What is done to the global equations because the bar is fixed at both ends?

Rows and columns 1 and 5 are deleted since u_1 = u_5 = 0

What are the support reactions and their directions?

R_1 = 0.8P and R_5 = 0.2P, both opposite to P

R_1 = 0.5P and R_5 = 0.5P, along +x

R_1 = 0.2P and R_5 = 0.8P, both opposite to P

R_1 = P and R_5 = 0, opposite to P

2071 Bhadra · 8 marks

Stepped bar displacements and support forces

Short answer

Use four bar elements of length l/4 with areas A, 1.5A, 0.5A, 0.75A and fixed ends. Nodal displacements are u2=Pl/(5AE), u3=Pl/(6AE), u4=Pl/(15AE). Support reactions are 0.8P and 0.2P, both opposite to the applied load.

For the given stepped bar obtain the nodal displacements at nodes 2, 3 and 4. Also obtain the forces developed at the supports. Take $E$ = constant and cross-sectional areas as indicated in the figure.

[Figure: bar fixed at both ends (nodes 1 and 5), four equal elements each of length $l/4$ between nodes 1-2-3-4-5; cross-sectional areas $A$ (element 1), $3A/2$ (element 2), $A/2$ (element 3), $3A/4$ (element 4); a point load $P$ acts at node 2 in the +x direction]

Take $k=AE/l$ as the unit of stiffness. Each element has length $l/4$ , so $k_e = A_eE/(l/4)=4A_eE/l$ .

Element stiffnesses

Element	Nodes	Area	$k_e$
1	1-2	$A$	$4k$
2	2-3	$3A/2$	$6k$
3	3-4	$A/2$	$2k$
4	4-5	$3A/4$	$3k$

Each element matrix is $k_e\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$ .

Assembled global equations

k\begin{bmatrix}4&-4&0&0&0\\-4&10&-6&0&0\\0&-6&8&-2&0\\0&0&-2&5&-3\\0&0&0&-3&3\end{bmatrix}\begin{bmatrix}u_1\\u_2\\u_3\\u_4\\u_5\end{bmatrix}=\begin{bmatrix}R_1\\P\\0\\0\\R_5\end{bmatrix}

Boundary conditions: $u_1=u_5=0$

Deleting rows and columns 1 and 5:

\frac{AE}{l}\begin{bmatrix}10&-6&0\\-6&8&-2\\0&-2&5\end{bmatrix}\begin{bmatrix}u_2\\u_3\\u_4\end{bmatrix}=\begin{bmatrix}P\\0\\0\end{bmatrix}

From row 3: $u_4=0.4u_3$ . Row 2: $-6u_2+8u_3-2(0.4u_3)=0\Rightarrow u_3=\dfrac{6}{7.2}u_2=\dfrac{5}{6}u_2$ . Row 1: $10u_2-6\cdot\dfrac56u_2=5u_2=\dfrac{Pl}{AE}$ .

u_2=\frac{Pl}{5AE},\qquad u_3=\frac{Pl}{6AE},\qquad u_4=\frac{Pl}{15AE}

Support reactions

R_1=4k(u_1-u_2)=-\frac{4P}{5}=-0.8P,\qquad R_5=3k(u_5-u_4)=-\frac{P}{5}=-0.2P

Both act in the $-x$ direction (towards the left). Check: $P-0.8P-0.2P=0$ .

Element forces (check)

$F_1=k_1(u_2-u_1)=+0.8P$ (tension); $F_2=6k(u_3-u_2)=-0.2P$ ; $F_3=2k(u_4-u_3)=-0.2P$ ; $F_4=3k(u_5-u_4)=-0.2P$ (compression). Equilibrium at node 2: $P-0.8P-0.2P=0$ .

Answer: $u_2=\dfrac{Pl}{5AE},\ u_3=\dfrac{Pl}{6AE},\ u_4=\dfrac{Pl}{15AE}$ ; $R_1=0.8P$ and $R_5=0.2P$ , both acting opposite to $P$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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