Question 1 of 3: Three-bar truss with a heated member

Question 1: Three-bar truss with a heated member (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

From Figure 1 the inclined bars 1–2 and 1–4 make 45° with the vertical bar 1–3, and $L$ is the vertical distance from joint 1 up to the line of supports. Their length is therefore $L\sqrt{2} = 2828.4$ mm while the vertical bar is $L = 2000$ mm long; all three share the same $EA$, so they do not share the same axial stiffness $EA/L_b$.

Find. The two displacement components of the single free joint and the axial force in each of the three bars, for the combined action of the mechanical load $P$ and the temperature rise in bar 1–2; then a statement of how the same procedure is carried into three dimensions.

[Figure not reproduced: Figure 1 redrawn from the examination paper: joint 1 is the only free node; supports 2, 3 and 4 lie on one horizontal line a distance L above it. Bars 1–2 and 1–4 stand at 45° either side of the vertical bar 1–3, so each is L√2 long. Only bar 1–2 (red) is heated. See the official exam paper.]

Approach. Assemble the 2×2 stiffness of the one free joint from the three bar contributions, convert the temperature rise in bar 1–2 into an equivalent nodal load, superpose it on $P$, solve for $\{u_1, v_1\}$, and recover each bar force from the axial strain less the free thermal strain of that bar.

1. Part 1.1 — set up the geometry and the direction cosines. Put the origin at joint 1 with $x$ to the right and $y$ upward, so the nodes are $1(0,0)$, $2(-L, L)$, $3(0, L)$ and $4(L, L)$. Measuring the orientation angle from the positive $x$-axis toward the far end of each bar, $$\begin{aligned} \text{bar }1\text{-}2:&\quad C=-\tfrac{1}{\sqrt2},\ S=\tfrac{1}{\sqrt2},\ L_b=L\sqrt2 \\ \text{bar }1\text{-}3:&\quad C=0,\ S=1,\ L_b=L \\ \text{bar }1\text{-}4:&\quad C=\tfrac{1}{\sqrt2},\ S=\tfrac{1}{\sqrt2},\ L_b=L\sqrt2 \end{aligned}$$ Nodes 2, 3 and 4 are pinned, so their four displacement components are prescribed zero and only the two components at joint 1 survive the assembly.
2. Assemble the active 2×2 stiffness. The upper-left 2×2 block of the supplied $[k]$ is what a bar contributes to its own first node, so $$[K] = \sum_b \frac{EA}{L_b}\begin{bmatrix} C^2 & CS \\ CS & S^2 \end{bmatrix} = \frac{EA}{L}\begin{bmatrix} \dfrac{1}{\sqrt2} & 0 \\\ 0 & 1+\dfrac{1}{\sqrt2}\end{bmatrix}.$$ The off-diagonal terms of the two inclined bars carry opposite signs and cancel, which is the algebraic statement of the geometric symmetry of the frame. With $EA/L = 6.00\times10^{8}/2000 = 3.000\times10^{5}$ N/mm, $$K_{xx}= 212\,132\ \text{N/mm}, \qquad K_{yy}= 512\,132\ \text{N/mm}.$$
3. Convert the temperature rise into an equivalent nodal load. A bar free to expand would lengthen by $\alpha\,\Delta T\,L_b$. Restraining that elongation requires the compressive force $EA\alpha\Delta T$, so in the finite element equations the heated bar is replaced by an unheated bar plus the pair of self-equilibrated nodal forces that push its two ends apart: $$\{f\}^{\text{th}} = EA\,\alpha\,\Delta T\begin{Bmatrix} -C \\ -S \\ C \\ S \end{Bmatrix}, \qquad EA\,\alpha\,\Delta T = (6.00\times10^{8})(12\times10^{-6})(30) = 216.0\ \text{kN}.$$ Note that this force does not depend on the bar's length. For bar 1–2, whose first node is joint 1, the load delivered to joint 1 is $$\{f_1\}^{\text{th}} = 216.0\begin{Bmatrix} +1/\sqrt2 \\ -1/\sqrt2\end{Bmatrix} = \begin{Bmatrix} +152.74 \\ -152.74 \end{Bmatrix}\ \text{kN},$$ i.e. the expanding member shoves joint 1 down and to the right, away from support 2.
4. Solve the two uncoupled equations. The applied load contributes $\{0,\,-P\}$, so with $[K]$ diagonal the system separates: $$u_1 = \frac{EA\alpha\Delta T/\sqrt2}{EA/(L\sqrt2)} = \alpha\,\Delta T\,L, \qquad v_1 = -\,\frac{L}{EA}\,\frac{P + EA\alpha\Delta T/\sqrt2}{1+1/\sqrt2}.$$ The horizontal result is worth pausing over: the geometry cancels completely and joint 1 simply moves the free thermal elongation of the heated bar measured horizontally. Putting in the numbers, with $P$ in kN and the answer in mm, $$\boxed{\begin{aligned} u_1 &= \alpha\Delta T\,L = 0.720\ \text{mm (to the right)} \ v_1 &= -\left(0.2982 + 1.9526\times10^{-3}P\right)\ \text{mm (downward)} \end{aligned}}$$ The temperature alone therefore drops joint 1 by 0.298 mm; every additional kilonewton of $P$ adds a further 1.95 micrometres.
5. Recover the bar forces from the strain, less the thermal strain. The elongation of a bar follows from the printed formula, and the axial force is the elastic strain only — the free thermal part carries no stress: $$e_b = C(u_j-u_i)+S(v_j-v_i), \qquad N_b = \frac{EA}{L_b}\,e_b \;-\; EA\,\alpha\,\Delta T_b .$$ With nodes 2, 3, 4 fixed, $e_b = -(C u_1 + S v_1)$. Taking each bar in turn and writing $P$ in kN, $$\begin{aligned} N_{12} &= \frac{EA}{2L}\,(u_1-v_1) - EA\alpha\Delta T = -63.26 + 0.2929\,P \ \text{kN},\\ N_{13} &= \frac{EA}{L}\,(-v_1) = +89.47 + 0.5858\,P \ \text{kN},\\ N_{14} &= \frac{EA}{2L}\,(-u_1-v_1) = -63.26 + 0.2929\,P \ \text{kN}. \end{aligned}$$ The two inclined bars come out with identical force even though only one of them is heated. That is not a coincidence: horizontal equilibrium of joint 1 requires $(N_{14}-N_{12})/\sqrt2 = 0$, and the heated bar sheds exactly enough of its blocked thermal force through the sideways movement $u_1$ to make it so.
6. Check joint equilibrium. A tensile force $N_b$ pulls joint 1 toward the far support, i.e. along $(C, S)$. Summing for the thermal-only case ($P=0$), $$\begin{aligned} \sum F_x &= \tfrac{1}{\sqrt2}\bigl(-(-63.26) + (-63.26)\bigr) = 0, \\ \sum F_y &= \tfrac{1}{\sqrt2}\bigl(-63.26-63.26\bigr) + 89.47 = -89.47+89.47 = 0 . \end{aligned}$$ Both close to the digits carried, and the same check closes for any $P$ because $0.2929\sqrt{2} + 0.5858 = 1$ exactly. A worked numerical case is given in the callout below.

