Strength of materials |
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Page 91
In this case, the external load is vertical, so the condition ZX = 0 (the X axis is
horizontal) is automatically satisfied. To satisfy 2F = 0, the vertical unbalance
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 shearing force. For the loading shown, VT
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 vertical unbalance (which is
equal but ...
In this case, the external load is vertical, so the condition ZX = 0 (the X axis is
horizontal) is automatically satisfied. To satisfy 2F = 0, the vertical unbalance
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 shearing force. For the loading shown, VT
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 vertical unbalance (which is
equal 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 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 ...
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 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 not
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 not
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 ...
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Bibliographic information
| Title | Strength of materials |
| Author | Ferdinand Leon Singer |
| Edition | 2 |
| Publisher | Harper, 1962 |
| Original from | the University of Michigan |
| Digitized | 29 Nov 2007 |
| Length | 590 pages |
| Subjects | Strength of materials |
| Export Citation | BiBTeX EndNote RefMan |