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).

\frac{d \theta}{d x}=\frac{\alpha\left(T_{2}-T_{1}\right)}{h}           (9-137)

\frac{d^{2} v}{d x^{2}}=\frac{\alpha\left(T_{2}-T_{1}\right)}{h}            (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:

\frac{d}{d x} v(x)=\frac{\alpha}{h}\left(T_{2}-T_{1}\right) x+C_{1}           (a)

v(x)=\frac{\alpha}{h}\left(T_{2}-T_{1}\right) \frac{x^{2}}{2}+C_{1} x+C_{2}            (b)

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                           (c)

C_{1}=\frac{1}{L}\left[\frac{-\alpha L^{2}}{2 h}\left(T_{2}-T_{1}\right)\right]=-\left[\frac{L \alpha\left(T_{2}-T_{1}\right)}{2 h}\right]                    (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

v(x)=\frac{\alpha x\left(T_{2}-T_{1}\right)(x-L)}{2 h}             (e)

If x = L + a in Eq. (e), an expression for the deflection of the beam at C is

\delta_{C}=v(L+a)=\frac{\alpha(L+a)\left(T_{2}-T_{1}\right)(L+a-L)}{2 h}=\frac{\alpha\left(T_{2}-T_{1}\right) a(L+a)}{2 h}                   (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 W 30 × 211 [see Table F-1(a)] with a length of L = 30 ft and with an overhang a = L / 2, compare the deflection at C due to self-weight (see Example 9-9; let q = 211 lb/ft) to the deflection at C due to temperature differential (T_{2} – T_{1}) = 5°F. From Table I-4, the coefficient of thermal expansion for structural steel is \alpha=6.5 \times 10^{-6} /{ }^{\circ} F. The modulus for steel is 30,000 ksi.

From Eq. (9-68), the deflection at C due to self-weight is

= 7.467 \times 10^{-3}  in.

where a = 15 ft = 180 in. and L = 30 ft = 360 in.

The deflection at C due to a temperature differential of only 5°F is from Eq. (f):

= \frac{\left(6.5 \times 10^{-6}\right)(5)(180  in .)(360  in .+180  in .)}{2(30  in .)}=0.053  in .                (h)

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