(a) A steel beam of rectangular section, 80 mm deep by 30 mm wide, is simply supported over a span of 1.4 m and carries a u.d.1. w. If the yield stress of the material is 240 MN/m², determine the value of w when yielding of the beam material has penetrated to a depth of 20 mm from each surface of

Question

(a) A steel beam of rectangular section, 80 mm deep by 30 mm wide, is simply supported over a span of 1.4 m and carries a u.d.1. w. If the yield stress of the material is [latex] 240 MN/m^{2}[/latex], determine the value of w when yielding of the beam material has penetrated to a depth of 20 mm from each surface of the beam.
(b) What will be the magnitudes of the residual stresses which remain when load is removed?
(c) What external moment must be applied to the unloaded beam in order to return it to its undeformed (straight) position?

Accepted Answer

(a) From eqn. (3.2) the partially plastic moment carried by a rectangular section is given by

[latex]M_{pp} =\frac{B\sigma _{y} }{12} \left[3D^{2} -d^{2} ight][/latex] (3.2)

Thus, for the simply supported beam carrying a u.d.l., the maximum B.M. will be at the centre of the span and given by

[latex]BM_{\max } =\frac{wL^{2} }{8}=\frac{B\sigma _{y} }{12} \left[3D^{2} -d^{2} ight][/latex]
[latex]w=\frac{8 imes 30 imes 10^{-3} imes 240 imes 10^{6} }{1.4^{2} imes 12 } \left[3 imes 80^{2}-40^{2} ight] 10^{-6}[/latex]
[latex]=43.1 kN/m[/latex]

(b) From the above working

[latex]M_{pp} =\frac{B\sigma _{y} }{12} \left[3D^{2} -d^{2} ight]=\frac{wL^{2} }{8}[/latex]
[latex]=43.1 imes 10^{3} imes \frac{1.4^{2} }{8}=10.6 kN/m[/latex]

During the unloading process a moment of equal value but opposite sense is applied to the beam assuming it to be completely elastic. Thus the equivalent maximum elastic stress [latex] \sigma ^{′}[/latex] introduced at the outside surfaces of the beam by virtue of the unloading is given by the simple bending theory with [latex] M = M_{pp} = 10.6 kNm [/latex],

i.e.

[latex] \sigma ^{′}=\frac{My}{I}=\frac{10.6 imes 10^{3} imes 40 imes 10^{-3} imes 12 }{30 imes 80^{3} imes 10^{-12} }[/latex]
[latex]=0.33 imes 10^{9} =330 MN/m^{2}[/latex]

The unloading, elastic stress distribution is then linear from zero at the A. to ±330 [latex]MN/m^{2}[/latex] at the outside surfaces, and this may be subtracted from the partially plastic loading stress distribution to yield the residual stresses as shown in Fig. 3.40.

(c) The residual stress distribution of Fig. 3.40 indicates that the central portion of the beam, which remains elastic throughout the initial loading process, is subjected to a residual stress system when the beam is unloaded from the partially plastic state. The beam will therefore be in a deformed state. In order to remove this deformation an external moment must be applied of sufficient magnitude to return the elastic core to its unstressed state. The required moment must therefore introduce an elastic stress distribution producing stresses of ±75 [latex]MN/m^{2}[/latex] at distances of 20 mm from the N.A. Thus, applying the bending theory,

[latex]M=\frac{\sigma I}{y} =\frac{75 imes 10^{6} }{20 imes 10^{-3} } imes \frac{30 imes 80^{3} imes 10^{-12} }{12}[/latex]
[latex]=4.8 kNm[/latex]

Alternatively, since a moment of 10.6 kNm produces a stress of [latex]165 MN/m^{2}[/latex] at 20 mm from the N.A., then, by proportion, the required moment is

[latex]M=10.6 imes \frac{75}{165}=4.8 kNm[/latex]

Question 3.3: (a) A steel beam of rectangular section, 80 mm deep by 30 mm...

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