Deflection of Beam MCQ Quiz - Objective Question with Answer for Deflection of Beam - Download Free PDF

Explanation:

1. A fixed beam has its ends fixed, meaning it cannot rotate or displace at those ends. Due to the constraints:

2. The deflection at the fixed ends is zero because the beam is not allowed to move vertically at the supports.

3. The slope at the fixed ends is also zero because there is no rotation allowed at the fixed supports.

Additional Information

Simply Supported Beam

Definition: A simply supported beam is one that is supported at both ends, with one end typically resting on a roller and the other on a hinge or pin.

Key Characteristics:

1. The beam is free to rotate at the supports, meaning there is no moment resistance at the supports.

2. It experiences vertical reactions at the supports but no bending moments at the supports.

3. The deflection is present along the length of the beam, and it is typically greatest at the midpoint.
4. Typical Applications: Bridges, building floors, and roof structures.

Concept:

Moment Area Method

$\frac{d^{2}y}{d{x}^2}=\int \frac{Mx}{E{I}_x}dx$

${\int }_A^B\theta ={\int }_A^B\frac{Mx}{E{I}_x}dx$

Mohr's Theorem No.1

${\theta }_B-{\theta }_A=[Areaof\frac{BMD}{EI}diagrambetweenAandB]$

Mohr's Theorem No.2

​$Y_B-Y_A=[MomentofAreaof\frac{BMD}{EI}diagrambetweenBandA]$

where A = Reference point where slope zero, B = Origin point where slope and deflection are to be determined.

The distance of centroid of the area should be measured from the origin.

Calculation:

As we have to find the deflection at the midpoint (C) the area of half the bending moment diagram can be considered. 

Area of the bending moment diagram is given by:

$A=\frac{1}{2}\times \frac{WL}{4}\times \frac{L}{2}$

$A=\frac{W{L}^2}{16}$

Area of the $\frac{BMD}{EI}$ diagram = $\frac{W{L}^2}{16EI}$

x̅ = Distance of the centroid of the Bending Moment Diagram from the origin

$\overline{x}=\frac{2}{3}\frac{L}{2}=\frac{L}{3}$

$Deflection=\frac{A\overline{x}}{EI}=\frac{W{L}^2}{16EI}\times \frac{L}{3}=\frac{W{L}^3}{48EI}$

Additional Information

Deflection and slope of various beams are given by:

Where, y = Deflection of beam, θ = Slope of beam

Concept:

For a simply supported beam carrying a uniformly distributed load (UDL) of intensity $W$ across the full span $L$, the slope at the supports is given by:

$\theta =\frac{W{L}^3}{24EI}$

Where:

• $W$ = Load per unit length
• $L$ = Length of beam
• $E$ = Modulus of elasticity
• $I$ = Moment of inertia

Explanation:

Deflection and slope of various beams are given by:

Where, y = Deflection of beam, θ = Slope of beam

Explanation:

The deflection value for different conditions are as tabulated below:

Concept:

Deflection at centre of a simply supported beam due to point load (W):

${\delta }_1=\frac{W{L}^3}{48EI}$

Deflection at centre of a simply supported beam due to uniformly distribute load (W = wL):

${\delta }_2=\frac{5w{L}^4}{384EI}=\frac{5W{L}^3}{384EI}(\because w=\frac{W}{L})$

Calculation:

Given:

$\frac{{\delta }_2}{{\delta }_1}=\frac{5W{L}^3}{384EI}\times \frac{48EI}{W{L}^3}=\frac{5}{8}$

Hence,

The deflection will be reduced by

$={\delta }_1-{\delta }_2=\left(1-\frac{5}{8}\right){\delta }_1=\frac{3}{8}{\delta }_1$

Concept:

Slope for cantilever beam with point load,

$\theta =\frac{P{L}^2}{2EI}=0.001$

∴ $\frac{L^2}{2EI}=\frac{0.001}{P}$

Now, deflection for cantilever beam with moment,

$\mathrm{\Delta }=\frac{M{L}^2}{2EI}=M\times \frac{0.001}{P}=100\times \frac{0.001}{10}=0.01m=10mm$

Concept:

Area moment method

Theorem 1: The difference of slope of any two points of a beam is equal to the area of$\frac{M}{EI}$ diagram between those points.

θ_B – θ_A = Area of $\frac{M}{EI}$ diagram between B and A.

Theorem 2: The difference of deflection of two points of a beam is equal to the moment of area of$\frac{M}{EI}$diagram between those points.

Y_B – Y_A = Moment of area of$\frac{M}{EI}$diagram between B and A.

