## Strength of materials |

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Page 91

In this case, the external load is

horizontal) is automatically satisfied. To satisfy 2F = 0, the

caused by Ri requires the fibers in section a-a to create a resisting force. This is

shown as Vr, and is called the resisting

is numerically equal to Ri; but if additional loads had been applied between Ri

and section a-a (as in Figs. 4-5 and 4-6), the net

equal but ...

In this case, the external load is

**vertical**, so the condition ZX = 0 (the X axis ishorizontal) is automatically satisfied. To satisfy 2F = 0, the

**vertical**unbalancecaused by Ri requires the fibers in section a-a to create a resisting force. This is

shown as Vr, and is called the resisting

**shearing**force. For the loading shown, VTis numerically equal to Ri; but if additional loads had been applied between Ri

and section a-a (as in Figs. 4-5 and 4-6), the net

**vertical**unbalance (which isequal but ...

Page 163

5-21 we note that dF = S,b dx, where S, is the average shearing stress over the

differential area of width b and length dx; also that M2 — Mi represents the

differential change in bending moment dM in the distance dx; hence the above

relation is rewritten as C dM f A A S' = -IbdxlUiydA dM From Art. 4-4 we recall that

— — = V, the

4) We have replaced the integral yd A, which means the sum of the moments of

the ...

5-21 we note that dF = S,b dx, where S, is the average shearing stress over the

differential area of width b and length dx; also that M2 — Mi represents the

differential change in bending moment dM in the distance dx; hence the above

relation is rewritten as C dM f A A S' = -IbdxlUiydA dM From Art. 4-4 we recall that

— — = V, the

**vertical shear**; so we obtain for dx the horizontal shearing stress, (5-4) We have replaced the integral yd A, which means the sum of the moments of

the ...

Page 164

It is this

feasible to determine S,v directly, we have resorted to deriving the numerically

equal value of S,h. To prove the equivalence of S,h and S,v, consider their effect

on a free-body diagram of a typical element in Fig. 5-23. A pictorial view of this

element is shown in Fig. 5-24a; a front view, in Fig. 5-24b. For equilibrium of this

element, the ...

It is this

**vertical shearing**stress S,v, shown in Fig. 5-23, that forms the resisting**vertical shear**Vr = fS, dA which balances the**vertical shear**V. Since it is notfeasible to determine S,v directly, we have resorted to deriving the numerically

equal value of S,h. To prove the equivalence of S,h and S,v, consider their effect

on a free-body diagram of a typical element in Fig. 5-23. A pictorial view of this

element is shown in Fig. 5-24a; a front view, in Fig. 5-24b. For equilibrium of this

element, the ...

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