Question 4.SP.7: For the beam of Sample Prob. 4.5, determine the residual str......
For the beam of Sample Prob. 4.5, determine the residual stresses and the permanent radius of curvature after the 10,230-kip⋅in. couple M has been removed.
STRATEGY: Start with the moment and stress distribution when the flanges have just become plastic. The beam is then unloaded by a couple that is equal and opposite to the couple originally applied. During the unloading, the action of the beam is fully elastic. The stresses due to the original loading and those due to the unloading are superposed to obtain the residual stress distribution.
MODELING and ANALYSIS:
Loading. In Sample Prob. 4.5, a couple of moment M = 10,230 kip⋅in. was applied and the stresses shown in Fig. 1a were obtained.
Elastic Unloading. The beam is unloaded by the application of a couple of moment M = −10,230 kip⋅in. (which is equal and opposite to the couple originally applied). During this unloading, the action of the beam is fully elastic; recalling from Sample Prob. 4.5 that I=1524 in ^4
\sigma_m^{\prime}=\frac{M c}{I}=\frac{(10,230 \text{kip} \cdot \text{in} .)(8 in .)}{1524 in ^4}=53.70 ksi
The stresses caused by the unloading are shown in Fig. 1b.
Residual Stresses. We superpose the stresses due to the loading (Fig. 1a) and to the unloading (Fig. 1b) and obtain the residual stresses in the beam (Fig. 1c).
Permanent Radius of Curvature. At y = 7 in. the residual stress is σ = −3.01 ksi. Since no plastic deformation occurred at this point, Hooke’s law can be used, and \varepsilon_x=\sigma / E . Recalling Eq. (4.8), we write
\varepsilon_x=-\frac{y}{\rho} (4.8)
\rho=-\frac{y}{\varepsilon_x}=-\frac{y E}{\sigma}=-\frac{(7 \text { in. })\left(29 \times 10^6 psi \right)}{-3.01 ksi }=+67,400 \text { in. } \quad \rho=5620 \text{ft}
REFLECT and THINK: From Fig. 2, note that the residual stress is tensile on the upper face of the beam and compressive on the lower face, even though the beam is concave upward.
