E of concrete was not given. What was assumed?

E = 25,000 MPa (M25, 5000 sqrt(fck))

How was the mid-span deflection of 0.175 mm checked?

Using 7PL^3/(768EI) for a propped cantilever with central load

Using PL^3/(48EI) for a simply supported beam

Using PL^3/(3EI) for a cantilever

What are the end reactions and the moment at the fixed end?

R_A = 13.75 kN, R_C = 6.25 kN, M_A = 37.5 kN m

R_A = 6.25 kN, R_C = 13.75 kN, M_A = 31.25 kN m

2078 Chaitra · 10 marks

Beam BMD, SFD and mid-span deflection (FEM)

Short answer

Model as two 5 m beam elements with assumed M25 concrete giving EI=1.042e6 kN m^2; fixed and roller supports leave v2, theta2, theta3 free. Solving gives mid-span deflection 0.175 mm downward, matching 7PL^3/768EI. Support reactions are 13.75 kN at A, 37.5 kN m at A, and 6.25 kN at C; SFD is +13.75 then -6.25 kN and BMD has -37.5, +31.25, 0 kN m.

Draw Bending Moment and Shear force diagrams of the concrete beam shown in the following figure using the finite element method. Take the beam section $500\ \text{mm} \times 1000\ \text{mm}$ . Also determine the deflections at the mid span.

[Figure: beam of total length 10 m (5 m + 5 m), fixed at the left end and roller supported at the right end, with a 20 kN point load at mid-span]

Data: $L=10$ m, two elements of 5 m, fixed at A (left), roller at C (right), 20 kN at mid-span B. Section $b\times h=500\times1000$ mm: $I=\dfrac{bh^3}{12}=\dfrac{0.5\times1^3}{12}=0.041667$ m $^4$ . $E$ of concrete is not given; assume M25 concrete, $E_c=5000\sqrt{f_{ck}}=25\,000$ MPa (IS 456, cl. 6.2.3.1), so

EI=25\times10^6\times0.041667=1.042\times10^{6}\ \text{kN m}^2

Nodes: 1 = A, 2 = B (mid-span), 3 = C. Units: kN, m.

Element stiffness

[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T

(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.) with $L=5$ m for both elements.

Assembly and boundary conditions

Boundary conditions: $v_1=\theta_1=0$ (fixed), $v_3=0$ (roller). Free DOFs: $v_2,\theta_2,\theta_3$ . Load vector: $F_{v2}=-20$ kN.

EI\,\begin{bmatrix}0.1920 & 0 & 0.2400 \\ 0 & 1.6000 & 0.4000 \\ 0.2400 & 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}v_2\\\theta_2\\\theta_3\end{Bmatrix}=\begin{Bmatrix}-20.0 \\ 0 \\ 0\end{Bmatrix}

(the matrix shown is $[K]/EI$ ). Solving:

v_2=-0.1750\ \text{mm},\quad \theta_2=-1.5\times10^{-5}\ \text{rad},\quad \theta_3=6\times10^{-5}\ \text{rad}

Mid-span deflection = 0.175 mm downward. Check with the standard result for a propped cantilever with a central load: $\dfrac{7PL^3}{768EI}=\dfrac{7\times20\times1000}{768\times1.042\times10^{6}}=0.175$ mm. It agrees.

End forces

Element	$V_i$ (kN)	$M_i$ (kN m)	$V_j$ (kN)	$M_j$ (kN m)
AB	13.75	37.50	-13.75	31.25
BC	-6.25	-31.25	6.25	0.00

Reactions: $R_A=13.75$ kN, $M_A=37.5$ kN m (counter-clockwise), $R_C=6.25$ kN. Check: $13.75+6.25=20$ kN.

Shear force diagram (kN)

A to B: $+13.75$ (constant)
B to C: $-6.25$ (constant)

Bending moment diagram (kN m, sagging +)

A: $-37.50$ (hogging); B: $+31.25$ (13.75 x 5 - 37.5); C: $0$ .
Zero moment at $x=37.5/13.75=2.73$ m from A.

 A(fixed)                    B(mid)                C(roller)
 |---------------------------|---------------------|
 BMD: -37.5                  +31.25                0   (kN m)
 SFD: +13.75 (constant)      -6.25 (constant)

✓ Answer checked by Claude Opus 5.5 (high effort)

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