[Solved] Double Integration Method MCQ [Free PDF] - Objective Question Answer for Double Integration Method Quiz - Download Now! (2024)

1. deflection
2. slope
3. twisting angle
4. diameter

Explanation:

Double Integration Method

The double integration method is a method used tocalculate the deflection and slope of a beam at any point due to bending.

In calculus, the radius of curvature of a curve y = f(x) is given by

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

In the derivation of theflexure or bending formula, the radius of curvature of a beam is given as

\(R = \frac{{EI}}{M}\)

Deflection of beams is so small, such that the slope of the elastic curve dy/dx is very small, and squaring this expression the value becomes practically negligible, hence

\(R = \frac{1}{{\frac{{{d^2}y}}{{d{x^2}}}}} = \frac{1}{{y''}}\)

\(y'' = \frac{M}{{EI}}\)

If EI is constant, the equation may be written as:\(EIy'' = M\)

• Thefirst integration y' yields the slopeof the elastic curve and thesecond integration y gives the deflectionof the beam at any distance x.
• The resulting solution must contain two constants of integration sinceEI y" = M is of second order.
• These two constants must be evaluated fromknown conditions concerning the slope deflection at certain points of the beam.

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1. ML/(2EI)
2. ML/(3EI)
3. ML/(6EI)
4. ML/(12EI)

Given equation of deflection:

\(w = - \frac{{Mx}}{{12EIL}}\left( {L - x} \right)\left( {x + c} \right)\)

\( = - \frac{{Mx}}{{12EIL}}\left[ {Lx + Lc - {x^2} - cx} \right]\)

\(w = \; - \frac{{M}}{{12EIL}}\left[ {L{x^2} + Lcx - {x^3} - c{x^2}} \right]\)

\(\frac{{dw}}{{dx}} = \; - \frac{{M}}{{12EIL}}\left[ {2Lx + Lc - 3{x^2} - 2cx} \right]\)

\(\frac{{{d^2}w}}{{d{x^2}}} = - \frac{M}{{12EIL}}\left[ {2L - 6x - 2c} \right]\;\)

\( \Rightarrow EI\frac{{{d^2}w}}{{d{x^2}}} = - \frac{M}{{12L}}\left[ {2L - 6x - 2c} \right]\)

As we know:

\( \Rightarrow EI\frac{{{d^2}w}}{{d{x^2}}} = - M_x\)

Boundary condition

X = L, M_x = 0

\(- \frac{M}{{12L}}\left[ {2L - 6L - 2c} \right]=0\)

∴ c= -2L

\(\therefore Slope, \frac{{dw}}{{dx}} = - \frac{M}{{12EIL}}\left[ {2Lx - 3{x^2} + Lc - 2{cx}} \right]\)

Put x = 0 & c = -2L

\({\left| {\frac{{dw}}{{dx}}} \right|_{x - 0}} = \frac{{ML}}{{6EI}}\)

Points to remember:

• \(Slope\;\theta = \frac{{dw}}{{dx}}.\;\)Where wis deflection

• Boundary Moment.\(M = EI\frac{{{d^2}w}}{{d{x^2}}}\)

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1. \(EI\frac{{{d^2}y}}{{d{x^2}}} = - M\)
2. \(EI\frac{{{d}y}}{{d{x}}} = M^2\)
3. \(EI\frac{{{d^2}y}}{{d{x^2}}} = M^2\)
4. \(M\frac{{{d^2}y}}{{d{x^2}}} = -EI\)

We know curvature of the beam is

\(K = \frac{1}{\rho } = \frac{{{d^2}y}}{{d{x^2}}}\)

From bending equation

\(K = \frac{1}{\rho } = \frac{M}{{E1}}K = \frac{1}{\rho } = \frac{M}{{E1}}\)

\(∴ \frac{{{d^2}y}}{{d{x^2}}} = \frac{M}{{E1}}\)

For Hogging moment, use M = -M in the above equation, Hence final equation becomes:

\(\frac{{{d^2}y}}{{d{x^2}}} = \frac{-M}{{E1}}\)

∴\(EI\frac{{{d^2}y}}{{d{x^2}}} = - M\)

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1. δ₁ = δ₂
2. δ₁ < δ₂
3. δ₁ > δ₂
4. δ₁ ≫ δ₂

