Question 16.17: The beam section of P.16.4 is subjected to a temperature ris......
The beam section of P.16.4 is subjected to a temperature rise of 4 T_{0} in its upper flange 12, a temperature rise of 2T_{0} in both vertical webs, and a temperature rise of T_{0} in its lower flange 34. Determine the changes in axial force and in the bending moments about the x and y axes. Young’s modulus for the material of the beam is E and its coefficient of linear expansion is α.
Answer: N_{T}=9E \alpha d t T_{0},M_{x T}=3E\alpha\,d^{2}t\;T_{0}/2,\,M_{y T}=3E\alpha\,d^{2}t\;T_{0}/4
From Eq. (16.48) and Fig. P.16.4,
M_{x T}=\int_{}^{}{\int_{A}E \alpha\Delta T y\;\mathrm{d}A} (16.48)
N_{T}=E\alpha(4T_{0}d t+2\times2T_{0}d t+T_{0}d t)
i.e.,
N_{T}=9E\alpha\,d t T_{0}From Eq. (16.49),
M_{y T}=\int_{}^{}{\int_{A}E \alpha\Delta T x\;\mathrm{d}A} (16.49)
M_{x T}=E\alpha\left[4T_{0}d t\Bigl({\frac{d}{2}}\Bigr)+2\times2T_{0}d t(0)+T_{0}d t\Bigl(-{\frac{d}{2}}\Bigr)\right]
i.e.,
M_{x T}={\frac{3E\alpha d^{2}t T_{0}}{2}}From Eq. (16.50),
N_{T}=\Sigma E \alpha\Delta T A_{i} (16.50)
M_{y T}=E\alpha\Bigl[4T_{0}d t\Bigl({\frac{d}{4}}\Bigr)+2T_{0}d t\Bigl({\frac{d}{4}}\Bigr)+2T_{0}d t\Bigl(-{\frac{d}{4}}\Bigr)+T_{0}d t\Bigl(-{\frac{d}{4}}\Bigr)\Bigr]
i.e.
M_{y T}={\frac{3E\alpha d^{2}t T_{0}}{4}}