Question 10.7: The two-span beam ABC in Fig. 10-22 has a pin support at A, ...

Use a four-step problem-solving approach.
Part (a): Roller support at C.
1. Conceptualize:
Roller support at C: This beam (Fig. 10-22a) is statically indeterminate to the first degree (see discussion in Example 10-3 solution). Select reaction R_{C} as the redundant in order to use the analyses of the released structure (with the support at C removed) presented in Examples 9-5 (concentrated load applied at C) and 9-19 (subject to temperature differential). Use the method of superposition, also known as the force or flexibility method, to find the solution.

2. Categorize:
Superposition: The superposition process is shown in the Figs. 10-22b and c in which redundant R_{C} is removed to produce a released (or statically determinate) structure.

First apply the “actual loads” [here, temperature differential (T_{2} – T_{1}) ], and then apply the redundant R_{C} as a load to the second released structure.

3. Analyze:
Equilibrium: Sum forces in the y direction in Fig. 10-22a (using a statics sign convention in which upward forces in the y direction are positive) to find that

R_{A}+R_{B}=-R_{C}             (a)

Sum moments about B (again using a statics sign convention in which counterclockwise is positive) to find

so

R_{A}=\left(\frac{a}{L}\right) R_{C}            (b)

which can be subsitituted back into Eq. (a) to give

R_{B}=-R_{C}-\left(\frac{a}{L}\right) R_{C}=-R_{C}\left(1+\frac{a}{L}\right)              (c)

(Note that you could also find reactions R_{A}  and  R_{B} using superposition of the reactions shown in Figs. 10-22b and c: R_{A} = R_{A1} + R_{A2}  and  R_{B} = R_{B1} + R_{B2}, where R_{A1}  and  R_{B1} are known to be zero.)

Compatibility: Displacement \delta_{C} = 0 in the actual structure (Fig. 10-22a), so compatibility of displacements requires that

\delta_{C 1}+\delta_{C 2}=\delta_{C}=0            (d)

where \delta_{C1}  and  \delta_{C2} are shown in Figs. 10-22b and c for the released structures subject to temperature differential and applied redundant force R_{C}, respectively. Initially, \delta_{C1}  and  \delta_{C2} are assumed positive (upward) when using a statics sign convention, and a negative result indicates that the reverse is true.

Force-displacement and temperature-displacement relations: Now use the results of Examples 9-5 and 9-19 to find displacements \delta_{C1}  and  \delta_{C2}. First, from Eq. (f) in Example 9-19,

\delta_{C 1}=\frac{\alpha\left(T_{2}-T_{1}\right) a(L+a)}{2 h}           (e)

and from Eq. (9-55) (modified to include variable a as the length of member BC and replacing load P with redundant force R_{C} – see solution to Problems 9.8-5(b) or 9.9-3),

\delta_{C}=-v\left(\frac{3 L}{2}\right)=\frac{P L^{3}}{8 E I}              (9-55)

\delta_{C 2}=\frac{R_{C} a^{2}(L+a)}{3 E I}             (f)

Reactions: Now substitute Eqs. (e) and (f) into Eq. (d) and then solve for redundant R_{C}:

so

R_{C}=\frac{-3 E I \alpha\left(T_{2}-T_{1}\right)}{2 a h}               (g)

noting that the negative result means that reaction force R_{C} is downward [for positive temperature differential (T_{2} – T_{1})]. Now substitute the expression for R_{C} into Eqs. (b) and (c) to find reactions R_{A} and R_{B} as

R_{A}=\left(\frac{a}{L}\right) R_{C}=\left(\frac{a}{L}\right)\left[\frac{-3 E I \alpha\left(T_{2}-T_{1}\right)}{2 a h}\right]=\frac{-3 E I \alpha\left(T_{2}-T_{1}\right)}{2 L h}             (h)

\begin{aligned}R_{B} &=-R_{C}\left(1+\frac{a}{L}\right)=\frac{3 E I \alpha\left(T_{2}-T_{1}\right)}{2 a h}\left(1+\frac{a}{L}\right) \\&=\frac{3 E I \alpha\left(T_{2}-T_{1}\right)(L+a)}{2 L a h}\end{aligned}             (i)

where R_{A} acts downward and R_{B} acts upward.