\(\begin{array}{l}{{\rm{\delta }}_1} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\\= \frac{5}{{384}} \times \frac{{6 \times {4^4}}}{{10000\;m}} = 20\;mm\end{array}\)

\(\begin{array}{l}{{\rm{\delta }}_{\rm{L}}} = \frac{{P{L^3}}}{{48EI}}\\= \frac{{120 \times {2^3}}}{{48 \times 10000\;m}}\end{array}\)

= 20 mm

∴ δ₁ = δ₂

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1. δ_max1 ≠ δ_max2 and θ_max1 ≠ θ_max2
2. δ_max1 = δ_max2 and θ_max1 ≠ θ_max2
3. δ_max1 ≠ δ_max2 and θ_max1 = θ_max2
4. δ_max1 = δ_max2 and θ_max1 = θ_max2

Beam XYZ can be analyzed as half beam YZ or XY due to symmetry of load. Hence, beam YZ or XY can be considered similar to beam PQ which is free to move at the end. So, rotation and deflection will be same in both.

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Double Integration Method Question 6

1. \(\frac{{M{L^2}}}{{2EI}}\)
2. \(\frac{{M{L^2}}}{{EI}}\)
3. \(\frac{{2M{L^2}}}{{EI}}\)
4. \(\frac{{4M{L^2}}}{{EI}}\)

Double Integration Method Question 6 Detailed Solution

Concept:

\({EI}\frac{{{d^2}y}}{{d{x^2}}} = M\)

Calculation:

Byintegratingthe above equation we get,

\(EI\frac{{dy}}{{dx}} = Mx + {C_1}\)

Once again integrating, we get

\(EI\;y = \frac{{M{x^2}}}{2} + {C_1}x + {C_2}\)

For cantilever beam at\(x = 0,\frac{{dy}}{{dx}} = 0\;\& \;x = 0,~y = 0\)(x taken from fixed-end).

From this we get C₁ =C₂= 0, Hence

\(y = \frac{{M{x^2}}}{{2EI}}\)

Deflection will be maximum when x = L

∴ Maximum deflectionis

\({y_{max}} = \frac{{M{L^2}}}{{2EI}}\)

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Double Integration Method Question 7

1. ML/(2EI)
2. ML/(3EI)
3. ML/(6EI)
4. ML/(12EI)

Double Integration Method Question 7 Detailed Solution

Given equation of deflection:

\(w = - \frac{{Mx}}{{12EIL}}\left( {L - x} \right)\left( {x + c} \right)\)

\( = - \frac{{Mx}}{{12EIL}}\left[ {Lx + Lc - {x^2} - cx} \right]\)

\(w = \; - \frac{{M}}{{12EIL}}\left[ {L{x^2} + Lcx - {x^3} - c{x^2}} \right]\)

\(\frac{{dw}}{{dx}} = \; - \frac{{M}}{{12EIL}}\left[ {2Lx + Lc - 3{x^2} - 2cx} \right]\)

\(\frac{{{d^2}w}}{{d{x^2}}} = - \frac{M}{{12EIL}}\left[ {2L - 6x - 2c} \right]\;\)

\( \Rightarrow EI\frac{{{d^2}w}}{{d{x^2}}} = - \frac{M}{{12L}}\left[ {2L - 6x - 2c} \right]\)

As we know:

\( \Rightarrow EI\frac{{{d^2}w}}{{d{x^2}}} = - M_x\)

Boundary condition

X = L, M_x = 0

\(- \frac{M}{{12L}}\left[ {2L - 6L - 2c} \right]=0\)

∴ c= -2L

\(\therefore Slope, \frac{{dw}}{{dx}} = - \frac{M}{{12EIL}}\left[ {2Lx - 3{x^2} + Lc - 2{cx}} \right]\)

Put x = 0 & c = -2L

\({\left| {\frac{{dw}}{{dx}}} \right|_{x - 0}} = \frac{{ML}}{{6EI}}\)

Points to remember:

• \(Slope\;\theta = \frac{{dw}}{{dx}}.\;\)Where wis deflection

• Boundary Moment.\(M = EI\frac{{{d^2}w}}{{d{x^2}}}\)

