## Q. 3.11

Two elements are stress-free when they are welded together at B and welded to rigid walls at A and C (Fig. 1). Subsequently, element (1) is heated by an amount ΔT, while element (2) is held at the reference temperature.
Determine an expression for the axial force induced in each element, and determine an expression for the displacement $u_B$ of the joint at B. ## Verified Solution

Plan the Solution This problem can be solved by using the same Basic- Force-Method steps that were employed in Section 3.5. To incorporate the thermal strain, we just need to use Eq. 3.27 instead of Eq. 3.14. Since element (1) is heated, it will push on element (2), and both elements will therefore have an induced compressive stress.

$e = fF = αLΔT, f = \frac{L}{AE}$           (3.27)

$e = fF, \text{where} f ≡ \frac{L}{AE}$        (3.14)

Equilibrium: Although there are no external loads, that is, no external forces other than the reactions at A and C, we still need an equilibrium equation of the joint at B in order to relate the element forces to each other (Fig. 2). Note that tension is assumed positive.

$\underrightarrow{+} \sum{F_x}=0: -F_1 + F_2 = 0$    Equilibrium      (1)

Since there are two unknown forces but only one equilibrium equation, this is a statically indeterminate system of bars. Therefore, we must turn to the geometry of deformation for an appropriate compatibility equation. To prepare for this, however, we first establish the correct form for relating the element forces to the total elongations of the elements, including thermal effects.
Element Force-Temperature-Deformation Behavior: These are the equations through which the temperature effect directly enters the solution, and they also relate the forces in the equilibrium equation to the elongations in the deformation-geometry equation. Equation 3.27 is the appropriate equation to use. We write this equation for each element, noting that $ΔT_1 = ΔT$ and $ΔT_2 = 0.$

$e_1 = f_1F_1 + α_1L_1ΔT$      Element Force-Temperature-Deformation Behavior  (2a,b)

$e_2 = f_2F_2$

where $f_i = (L/AE)_i.$

Geometry of Deformation: The elongations $e_1$ and $e_2$ in Eqs. (2) are the $\underline{\text{total elongations}}$ of the respective elements, due to both stress and temperature.
Therefore, the compatibility equation is simply

$e_{AC} = e_1 + e_2$ = 0    Geometry of Deformation  (3)

Solution for Element Forces: Following the procedure of the Basic Force Method, we can eliminate the e’s by substituting Eqs. (2) into Eq. (3), giving the following compatibility equation in terms of forces:

$f_1F_1 + f_2F_2 = -α_1L_1ΔT$    Compatibility in Terms of Element Forces  (4)

Inserting expressions for the $f_i$’s into Eq. (4), and solving Eqs. (1) and (4) simultaneously, we get

$F_1 = F_2 = \frac{-A_1A_2Eα_1L_1ΔT} {A_1L_2 + A_2L_1}$     Ans. (5)

Solution for the Displacement of B: To obtain an expression for the displacement $u_B$ we can see from Fig. 1 that element (1) is elongated by an amount $u_B$ while element (2) is shortened by an amount $u_B$, which leads to Eq. (3). Since Eq. (2b) is shorter than Eq. (2a), we will use

$e_2 = -u_B$    Geometry of Deformation  (6)

so

$u_B = -f_2F_2 = \left( \frac{A_1L_2} {A_1L_2 + A_2L_1}\right)(α_1L_1ΔT)$      Ans. (7)

Review the Solution As we would expect, heating element (1) causes a compression to be induced in both elements and, since there is no external force on the joint connecting the two elements, the same compressive force is induced in both elements. The right-hand side of Eq. (5) has the proper dimensions of force, and the right-hand side of Eq. (7) has the proper dimensions of displacement. Also, it can be seen that the displacement $u_B$ is less than the free expansion of element (1) would be if element (2) were not there. 