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 RC 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
\quad\quad\quad\quad R_{A}+R_{B} {=}-R_{C}\quad\quad(a)
Sum moments about B (again using a statics sign convention in which counterclockwise is positive) to find
\quad\quad\quad\quad -R_{A}L+R_{C}a=0
so
\quad\quad\quad\quad R_{A}=\left({\frac{a}{L}}\right)R_{C}(b)
which can be subsitituted back into Eq. (a) to give
\quad\quad\quad\quad R_{B}=-R_{C}-{\Bigg\lgroup}{\frac{\alpha}{L}}{\Bigg\rgroup}R_{C}=-R_{C}{\Bigg\lgroup}1+{\frac{\alpha}{L}}{\Bigg\rgroup}\quad\quad (c)
(Note that you could also find reactions R_{A1} ~ and ~ R_{B1} 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
\quad\quad\quad\quad \delta_{C1}+\delta_{C2}=\delta_{C}=0\quad\quad (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,
\quad\quad\quad\quad \delta_{C1}={\frac{\alpha(T_{2}-T_{1})\alpha(L+a)}{2h}}\quad\quad (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),
\quad\quad\quad\quad \delta_{C2}={\frac{R_{C}a^{2}(L+\alpha)}{3E I}}\quad\quad (f)
Reactions: Now substitute Eqs. (e) and (f) into Eq. (d) and then solve for redundant R_C:
\quad\quad\quad\quad{\frac{\alpha(T_{2}-T_{1})}{2h_{-}}}(a)(L+a)+{\frac{R_{C}a^{2}(L+a)}{3E l}}= {0}
so
\quad\quad\quad\quad R_{C}={\frac{-3E I\alpha(T_{2}-T_{1})}{2\alpha h}}\quad\quad (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

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, HE 700B (see Table F-1), with a length L = 9 m, an overhang a = L/2, and subject 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. Now find numerical values of reactions R_A, R_B, ~ and ~ R_C using Eqs. (g), (h), and (i):
\quad\quad\quad\quad R_A = \frac{- 3{E}{I}\alpha({T}_{2} – {T}_{1})}{2Lh} = \frac{-3(210~GPa)(256900~ cm^4)(12 \times 10^{-6})(3)}{2(9 ~M)(700 ~mm)} \\ \quad\quad\quad\quad = -4.62 kN (downward) \\ \quad\quad\quad\quad R_B = \frac{ 3{E}{I}\alpha({T}_{2} – {T}_{1})(L + \alpha)}{2L\alpha h} \\ \quad\quad\quad\quad\frac{3(210~GPa)(256900~ cm^4)(12 \times 10^{-6})(3)(9~m + 4.5~m)}{2(9 ~M)(4.5~m)(700 ~mm)} \\ \quad\quad\quad\quad = 13.87 kN ~(upward) \\ \quad\quad\quad\quad R_C = \frac{- 3{E}{L}\alpha({T}_{2} – {T}_{1})}{2\alpha h} = \frac{-3(210~GPa)(256900~ cm^4)(12 \times 10^{-6})(3)}{2(4.5 ~M)(700 ~mm)} \\ \quad\quad\quad\quad = -9.25 kN (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:
\quad\quad\quad\quad \delta_{1}+\delta_{2}= \delta =0\quad\quad (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
\quad\quad\quad\quad \delta_{2}={\frac{R_{C}}{k}}\,+\,\delta_{C2}\,=\,{\frac{R_{C}}{k}}\,+\,{\frac{R_{C}a^{2}(L\,+\,a)}{3E I}}\quad \quad(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:
\quad\quad\quad\quad \frac{\alpha(T_{2}-T_{1})}{2h}(a)(L+a)+\frac{{R}_{C}a^{2}(L+a)}{3E I}+\frac{{R}_{C}}{k}=0
so
\quad\quad\quad\quad R_{C}=\frac{-a\alpha(T_{2}-T_{1})(L+a)}{2h\left[\frac1k+\frac{a^{2}(L+a)}{3E I}\right]},\quad\quad(l)
From statics [Eqs. (b) and (c)], reactions at A and B are
\quad\quad\quad\quad R_{A} = {\Bigg\lgroup}\frac{a}{{L}}{\Bigg\rgroup}{R}_{C} = \frac{-\alpha\alpha(T_{2}-T_{1})\alpha(L+a)}{2Lh\left[\frac1k+\frac{a^{2}(L+a)}{3E I}\right]}\quad\quad(m)
\quad\quad\quad\quad R_{B} = -{R}_{C} {\Bigg\lgroup}1 + \frac{a}{{L}}{\Bigg\rgroup} = \frac{\alpha\alpha(T_{2}-T_{1})(L+a)^2}{2Lh\left[\frac1k+\frac{a^{2}(L+a)}{3E I}\right]}\quad\quad(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).