## 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 horizontal.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

acting actual allowable angle applied assumed axes axis beam shown bending bending moment cantilever carries caused centroid circle CN CN column compressive compressive stress compute concentrated consider constant couple cross section deflection deformation Determine developed diameter direction distance distributed distributed load effect elastic curve element equal equation equivalent expressed flange flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia lb/ft length limit load loaded as shown material maximum method midspan moments negative neutral axis obtain occurs plane plate positive Prob PROBLEMS produce reaction reference relation resisting respect restrained resultant rivet segment shear diagram shearing stress shown in Fig shows simply supported slope Solution Solve span steel strain strength supported Table tangent tensile thickness varies vertical wall weight weld yield zero