Part 1.2 — extending the procedure to a three-dimensional structure with a temperature gradient. Nothing in the method above is two-dimensional in principle; what changes in three dimensions is the size of the arrays and the number of direction cosines. Each space-truss bar now has three direction cosines $C_x = (x_j-x_i)/L_b$, $C_y = (y_j-y_i)/L_b$ and $C_z = (z_j-z_i)/L_b$, and its 6×6 global stiffness matrix is the outer product $[k] = \dfrac{EA}{L_b}\begin{bmatrix} \mathbf{c}\mathbf{c}^{T} & -\mathbf{c}\mathbf{c}^{T} \\ -\mathbf{c}\mathbf{c}^{T} & \mathbf{c}\mathbf{c}^{T}\end{bmatrix}$ with $\mathbf{c} = \{C_x, C_y, C_z\}^{T}$ — the exact three-dimensional analogue of the matrix printed on the examination paper. For frame members one adds torsion and bending about two axes and the transformation becomes a 3×3 rotation applied to each of four sub-blocks, but the assembly rule is unchanged.

The temperature field is handled as an initial strain, and this is the part of the procedure worth stating carefully. One does not attempt to add temperature as an unknown. Instead, for every element the free thermal strain $\{\varepsilon_0\} = \alpha\,\Delta T\{1,1,1,0,0,0\}^{T}$ — isotropic, with no shear component — is evaluated from the temperature rise at that element, and the work-equivalent nodal load

is computed and added to the mechanical load vector. For the two-node bar this integral collapses to the pair of forces $\pm EA\alpha\Delta T$ along the member axis used in step 3 above; for a solid or plate element it is a genuine integral, usually evaluated at the same Gauss points as the stiffness. A temperature gradient rather than a uniform rise simply means $\Delta T$ varies with position: interpolate the nodal temperatures with the same shape functions used for displacement, $\Delta T(x,y,z) = \sum_i N_i \Delta T_i$, so that the thermal load is integrated consistently element by element. Through the thickness of a beam or plate the linear part of the gradient produces a thermal curvature and hence equivalent end moments $M^{\text{th}} = EI\alpha\,(\Delta T_{\text{top}}-\Delta T_{\text{bot}})/h$, alongside the axial force from the mean rise.

The solution sequence is then exactly the one used in 1.1. Assemble $[K]\{D\} = \{R\}_{\text{mech}} + \{R\}_{\text{th}}$; apply the displacement boundary conditions; solve for the free displacements; and finally recover element stresses from $\{\sigma\} = [D]\bigl([B]\{d\} - \{\varepsilon_0\}\bigr)$, subtracting the free thermal strain before multiplying by the constitutive matrix. Two consequences deserve to be stated to a client or a checker. First, the thermal load vector is self-equilibrated: on a statically determinate structure it produces displacements but no member forces at all, so any computed thermal stress is a measure of redundancy, not of temperature as such. Second, because the analysis is linear, mechanical and thermal effects superpose, which is precisely why the answers in 1.1 could be written as a constant plus a multiple of $P$. If the temperature change is large enough to move $E$ or $\alpha$ appreciably, or if gaps open and close, that superposition fails and the problem must be stepped incrementally.