Question 16.18: The beam section shown in Fig. P.16.18 is subjected to a tem......
The beam section shown in Fig. P.16.18 is subjected to a temperature change which varies with y such that T=T_{0}y/2a. Determine the corresponding changes in the stress resultants. Young’s modulus for the material of the beam is E, while its coefficient of linear expansion is α.
Answer: N_{T}=0,\;M_{x T}=5E\alpha 7a^{2}t\,T_{0}/3,\;M_{y T}=E \alpha\ a^{2}t\,T_{0}/6
Referring to Fig. S.16.18 and taking moments of areas about the upper flange,
(a t+2a t){\overline{{y}}}=2a t awhich gives
{\overline{{y}}}={\frac{2}{3}}aNow taking moments of areas about the vertical web,
3a t\overline{{x}}=a t\frac{a}{2}so that
{\bar{x}}={\frac{a}{6}}From Eq. (16.51),
M_{x T}=\Sigma E \alpha\Delta T{\bar{y}}_{i}A_{i} (16.51)
N_{T}=\int_{A}^{}{E\alpha{\frac{T_{0}y}{2a}}t\ \mathrm{d}s={\frac{E\alpha T_{0}t}{2a}}\int_{A}^{}{yds} }
But t\int_{A}^{}{y} ds is the first moment of area of the section about the centroidal axis Cx, i.e., t\int_{A}^{}{y} ds = 0.
Therefore,
From Eq. (16.52),
M_{y T}=\Sigma E \alpha\Delta T\bar{x}_{i}A_{i} (16.52)
M_{xT}=\int_{A}^{}{E\alpha{\frac{T_{0}}{2a}}t y^{2}\,\mathrm{d}s={\frac{E\alpha T_{0}}{2a}}\int_{A}^{}{ty^{2}ds} }
But
\int_{A}^{}{ty^{2}\,\mathrm{d}s=I_{x x}=a t\left(\frac23a\right)^{2}+t\frac{(2a)^{3}}3+2a t\bigl(\frac{a}{3}\bigr)^{2}}i.e.,
I_{x x}={\frac{10a^{3}t}{3}}Therefore,
M_{x T}={\frac{5E\alpha a^{2}t T_{0}}{3}}From Eq. (16.53),
N_{T}=\int_{A}E \alpha{\Delta T}(x,y)t\,{\mathrm{d}}s (16.53)
M_{yT}=\int_{A}^{}{E\alpha{\frac{T_{0}}{2a}}t x y\ \mathrm{d}s={\frac{E\alpha T_{0}}{2a}}\int_{A}^{}{txyds} }
But
\int_{A}^{}{t x y\mathrm{d}s=I_{x y}=a t\left({\frac{a}{3}}\right)\left({\frac{2}{3}}a\right)+2a t\left(-{\frac{a}{6}}\right)\left(-{\frac{a}{3}}\right)}i.e.,
I_{xy}={\frac{a^{3}t}{2}}Then
M_{y T}={\frac{E\alpha a^{2}t T_{0}}{6}}