Learning Objectives:

• Determine the deformations and/or normal stress in a member due to application of forces.

Elastic Deformation wherein body recovers its original shape after removal of force whereas Plastic Deformation permanent deformation (body does not recover its shape after forces are removed.

Solved Example: 47-1-01

A ductile material is defined as one, for which the plastic deformation before fracture:

A. Is smaller than the elastic deformation

B. Vanishes

C. Is equal to the elastic deformation

D. Is much larger than the elastic deformation

Correct Answer: D

Solved Example: 47-1-02

In a ductile material, post elastic strain is:

A. 1%

B. 2 - 3%

C. 3 - 5%

D. Greater than 5%

Correct Answer: D

Learning Objectives:

• Calculate deflection using the equation of the elastic curve.
• Derive the deflection and slope curves for a beam through integration of the moment-curvature relationship.

Differential Equation of deflection curve,

EI $\dfrac{d^2y}{dx^2}$= M

Integrating once, we get

$EI \dfrac{dy}{dx} = \displaystyle \int M dx$       Page 136

Integrating once more, we get

$EI y = \displaystyle \int \displaystyle \int M dx dx$       Page 136

The constants of integration can be eliminated by applying boundary conditions, such as: At (unyielding) supports, \[y = 0\] At fixed supports \[\dfrac{dy}{dx} = \mathrm{slope\ of\ deflected\ curve} = 0\]

Hermanoere, CC BY-SA 4.0, via Wikimedia Commons

Beam Deflection Formulae:

1. Cantilever Beam – Concentrated load P at the free end

   \[\delta_{max} = \dfrac{PL^3}{3EI}\]

2. Cantilever Beam – Concentrated load P at any point

   $\delta_{max} = \dfrac{Pa^2}{6EI}(3l-a)$       Page 141
   
   

3. Cantilever Beam – Uniformly distributed load w(N/m)

   $\delta_{max} = \dfrac{wL^4}{8EI}$       Page 141
   
   

4. Cantilever Beam – Uniformly varying load: Maximum intensity \(w_0\)(N/m)

   $\delta_{max} = \dfrac{w_0L^4}{30EI}$       Page 141
   
   

5. Cantilever Beam – Couple Moment M at the free end

   $\delta_{max} = \dfrac{ML^2}{2EI}$       Page 141
   
   

Solved Example: 47-2-01

Slope at a point in a beam is the:

A. Vertical displacement

B. Angular displacement

C. Horizontal displacement

D. None of these.

Correct Answer: B

Solved Example: 47-2-02

Which of the following is an elastic curve equation for shear force? (EI = flexural rigidity)

A. $S = EI \dfrac{dy}{dx}$

B. $S = EI \dfrac{d^2y}{dx^2}$

C. $S = EI \dfrac{d^3y}{dx^3}$

D. $S = EI \dfrac{d^4y}{dx^4}$

Correct Answer: C

Solved Example: 47-2-03

Which of the following statements is/are true for a simply supported beam?

A. Deflection at supports in a simply supported beam is maximum

B. Deflection is maximum at a point where slope is zero

C. Slope is minimum at supports in a simply supported beam

D. All of the above.

Correct Answer: B

Solved Example: 47-2-04

Which of the following is a differential equation for deflection?

A. $\dfrac{dy}{dx} = \dfrac{M}{EI}$

B. $\dfrac{dy}{dx} = \dfrac{MI}{E}$

C. $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$

D. $\dfrac{d^2y}{dx^2} = \dfrac{ME}{I}$

Correct Answer: C

Solved Example: 47-2-05

The equation of deformation is derived to be $y = x^3 - xL$ for a beam as shown in the figure. Curvature of the beam at the mid-span (in units, in integer) will be:

A. $2$

B. $2L$

C. $\dfrac{L}{2}$

D. $\dfrac{L}{3}$

\[y = x^2 - xL\] Curvature at the midsection is: \[\dfrac{1}{R} = \dfrac{d^2y}{dx^2}\] \[\dfrac{dy}{dx} = 2x - L\] \[\dfrac{d^2y}{dx^2} = 2\]

Correct Answer: A

Solved Example: 47-3-01

The simply supported beam 'A' of length l carries a central point load W. Another beam 'B' is loaded with a uniformly distributed load such that the total load on the beam is W. The ratio of maximum deflections between beams A and B is:

