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

(a) Roller support at C. This beam (Fig. 10-22a) is statically indeterminate to the first degree (see discussion in Example 10-3 solution). We select reac-tion R_{C} as the redundant which allows us 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 tem-perature differential). We will use the method of superposition, also known as the force or flexibility method, to find the solution.

Superposition. The superposition process is shown in the Figs. 10-22b and 10-22c in which redundant R_{C} is removed to produce a released (or statically determinate) structure. We first apply the “actual loads” [here, temperature differential (T_{2} – T_{1} )], and then we apply the redundant as a load to the second released structure.

Equilibrium. Summing forces in the y direction in Fig. 10-22a (using a statics sign convention in which forces in the positive y direction are positive), we find that

R_{A}+R_{B}=-R_{C}    (a)

Summing moments about B (again using a statics sign convention in which counterclockwise is positive), we find

so

R_{\mathrm{A}}=\left(\frac{\mathrm{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 we could also find reactions RA and RB using superposition of the reactions shown in Figs. 10-22b and c, as follows: R_{A}=R_{A 1}+R_{A 2} and R_{B}=R_{B 1}+R_{B 2} , where R_{A 1} and R_{B 1} 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_{\mathrm{C1}} and \delta_{\mathrm{C2}} are shown in Figs. 10-22b and 10-22c for the released structures subject to temperature differential and applied redundant force R_{C}, respectively. Initially, \delta_{\mathrm{C1}} and \delta_{\mathrm{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. We can now use the results of Examples 9-5 and 9-19 to find displacements \delta_{\mathrm{C1}} and \delta_{\mathrm{C1}} . First, from Eq. (f) in Example 9-19, we see that

\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}:

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

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

so

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

noting that the negative result means that reaction force R_{C} is down-ward [for positive temperature differential (T_{2} – T_{1} ) ]. The result for R_{C} is now substituted 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 l \alpha\left(T_{2}-T_{1}\right)}{2 a h}\right]=\frac{-3 E l \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 \text { El } \alpha\left(T_{2}-T_{1}\right)}{2 a h}\left(1+\frac{a}{L}\right) \\&=\frac{3 E l \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.

Numerical example. In Example 9-19, we computed the upward displace-ment at joint C [see Eq. (h), Example 9-19] assuming that beam ABC is a steel wide flange, HE 700b (see Table E-1a), with a length L = 9 m an overhang a = L/2 , and is subject to temperature differentia T_{2} – T_{1} = 3° Celsius. From Table H-4, the coefficient of therma  expan- sion for structural steel is \alpha=12 \times 10^{-6}{ }^{\circ} \mathrm{C} . The modulus for steel is 210 GPa. Now we can find numerical values of reactions R_{A}, R_{B}, and R_{C} using Eqs. (g), (h), and (i):

We note that the reactions sum to zero as required for equilibrium.

(b) Spring support at C. Once again, we select reaction R_{C} as the redundant. However, R_{C} is now at the base of the elastic spring support. When redundant reaction R_{C} is applied to the second released structure, it will first compress the spring and then be applied to the beam at C, causing upward deflection.
Superposition. The superposition solution approach (i.e., force or flexi-bility method) follows that used previously and is shown in Fig. 10-23.

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, we see that compatibility of displacements requires:

\delta_{1}+\delta_{2}=\delta=0   (i)

Force-displacement and temperature-displacement relations. The spring is assumed to be unaffected by the differential change, so we conclude that 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_{C 1} . However, we must include the compression of the spring in the expression for \delta_{2}, so we get

\delta_{2}=\frac{R_{C}}{k}+\delta_{C}=\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. We can now substitute Eqs. (e) and (k) into compatibility with Eq. (j) and then 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), (c)], we find:

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)

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).