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 × 10610^{6} psi and σY\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.

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a. Onset of Yield.     The centroidal moment of inertia of the section is

I=112(12 in. )(16 in.)3112(12 in.0.75 in.)(14 in.)3=1524 in4I=\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 σmax=σY\sigma_{max} = \sigma_{Y} = 50 ksi and c = 8 in., we have

MY=σYIc=(50 ksi)(1524 in4)8 in.M_{Y}=\frac{\sigma_{Y} I}{c}=\frac{(50  ksi )\left(1524  in ^{4}\right)}{8  in .}      MYM_{Y} = 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\epsilon_{Y} = \sigma_{Y} / E=(50  ksi ) /\left(29 \times 10^{6}  psi \right)= 0.001724, we have from Eq. (4.41)

c=ϵYρY8 in. =0.001724 ρYc=\epsilon_{Y} \rho_{Y} \quad 8   in.  = 0.001724  \rho_{Y}      ρY=4640 in.\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 R1\mathbf{R} _{1} and R2\mathbf{R} _{2}, and similarly replace the tensile forces by R3\mathbf{R} _{3} and R4\mathbf{R} _{4}.

R1=R4=(50 ksi)(12 in.)(1 in.)=600 kipsR_{1}=R_{4}=(50  ksi )(12   in. )(1   in. )=600  kips

R2=R3=12(50 ksi)(7 in.)(0.75 in.)=131.3 kipsR_{2}=R_{3}=\frac{1}{2}(50  ksi )(7   in. )(0.75   in. )=131.3  kips

Bending Moment.  Summing the moments of R1\mathbf{R} _{1}, R2\mathbf{R} _{2}, R3\mathbf{R} _{3}, and R4\mathbf{R} _{4} about the zz axis, we write

M=2[R1(7.5 in.)+R2(4.67 in.)]M =2\left[R_{1}(7.5   in. )+R_{2}(4.67   in. )\right]

=2[(600)(7.5)+(131.3)(4.67)]=2[(600)(7.5)+(131.3)(4.67)]               M = 10,230 kip • in.

Radius of Curvature.     Since yY=7y_{Y}=7 in. for this loading, we have from Eq. (4.40)

yY=ϵYρ7 in.=(0.001724) ρy_{Y}=\epsilon_{Y} \rho \quad 7   in. =(0.001724)  \rho          ρ=4060 in.=338 ft\rho=4060   in. =338  ft

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