Question 6.13: A wide-flange beam is subjected to a fully plastic moment, a......

We can follow the superposition-of-stresses procedure that was illustrated in Fig. 6.38 for a rectangular beam. We first calculate M_P using Figs. 2a and 2d and the beam dimensions given in Fig. 1.

C_1 = T_1 = (200 MPa)(0.150 m)(0.010 m) = 300 kN      (1)

C_2 = T_2 = (200 MPa)(0.010 m)(0.075 m) = 150 kN

M_P = (300 kN)(0.160 m) + (150 kN)(0.075 m) = 59.25 kN · m      (2)

The maximum elastic recovery stress, or modulus of rupture, is obtained by setting M = M_P in the flexure formula, Eq. 6.13, giving.

σ_x =\frac{-My} {I}    Flexure Formula        (6.13)

(σ_{er})_{max} = \frac{M_P \left(\frac{h}{2}\right)} {I} = \frac{(59.25  kN · m)(0.085  m)} {2.204(10^{-5})  m^4} = 229 MPa       (3)

To sketch the residual stress state in Fig. 2c we add the stress blocks in Figs. 2a and 2b. The point of zero stress in Fig. 2c may be determined by using similar triangles in Fig. 2b.

\frac{σ_Y} {d_Y} = \frac{(σ_{er})_{max}} {(h/2)} → d_Y = \left(\frac{200} {229}\right) (85 mm) = 74.4 mm        (4)

Therefore, the residual stress distribution has the form shown in Fig. 2c.