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Unformatted text preview: otation uA at the left-hand end, the deflection
d1 under the downward load, and the deflection d2 at the midpoint
of the beam. A B a a
L Solution 9.6-10 Simple beam with two loads
Because the beam is symmetric and the load is
antisymmetric, the deflection at the midpoint is zero.
‹ d2 0 ; ANGLE OF ROTATION uA AT END A
A1 a first moment of area between A and C
with respect to C L
2 a+ a
b + A2 a b a
2 Pa(L uA a)(L
12EI 2a) CC1
L/2 Pa(L a)(L
6LEI ab 2a) (clockwise) DEFLECTION d1 UNDER THE DOWNWARD LOAD
Distance a DD1 a
L/2 Pa2(L M1
A1 1 M1
2 EI A2 1 M1 L
2 tD2/A Pa(L
4LEI D2D1 DD1 2a) first moment of area between A
and D with respect to D a
A1 a b
6LEI D2D1 2 2a)2
6LEI (Downward) ; ;...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.
- Fall '11