A. 5:8

B. 8:5

C. 5:4

D. 4:5

Max. deflection of beam A = $\dfrac{PL^3}{48EI}$
Max. deflection of beam B = $\dfrac{5wL^4}{384EI}$
but P = wL
\begin{align*} \mathrm{Ratio} = \dfrac{\left(\dfrac{PL^3}{48EI}\right)}{\left(\dfrac{5wL^4}{384EI}\right)} &= \dfrac{\left(\dfrac{PL^3}{48EI}\right)}{\left(\dfrac{5PL^3}{384EI}\right)}\\ &= \dfrac{\left(\dfrac{1}{48}\right)}{\left(\dfrac{5}{384}\right)}\\ &= \dfrac{384}{(5 \times 48)}\\ &= 8:5 \end{align*}

Correct Answer: B

Solved Example: 47-3-02

A cantilever beam with length 100 mm and flexural rigidity of 200 N.m$^2$ is loaded with a point load of 500 N at the midpoint. The deflection in (mm) at the tip of the beam is :

A. 0.26 mm

B. 2.12 mm

C. 0.06 mm

D. 0.97 mm

\begin{align*} v_{max} &= \dfrac{Pa^2}{6EI}(3L - a)\\ &= \dfrac{(500)(50 \times 10^{-3})^2}{6(200)}(3(100) - 50)\times 10^{-3}\\ &= 0.26\ \mathrm{mm} \end{align*}

Correct Answer: A

Solved Example: 47-3-03

A cantilever of length 2 m carries a uniformly distributed load of 2.5 kN/m for a length of 1 m from the fixed end and a point load of 1 kN at the free end. Find the deflection at the free end if the section is rectangular 12 cm wide and 24 cm deep and E = 1 $\times$ 10$^4$ N/mm$^2$.

A. 2.46 mm

B. 0.99 mm

C. 1.92 mm

D. 2.84 mm

L = 2 m, w = 2.5 kN/m, W = 1 kN,
a = 1 m, b = 0.12 m, d = 0.24 m,
E = $1 \times 10^8\ \mathrm{N/mm}^2$ \[I = \dfrac{bd^3}{12} = \dfrac{(0.12) \times (0.24)^3}{12} = 1.3824 \times 10^{-4}\ \mathrm{m}^4\] Deflection at free end due to point load of 1 kN \[y_1 = \dfrac{WL^3}{3EI} = \dfrac{1000 \times 2000^3}{3 \times 10^4 \times 1.3824 \times 10^8} = 1.929\ \mathrm{mm}\] Deflection of the free end due to uniformly distributed load of 2.5 N/mm on a length of 1 m \[y_2 = \dfrac{7 wa^4}{384 EI} = 0.531\ \mathrm{mm}\] Total deflection = 1.929 + 0.531 = 2.46 mm

Correct Answer: A

Solved Example: 47-3-04

Determine the load capacity in kN on a 25 mm diameter $\times$ 1200 mm long steel shaft if its maximum elongation shall not exceed 1 mm. Assume E = 200,000 MPa.

A. 88.2 kN

B. 78.3 kN

C. 83.2 kN

D. 81.8 kN

\[A = \dfrac{\pi}{4}D^2 = 4.91 \times 10^{-4}\]

Correct Answer: D

Solved Example: 47-3-06

The deflection of a cantilever beam under load W is 8. If its width is halved, then the deflection under load W will be:

A. 28

B. 8/2

C. 48

D. 8/4

Correct Answer: C

Solved Example: 47-3-07

A non-yielding support implies that the:

A. Support is frictionless

B. Support can take any amount of reaction

C. Support holds member firmly

D. Slope of the beam at the support is zero

Correct Answer: D

Solved Example: 47-3-08

The ratio of elongation in a prismatic bar due to its own weight (W) as compared to another similar bar carrying an additional weight (W) will be:

A. 1 : 2

B. 1 : 3

C. 1 : 4

D. 1 : 2.5

Correct Answer: B

Solved Example: 47-3-09

In a prismatic member made of two materials so joined that they deform equally under axial stress, the unit stresses in two materials are:

A. Equal

B. Proportional to their respective moduli of elasticity

C. Inversely proportional to their moduli of elasticity

D. Average of the sum of moduli of elasticity

Correct Answer: B

Solved Example: 47-3-10

The deformation of a bar under its own weight compared to the deformation of same body subjected to a direct load equal to weight of the body is:

A. Same cross-section throughout the beam

B. Double

C. Half

D. Four times

Correct Answer: C