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Double Integration Method Question 8

Double Integration Method Question 8 Detailed Solution

Concept:

\({δ _A} = \frac{{p{l^3}}}{{3EI}},{θ _A} = \frac{{P{L^2}}}{{2EI}}\)

\(slope\;{θ _c} = \frac{{P{l^2}}}{{2EI}},{δ _C} = \frac{{P{l^3}}}{{3EI}}\)

Deflection at A =δ_c + θ_c × (AB - BC)

=δ_c+θ_c× l

\( = \frac{{P{l^3}}}{{3EI}} + \frac{{P{l^2}}}{{2EI}} \times l\)

Calculation:

Given P = 500N; EI = 200 Nm²

l = 50 mm = 0.05 m

2l = 100 mm = 0.10 m

\({\delta _A}\left( {at\;tip} \right) = \frac{{500 \times {{0.05}^3}}}{{3 \times 200}} + \frac{{500 \times {{0.05}^2}}}{{2 \times 200}} \times 0.05\)

= 2.604× 10^-4 m

= 0.26 mm

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Double Integration Method Question 9

1. δ₁ = δ₂
2. δ₁ < δ₂
3. δ₁ > δ₂
4. δ₁ ≫ δ₂

Double Integration Method Question 9 Detailed Solution

\(\begin{array}{l}{{\rm{\delta }}_1} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\\= \frac{5}{{384}} \times \frac{{6 \times {4^4}}}{{10000\;m}} = 20\;mm\end{array}\)

\(\begin{array}{l}{{\rm{\delta }}_{\rm{L}}} = \frac{{P{L^3}}}{{48EI}}\\= \frac{{120 \times {2^3}}}{{48 \times 10000\;m}}\end{array}\)

= 20 mm

∴ δ₁ = δ₂

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Double Integration Method Question 10

1. δ_max1 ≠ δ_max2 and θ_max1 ≠ θ_max2
2. δ_max1 = δ_max2 and θ_max1 ≠ θ_max2
3. δ_max1 ≠ δ_max2 and θ_max1 = θ_max2
4. δ_max1 = δ_max2 and θ_max1 = θ_max2

Double Integration Method Question 10 Detailed Solution

Beam XYZ can be analyzed as half beam YZ or XY due to symmetry of load. Hence, beam YZ or XY can be considered similar to beam PQ which is free to move at the end. So, rotation and deflection will be same in both.

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1. deflection
2. slope
3. twisting angle
4. diameter

Explanation:

Double Integration Method

The double integration method is a method used tocalculate the deflection and slope of a beam at any point due to bending.

In calculus, the radius of curvature of a curve y = f(x) is given by

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

In the derivation of theflexure or bending formula, the radius of curvature of a beam is given as

\(R = \frac{{EI}}{M}\)

Deflection of beams is so small, such that the slope of the elastic curve dy/dx is very small, and squaring this expression the value becomes practically negligible, hence

\(R = \frac{1}{{\frac{{{d^2}y}}{{d{x^2}}}}} = \frac{1}{{y''}}\)

\(y'' = \frac{M}{{EI}}\)

If EI is constant, the equation may be written as:\(EIy'' = M\)

• Thefirst integration y' yields the slopeof the elastic curve and thesecond integration y gives the deflectionof the beam at any distance x.
• The resulting solution must contain two constants of integration sinceEI y" = M is of second order.
• These two constants must be evaluated fromknown conditions concerning the slope deflection at certain points of the beam.

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1. \(EI\frac{{{d^2}y}}{{d{x^2}}} = - M\)
2. \(EI\frac{{{d}y}}{{d{x}}} = M^2\)
3. \(EI\frac{{{d^2}y}}{{d{x^2}}} = M^2\)
4. \(M\frac{{{d^2}y}}{{d{x^2}}} = -EI\)

We know curvature of the beam is

\(K = \frac{1}{\rho } = \frac{{{d^2}y}}{{d{x^2}}}\)

From bending equation

\(K = \frac{1}{\rho } = \frac{M}{{E1}}K = \frac{1}{\rho } = \frac{M}{{E1}}\)

\(∴ \frac{{{d^2}y}}{{d{x^2}}} = \frac{M}{{E1}}\)

