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A simply supported beam has a span of 15m UDL of 40KN/m, 5m long crosses the girder from left to right. Draw ILD for S.F and B.M at a section 6m from left hand.

Use this diagram to calculate to maximum S.F and B.M at this section.


1 Answer
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enter image description here

ILDforS.Fc

enter image description here


1.MaxS.Fc:

enter image description here


(i)Maximum+veSFatC:

0.69=y14

y1=0.267

Max+veS.Fc=40×(52(0.6+0.267))

Max+veS.Fc=86.7KN


(ii)MaximumveSFatC:

enter image description here


y21=0.46

y2=0.067

MaxveS.Fc=40×[52(0.4+0.067)]

=46.7KN


2.MaxBMatC:

I.L.DforBMc

enter image description here


3.69=y16=2.4

3.66=y14=2.4

Tofindxbysimilartriangle

x5x=69

x=2m

MaxBMc=40×[22(2.4+3.6)]+40×[32(3.6+2.4)]

MaxBMc=600KN.m

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