Question 9.19: An overhanging beam ABC of height h has a pin support at A a......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize: The displacement of this beam was investigated at selected points due to a concentrated load at C in Example 9-5, under a uniform load q in Example 9-9, and with uniform load q on AB and load P at C in Example 9-18. Now consider the effect of a temperature differential (T_2 – T_1) on the deflection v(x) of the beam using Eq. (9-137).
\quad\quad\quad \quad {\frac{d^{2}\nu}{d x^{2}}}={\frac{\alpha(T_{2}-{{T}}_{1})}{h}}\qquad\quad(9-138, repeated)
3. Analyze: Integration results in two constants of integration, C_1 ~ and ~ C_2, which must be determined using two independent bounday conditions:

The boundary conditions are v(0) = 0 and v(L) = 0. So v(0) = 0, which gives C_2 = 0.
Also, v(L) = 0, which leads to\quad\quad (c)
\quad\quad\quad \quad C_{1}=\frac{1}{L}\bigg[\frac{-\alpha L^{2}}{2h}(T_{2}-T_{1})\bigg]=-\bigg[\frac{L\alpha(T_{2}-T_{1})}{2h}\bigg]\quad\qquad\qquad (d)
Substituting C_1 ~ and ~ C_2 into Eq. (b) results in the equation of the elastic curve of the beam due to temperature differential (T_2 – T_1) as
\quad\quad\quad \quad \nu(x)={\frac{\alpha x(T_{2}-{T_{1}})(x-L)}{2h}}\quad\quad\quad (e)
If x = L + a in Eq. (e), an expression for the deflection of the beam at C is
\quad{\delta}_{C}=\nu(L+a)=\frac{\alpha(L+a)(T_{2}-T_{1})(L+a-L)}{2h}=\frac{\alpha(T_{2}-T_{1})a(L+a)}{2h}\quad(f)
4. Finalize: Linear elastic behavior was assumed here and in earlier examples, so (if desired) the principle of superposition can be used to find the total deflection at C due to simultaneous application of all loads considered in Examples 9-5, 9-9, and 9-18 and for the temperature differential studied here.
Numerical example: If beam ABC is a steel, wide flange HE 700B [see Table F-1] with a length of L = 9 m and with an overhang a = L / 2, compare the deflection at C due to self-weight (see Example 9-9; let q = 2.36 kN/m) to the deflection at C due to temperature differential (T_2  –  T_1) = 3°C.
From Table I-4, the coefficient of thermal expansion for structural steel is a = 12 \times 10^{-6} /°C. The modulus for steel is 210 GPa.
From Eq. (9-68), the deflection at C due to self-weight is \quad\qquad (g)

where a = 4.5 m and L = 9.0 m.
The deflection at C due to a temperature differential of only 3°C is from Eq. (f):

The deflection at C due to a small temperature differential is seven times that due to self-weight.