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 × 10^{6} psi and \sigma_{Y} = 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.
a. Onset of Yield. The centroidal moment of inertia of the section is
I=\frac{1}{12}(12 in. )(16 in. )^{3}-\frac{1}{12}(12 in. -0.75 in. )(14 in. )^{3}=1524 in^{4}
Bending Moment. For \sigma_{max} = \sigma_{Y} = 50 ksi and c = 8 in., we have
M_{Y}=\frac{\sigma_{Y} I}{c}=\frac{(50 ksi )\left(1524 in ^{4}\right)}{8 in .} M_{Y} = 9525 kip • in.
Radius of Curvature. Noting that, at c = 8 in., the strain is \epsilon_{Y} = \sigma_{Y} / E=(50 ksi ) /\left(29 \times 10^{6} psi \right)= 0.001724, we have from Eq. (4.41)
c=\epsilon_{Y} \rho_{Y} \quad 8 in. = 0.001724 \rho_{Y} \rho_{Y}=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 \mathbf{R} _{1} and \mathbf{R} _{2}, and similarly replace the tensile forces by \mathbf{R} _{3} and \mathbf{R} _{4}.
R_{1}=R_{4}=(50 ksi )(12 in. )(1 in. )=600 kips
R_{2}=R_{3}=\frac{1}{2}(50 ksi )(7 in. )(0.75 in. )=131.3 kips
Bending Moment. Summing the moments of \mathbf{R} _{1}, \mathbf{R} _{2}, \mathbf{R} _{3}, and \mathbf{R} _{4} about the z axis, we write
M =2\left[R_{1}(7.5 in. )+R_{2}(4.67 in. )\right]
=2[(600)(7.5)+(131.3)(4.67)] M = 10,230 kip • in.
Radius of Curvature. Since y_{Y}=7 in. for this loading, we have from Eq. (4.40)
y_{Y}=\epsilon_{Y} \rho \quad 7 in. =(0.001724) \rho \rho=4060 in. =338 ft
