Question 1 (50 marks total) (a) Konwing that I am 160 MPa and tal= 100 MPa, cons

Question

Question 1 (50 marks total) (a) Konwing that I am 160 MPa and tal= 100 MPa, consider the available section in the Table below, select the most economical metric wide-flange shape that should be used to support the loading shown in Figure 1. 22 kN/m 145 KN 3m1m1m Figure 1: Loading applied on a simply supported beam Shape Area (A), mm² Section Modulus (S), 103 mm W410x38.8 4990 637 W360x32.9 4170 474 W310x38.7 4940 549 W250x44.8 5720 535 W200x46.1 5860 448 Table 1: Cross section area and section modulus of wide-flange shape beams (b) Determine the values to be expected for om, T. and the principal stress omax at the junction of a flange and the web of the selected beam.

Explanation / Answer

Let us determine the reaction forces at the two supports.

Reaction at support A can be determined from considering moment equilibrium about support D

5*RA - (22*3*3.5)-45=0

RA = 55.2 kN

From vertical equilibrium of beam,RD = (22*3)+45-55.2=55.8 kN

Maximum shear force in the beam is between C and D and is equal to 55.8 kN

maximum moment will be where shear force is zero

Let shear force be zero at a distance x from support A

x=55.8/22=2.536 m

bending moment at that section = 55.2*2.536 - 22*2.5362/2 = 69.24 kNm=69.24*106 Nmm

normal bending stress allowable = 160 MPa=160 N/mm2

section modulus required = 69.24*106/160=432.768 mm3

sher stress allowable = 100 N/mm2

area of cross section required = 55.8*1000/100=558 mm2

Therefore, the required steel section is the one which has the above area of cross section and section modulus and also lightest

Therefore,the section is W360x32.9

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