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Steel Designers' Manual - 6th Edition (2003)
x 2
40
=
3
. m
Therefore MA ¥ + MB + MC
Since A and C are simple supports
MA = MC = 0
Ê 10 ¥ 1 200 ¥ 4 ˆ
2 2 ( 2 3)+ ¥ 3 = 6 +
Ë 3¥ 2 3¥ 3 ¯
Ê ˆ
Therefore M B = + = ( )= kN m
Ë ¯
SFA = + kN
SF B for span AB kN
SFC kN
SF B for span BC
-
6 10 800 6
90. 56 54. 33
10 6 9 10
0 54. 33
5 = 5- 2717 . = -2217
.
2
= 5+ 2717 . = 3217 .
200 ¥ 2 0-54. 33
= + = 133. 33 - 18. 11 = 115. 22
3 3
200
= + 18. 11 = 66. 67 + 18. 11 = 84.78
kN
3
Note that the negative reaction at A means that the end A will tend to lift off its
support and will have to be held down.
10.5 Plastic failure of single members
Plastic failure of single members 335
The concept of the plastic hinge, capable of undergoing large rotation once the
applied moment has reached the limiting value Mp, constitutes the basis of plastic
design. This concept may be illustrated by examining the development of the collapse
mode of a fixed-end beam subjected to a uniformly distributed load of increasing
intensity w (Fig. 10.9(a)). Such a member is statically indeterminate, having three
redundancies which however reduce to two unknowns if the axial thrust in the
member is assumed to be zero. It will be assumed that the two unknown quantities
are the fixing moments MA and MB.As the load increases, the beam initially behaves
in an elastic manner and the value of the redundant moments can be derived by
applying the three general conditions used in elastic structural analysis, namely
those of
(1) equilibrium (application of statics)
(2) moment–curvature (EI d 2 y/dx 2 = M)
(3) compatibility condition (continuity, including geometric conditions at the
supports).
For the beam in Fig. 10.9(a), the first condition (equilibrium) is satisfied by
drawing the bending moment diagram (shaded in Fig. 10.9(b)) as a superposition of