sm13 19

________________________________________________________________________
PROBLEM (13.19) A simple beam 1-3 of length L and flexural rigidity EI is subjected to a
uniformly distributed load of intensity w as shown in Fig. P13.19. Determine the deflection of the beam at
midpoint 2 by replacing the applied load with the equivalent nodal loads (see Table B15).
SOLUTION
Due to the symmetry only one-half of the beam need be considered. Referring to
Case 4 of Table B.13, equivalent nodal forces obtained (see Fig. a).
Boundary conditions are
12
0v
θ
==
We have 2
148MwL=− and
24
y
FwL
=
−
Equation (13.21a), with element
of length 12LL
=
.
2
21
11
32
11
46
48
612
4
LL
wL EI v
LL
wL
θ
⎡⎤
⎧⎫ −
⎧
⎫
=
⎨⎬ ⎨
⎬
⎢⎥
−
−⎩⎭
⎩⎭
⎣⎦
21
1
32
1
3
8312
LL
EI v
LL
θ
⎡⎤
−⎧⎫
=⎨⎬
⎢⎥
−⎩⎭
⎣⎦
Inverting
22
3
1
2
12 3 48
24 31
LLwL
L
vEI Lw
θ
⎡⎤⎧
⎧⎫
=4L
⎫
⎨ ⎬
⎢⎥
−
⎩⎭ ⎣⎦⎩⎭
⎨⎬
or
This is the exact solution (See Table B.14)
3
1
4
2
24
5384
wL EI
vwL EI
θ
⎧⎫
−
⎧⎫
=
⎨⎬⎨ ⎬
−
⎩⎭⎩⎭
wL/4
wL2/48 wL2/48
1 2
/4
wL
12
L
L=
Figure (a) Equivalent
nodal forces
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