4. Finalize:
Numerical example: In Example 9-19, the upward displacement at joint C was computed [see Eq. (h), Example 9-19] assuming that beam ABC is a steel wide flange, W 30 × 211 (see Table F-1a), with a length L = 30 ft, an overhang a = L/2, and subject 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} /°F . The modulus for steel is 30,000 ksi. Now find numerical values of reactions R_{A}, R_{B}, and R_{C} using Eqs. (g), (h), and (i):

= – 1.395 kips (downward)

= \frac{3(30,000  ksi )\left(10,300  in ^{4}\right)\left(6.5 \times 10^{-6}\right)(5)(360  in .+180  in .)}{2(360  in .)(180  in .)(30  in .)}

= 4.18 kips (upward)

= – 2.79 kips (downward)

Note that the reactions sum to zero as required for equilibrium.

Part (b): Elastic spring support at C.

1. Conceptualize:
Spring support at C: Once again, select reaction R_{C} as the redundant. However, R_{C} is now at the base of the elastic spring support (see Fig. 10-23). When redundant reaction R_{C} is applied to the second released structure (Fig. 10-23c), it will first compress the spring and then be applied to the beam at C, causing upward deflection.
2. Categorize:
Superposition: The superposition solution approach (i.e., force or flexibility method) follows that used previously and is shown in Fig. 10-23.
3. Analyze:
Equilibrium: The addition of the spring support at C does not alter the expressions of static equilibrium in Eqs. (a), (b), and (c).

Compatibility: The compatibility equation is now written for the base of the spring (not the top of the spring, where it is attached to the beam at C). From Fig. 10-23, compatibility of displacements requires:

\delta_{1}+\delta_{2}=\delta=0          (j)

Force-displacement and temperature-displacement relations: The spring is assumed to be unaffected by the temperature differential, so the top and base of the spring displace the same in Fig. 10-23b, which means that Eq. (e) is still valid and \delta_{1} = \delta_{C1}. However, the compression of the spring must be included in the expression for \delta_{2}, so

\delta_{2}=\frac{R_{C}}{k}+\delta_{C 2}=\frac{R_{C}}{k}+\frac{R_{C} a^{2}(L+a)}{3 E I}          (k)

where the expression for \delta_{C2} comes from Eq. (f).

Reactions: Now substitute Eqs. (e) and (k) into compatibility with Eq. (j) and solve for redundant R_{C}:

so

R_{C}=\frac{-a \alpha\left(T_{2}-T_{1}\right)(L+a)}{2 h\left[\frac{1}{k}+\frac{a^{2}(L+a)}{3 E I}\right]}            (l)

From statics [Eqs. (b) and (c)], reactions at A and B are

R_{A}=\left(\frac{a}{L}\right) R_{C}=\frac{-a \alpha\left(T_{2}-T_{1}\right) a(L+a)}{2 L h\left[\frac{1}{k}+\frac{a^{2}(L+a)}{3 E I}\right]}              (m)

R_{B}=-R_{C}\left(1+\frac{a}{L}\right)=\frac{a \alpha\left(T_{2}-T_{1}\right)(L+a)^{2}}{2 L h\left[\frac{1}{k}+\frac{a^{2}(L+a)}{3 E I}\right]}          (n)

4. Finalize: Once again, the minus signs for R_{A}  and  R_{C} indicate that they are downward [for positive (T_{2} – T_{1})], while R_{B} is upward. Finally, if spring constant k goes to infinity, the support at C is once again a roller support, as in Fig. 10-22, and Eqs. (l), (m), and (n) reduce to Eqs. (g), (h), and (i).