maximum bending moment of

20007T ft-lb. Determine the torque and hp that can also act simultaneously on the

shaft without exceeding a

normal ...

Be careful not to confuse stress trajectories with lines of constant stress. Stress

trajectories are lines of principal stress direction but of variable stress intensity.

ILLUSTRATIVE PROBLEMS 943. A shaft 4 in. in diameter that rotates at 1800

rpm is subjected to bending loads that produce a maximum bending moment of

20007T ft-lb. Determine the torque and hp that can also act simultaneously on the

shaft without exceeding a

**resultant**shearing stress S, = 12,000 psi or a**resultant**normal ...

### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Other editions - View all

### Common terms and phrases

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