Question 4.SP.5: Beam AB has been fabricated from a high-strength low-alloy s...
Beam AB has been fabricated from a high-strength low-alloy steel that is assumed to be elastoplastic with E = 29 × 106 psi and σY = 50 ksi. Neglecting the effect of fillets, determine the bending moment M and the corresponding radius of curvature (a) when yield first occurs, (b) when the flanges have just become fully plastic.

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a. Onset of Yield. The centroidal moment of inertia of the section is
I=121(12 in. )(16 in.)3−121(12 in.−0.75 in.)(14 in.)3=1524 in4
Bending Moment. For σmax=σY = 50 ksi and c = 8 in., we have
MY=cσYI=8 in.(50 ksi)(1524 in4) MY = 9525 kip • in.
Radius of Curvature. Noting that, at c = 8 in., the strain is ϵY=σY/E=(50 ksi)/(29×106 psi)=0.001724, we have from Eq. (4.41)
c=ϵYρY8 in. =0.001724 ρY ρY=4640 in.
b. Flanges Fully Plastic. When the flanges have just become fully plastic, the strains and stresses in the section are as shown in the figure below. We replace the elementary compressive forces exerted on the top flange and on the top half of the web by their resultants R1 and R2, and similarly replace the tensile forces by R3 and R4.
R1=R4=(50 ksi)(12 in.)(1 in.)=600 kips
R2=R3=21(50 ksi)(7 in.)(0.75 in.)=131.3 kips
Bending Moment. Summing the moments of R1, R2, R3, and R4 about the z axis, we write
M=2[R1(7.5 in.)+R2(4.67 in.)]
=2[(600)(7.5)+(131.3)(4.67)] M = 10,230 kip • in.
Radius of Curvature. Since yY=7 in. for this loading, we have from Eq. (4.40)
yY=ϵYρ7 in.=(0.001724) ρ ρ=4060 in.=338 ft