For Hogging moment, use M = -M in the above equation, Hence final equation becomes:

\(\frac{{{d^2}y}}{{d{x^2}}} = \frac{-M}{{E1}}\)

∴\(EI\frac{{{d^2}y}}{{d{x^2}}} = - M\)

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1. \(\frac{{M{L^2}}}{{2EI}}\)
2. \(\frac{{M{L^2}}}{{EI}}\)
3. \(\frac{{2M{L^2}}}{{EI}}\)
4. \(\frac{{4M{L^2}}}{{EI}}\)

Concept:

\({EI}\frac{{{d^2}y}}{{d{x^2}}} = M\)

Calculation:

Byintegratingthe above equation we get,

\(EI\frac{{dy}}{{dx}} = Mx + {C_1}\)

Once again integrating, we get

\(EI\;y = \frac{{M{x^2}}}{2} + {C_1}x + {C_2}\)

For cantilever beam at\(x = 0,\frac{{dy}}{{dx}} = 0\;\& \;x = 0,~y = 0\)(x taken from fixed-end).

From this we get C₁ =C₂= 0, Hence

\(y = \frac{{M{x^2}}}{{2EI}}\)

Deflection will be maximum when x = L

∴ Maximum deflectionis

\({y_{max}} = \frac{{M{L^2}}}{{2EI}}\)

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1. ML/(2EI)
2. ML/(3EI)
3. ML/(6EI)
4. ML/(12EI)

Given equation of deflection:

\(w = - \frac{{Mx}}{{12EIL}}\left( {L - x} \right)\left( {x + c} \right)\)

\( = - \frac{{Mx}}{{12EIL}}\left[ {Lx + Lc - {x^2} - cx} \right]\)

\(w = \; - \frac{{M}}{{12EIL}}\left[ {L{x^2} + Lcx - {x^3} - c{x^2}} \right]\)

\(\frac{{dw}}{{dx}} = \; - \frac{{M}}{{12EIL}}\left[ {2Lx + Lc - 3{x^2} - 2cx} \right]\)

\(\frac{{{d^2}w}}{{d{x^2}}} = - \frac{M}{{12EIL}}\left[ {2L - 6x - 2c} \right]\;\)

\( \Rightarrow EI\frac{{{d^2}w}}{{d{x^2}}} = - \frac{M}{{12L}}\left[ {2L - 6x - 2c} \right]\)

As we know:

\( \Rightarrow EI\frac{{{d^2}w}}{{d{x^2}}} = - M_x\)

Boundary condition

X = L, M_x = 0

\(- \frac{M}{{12L}}\left[ {2L - 6L - 2c} \right]=0\)

∴ c= -2L

\(\therefore Slope, \frac{{dw}}{{dx}} = - \frac{M}{{12EIL}}\left[ {2Lx - 3{x^2} + Lc - 2{cx}} \right]\)

Put x = 0 & c = -2L

\({\left| {\frac{{dw}}{{dx}}} \right|_{x - 0}} = \frac{{ML}}{{6EI}}\)

Points to remember:

• \(Slope\;\theta = \frac{{dw}}{{dx}}.\;\)Where wis deflection

• Boundary Moment.\(M = EI\frac{{{d^2}w}}{{d{x^2}}}\)

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Concept:

\({δ _A} = \frac{{p{l^3}}}{{3EI}},{θ _A} = \frac{{P{L^2}}}{{2EI}}\)

\(slope\;{θ _c} = \frac{{P{l^2}}}{{2EI}},{δ _C} = \frac{{P{l^3}}}{{3EI}}\)

Deflection at A =δ_c + θ_c × (AB - BC)

=δ_c+θ_c× l

\( = \frac{{P{l^3}}}{{3EI}} + \frac{{P{l^2}}}{{2EI}} \times l\)

Calculation:

Given P = 500N; EI = 200 Nm²

l = 50 mm = 0.05 m

2l = 100 mm = 0.10 m

\({\delta _A}\left( {at\;tip} \right) = \frac{{500 \times {{0.05}^3}}}{{3 \times 200}} + \frac{{500 \times {{0.05}^2}}}{{2 \times 200}} \times 0.05\)

= 2.604× 10^-4 m

= 0.26 mm

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