### Problem Statement1. For the W 21x 62 steel beam below, determine the actual value of \( \sigma_m \) in the beam and the maximum value of the principal stress \( \sigma_{max} \) at the junction of the flange and the web.### Diagram ExplanationThe diagram shows a simply supported steel beam labeled W 21x 62. The beam is subjected to:- Two point loads of 20 kips each at points A and D, located at the ends of the beam.- A uniformly distributed load of 2 kips/ft applied over the middle portion of the beam between points B and C.#### Beam Layout:- **Length of Beam:** Total length of 50 ft with supports placed at equal distances: - **10 ft** from A to the first support - **30 ft** between the supports - **10 ft** from the second support to D#### Load Distribution:- **Point Loads:** - At point A: 20 kips directed downward - At point D: 20 kips directed downward- **Uniformly Distributed Load:** - 2 kips/ft between points B and C, over a 30 ft length This setup represents a static loading condition where structural analysis can be used to find maximum fiber stress and principal stresses in the beam.

A simply supported beam at B and C, of length 30ft between B and C, with overhanging portion AB and CD of length 10 ft is given. The beam is made from W 21 x 62 steel beam. The specification for W 21 x 62 steel beam is mentioned below. DesignationImperial(In x lb/ft) Depthh(in) Widthw(in) Web Thicknesstw(in) Flange Thicknesstf(in) Sectional Area(in2) Weight(lbf/ft) Ix(in4) Iy(in4) Sx(in3) Sy W 21 x 62 21 8.24 0.400 0.615 18.3 62 1330 57.5 127 13.9 To determine Maximum bending stress, σm in the beam Maximum principal stress, σmax at the junction of flange and the webDue to symmetry, RC = RB = 50 kips The moment at B = -20×10 = -200 kip-ft The moment at mid of B and C =-20×25+50×15-2×1522 = 25 kip-ft Thus, the maximum moment, Mmax is 200kip-ft Again, The shear force at A and Bleft = -20 kips The shear force at Bright = 30 kips Thus, the maximum shear force is 30 kips Now, The maximum bending stress σm σm = MmaxSx=200×12127=18.897 kipin2 Thus, the maximum bending stress is 18.897 kipin2 The maximum moment, Mmax is 200kip-ft at B and the maximum shear force is 30 kips at Bright. We determine the maximum principal stress Bright of the given section. Therefore, the bending stress, σ and maximum shear stress, τ at the junction of the flange and the web, σ =MmaxIx×h2-tf=200×121330×212-0.615=17.84 kipin2 τ = V2Ixh24-h-2tf24×wtw=302×1330×2124-21-2×0.615248.240.4=2.913 kipin2 Thus, the bending stress σ and maximum shear stress, τ at the junction of the flange and the web are 17.84kipin2 and 2.913kipin2. Now, the Maximum principal stress is σmax σmax = σ2+σ22+τ2=17.842+17.8422+2.9132=8.92+9.384=18.304kipin2 Thus, the maximum principal stress, at the junction of the flange and the web = 18.304kipin2Maximum bending stress, σm in the beam = 18.897kipin2 Maximum principal stress, σmax at the junction of the flange and the web = 18.304 kipin2