Y_B – Y_A = (Ax̅)

Calculation:

δ_max = Ax̅ 

${\delta }_{max}=\frac{1}{3}\times L\times \frac{W{L}^2}{2EI}\times \frac{3}{4}L$

$\Rightarrow {\delta }_{max}=\frac{W{L}^4}{8EI}$

Concept: -

We know,

The radius of curvature is given by –

$R=\frac{{\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^{\frac{3}{2}}}{\frac{d^{2}y}{d{x}^2}}$

Also, from bending equation,

$R=\frac{EI}{M}$

Here, EI – flexural rigidity and M – moment

Assuming small deflection of beam, such that the slope of the elastic curve dy / dx is very small. Then the term (dy / dx)² can be neglected.

Hence,

$R=\frac{1}{\frac{d^{2}y}{d{x}^2}}=\frac{EI}{M}$

$\Rightarrow M=EI\frac{d^{2}y}{d{x}^2}$

Concept:

Deflection of cantilever beam due to point load = $\frac{P{l}^3}{3EI}$

Deflection of cantilever beam due to udl = $\frac{w{l}^4}{8EI}$

Deflection of SSB due to point load at mid point = $\frac{P{l}^3}{48EI}$

Solution:

Deflection at the end of cantilever beam is given by,

$=\frac{w{l}^4}{8EI}-\frac{R{l}^3}{3EI}$

Deflection at midpoint of SSB.

=$\frac{R{l}^3}{48EI}$

Equating both the cases,

$\frac{w{l}^4}{8EI}-\frac{R{l}^3}{3EI}=\frac{R{l}^3}{48EI}$

$\frac{w{l}^4}{8EI}=\frac{R{l}^3}{48EI}+\frac{R{l}^3}{3EI}$

$\frac{w{l}^4}{8EI}=(\frac{1+16}{48})\frac{R{l}^3}{EI}$

$R=\frac{6wl}{17}$

Concept:

The slope at the free end of a cantilever beam with varying flexural rigidity is determined using the moment-area theorem.

Given:

• Flexural rigidity up to length L/2 from the fixed end: $EI$
• Flexural rigidity for the remaining length: $EI/2$
• Moment applied at the free end: $M$

Calculation:

Using the moment-area theorem, the total slope at the free end is given by the sum of contributions from both segments.

For the first segment (0 to L/2) with flexural rigidity $EI$:

${\theta }_1=\frac{M(L/2)}{EI}$

For the second segment (L/2 to L) with flexural rigidity $EI/2$:

${\theta }_2=\frac{M(L/2)}{(EI/2)}=\frac{2M(L/2)}{EI}=\frac{ML}{EI}$

Total slope at the free end:

$\theta ={\theta }_1+{\theta }_2=\frac{ML}{2EI}+\frac{ML}{EI}$

$\theta =\frac{3ML}{2EI}$

Final Answer: $\frac{3ML}{2EI}$

Calculation:

Deflection of end B = 0

(Downward deflection due to uniformly distributed load) – (upward deflection due to RB) = 0

$\frac{w{l}^4}{8EI}-\frac{R_B{l}^3}{3EI}=0$

$\begin{array}{l}{R}_B=\frac{3}{8}wl\\ R_A=wl-\frac{3}{8}wl=\frac{5}{8}wl\end{array}$

Here R_B is the reaction of the prop.

$R_B=\frac{3}{8}\mathrm{W}\mathrm{l}\left(\mathrm{k}\mathrm{N}\right)$

Concept:

The standard deflection and slope formulas of simply supported beam under uniformly distributed load is given below:

Calculation:

Given:

L₁ = L, L₂ = 2L

Deflection under UDL at centre,

$y_c=\frac{5}{384}\frac{w{L}^4}{EI}$

⇒ $y_c\propto L^4$

$\frac{y_{c1}}{y_{c2}}=(\frac{L_1}{L_2}{)}^4$

$\frac{y_{c1}}{y_{c2}}=(\frac{L}{2L}{)}^4$

$\frac{y_{c1}}{y_{c2}}=\frac{1}{16}$

y_c2 = 16 y_c1

Important Points

Deflection and slope of various beams are given by: 

Where, y = Deflection of beam, θ = Slope of beam

Explanation:

Deflection of end B = 0

⇒ downward deflection due to udl – upward deflection due to RB = 0

$\begin{array}{l}\Rightarrow \frac{w{l}^4}{8EI}-\frac{R_B{l}^3}{3EI}=0\\ \Rightarrow R_B=\frac{3}{8}wl\\ \therefore R_A=wl-\frac{3}{8}wl=\frac{5}{8}wl\end{array}$

Hence, the reaction at the end B is $\frac{3}{8}wl$.

Explanation: