## Strength of Materials |

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Page 110

Finally between C and D, the intensity of loading is

slope of the shear diagram is

that the shear diagram consists of straight horizontal lines for intervals in which ...

Finally between C and D, the intensity of loading is

**zero**and the correspondingslope of the shear diagram is

**zero**(a horizontal line). We may conclude thereforethat the shear diagram consists of straight horizontal lines for intervals in which ...

Page 113

lb at D. From D to E, the loading is

. The concentrated load of 120 lb at E reduces the shear abruptly to

we locate the positions of

lb at D. From D to E, the loading is

**zero**, which means that the slope is hori- lontal. The concentrated load of 120 lb at E reduces the shear abruptly to

**zero**. Beforewe locate the positions of

**zero**shear at F and G on the shear diagram, consider ...Page 553

Product of Inertia Is

of symmetry, this axis together with any axis perpendicular to it will form a set of

axes for which the product of inertia is

Product of Inertia Is

**Zero**with Respect to Axes of Symmetry If an area has an axisof symmetry, this axis together with any axis perpendicular to it will form a set of

axes for which the product of inertia is

**zero**. Consideration of the symmetrical T ...### What people are saying - Write a review

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### Common terms and phrases

allowable stresses aluminum angle assumed axes axial load beam loaded beam shown bending bending moment bolt cantilever beam caused centroid CN CN CN column compressive stress Compute the maximum concentrated load concrete 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 Hooke's law horizontal ILLUSTRATIVE PROBLEMS lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan modulus Mohr's circle moment of area moment of inertia moments of inertia neutral axis obtain plane plastic product of inertia proportional limit radius 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