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You need to login to ask any Questions from chapter Cables-and-Arches of Structural-Engineering

Chapter:

Cables-and-Arches

1. A free end of a cantilever beam rotates by 0.001 radians under a point load 10 kN. Then deflection at free end due to a moment of 100 KN - m is:

10 mm 20 mm 25 mm 40 mm

10 mm

Concept:

Assignment 5 Shivam Set - 2 20Q Hindi.docx 18

Slope for cantilever beam with point load,

\(\ta = \frac{{P{L^2}}}{{2EI}} = 0.001\)

∴ \(\frac{{{L^2}}}{{2EI}} = \frac{{0.001}}{P}\)

Now, deflection for cantilever beam with moment,

F1 A.M Madhu 07.05.20 D7

\({\rm{\Delta }} = \frac{{M{L^2}}}{{2EI}} = M \times \frac{{0.001}}{P} = 100 \times \frac{{0.001}}{{10}} = 0.01\;m = 10\;mm\)

Hint Answer

2. For a simply supported beam of given material and cross section, if udl throughout is replaced by a central point load equal to total load due to udl, central deflection of beam would

would increases by 1.6 times would be doubled would decrease by 1.6 times remains same

would increases by 1.6 times

Concept:

The maximum deflection(at center) of a simple supported with uniformly distributed load(UDL) is given by

\(\delta = \frac{5wL^4}{384EI} \)

Where w is weight per length

When W is total load n, W = wl

\(\delta = \frac{5WL^3}{384EI} \)

Maximum deflections of a beam simply supported at its ends with an isolated central load

\(y = \frac{PL^3}{48EI} \)

Here, W = P

Let deflection increase by X times

\( \frac{PL^3}{48EI} = X\ \times \frac{5WL^3}{384EI} \)

X = \(\frac{384}{48\ \times \ 5}\)

X = 1.6

Therefore deflection would be increased by 1.6 times.

Hint Answer

3. Which of following represents bending at a section of beam?

EI d⁴y / dx⁴ EI d³y / dx³ EI d²y / dx² EI dy / dx

EI d²y / dx²

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 slope of elastic curve dy / dx is very small. Then term (dy / dx)² can be neglected.

Hence,

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

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

Hint Answer

4. The ratio of deflection at center of a fixed beam and simply supported beam under a concentrated load up at centre of space is.

0.5 0.25 0.2 0.75

0.25

Concept:

Case 1: Deflection at center of a fixed beam under a concentrated load, \({\delta _1} = \frac{{P{l^3}}}{{192EI}}\)

F2 A.M Madhu 09.04.20 D 9

Case 2: Deflection at center of a simply supported beam under a concentrated load, \({\delta _2} = \frac{{P{l^3}}}{{48EI}}\)

F2 A.M Madhu 09.04.20 D 5

Calculation:

Ratio = \(\frac{{{\delta _1}}}{{{\delta _2}}} = \frac{{\frac{{P{l^3}}}{{192EI}}}}{{\frac{{P{l^3}}}{{48EI}}}} = \frac{1}{4} = 0.25\)

quesImage111

Deflection and slope of various beams are given by:

	\({y_B} = \frac{{P{L^3}}}{{3EI}}\)	\({\ta _B} = \frac{{P{L^2}}}{{2EI}}\)
	\({y_B} = \frac{{w{L^4}}}{{8EI}}\)	\({\ta _B} = \frac{{w{L^3}}}{{6EI}}\)
	\({y_B} = \frac{{M{L^2}}}{{2EI}}\)	\({\ta _B} = \frac{{ML}}{{EI}}\)
	\({y_B} = \frac{{w{L^4}}}{{30EI}}\)	\({\ta _B} = \frac{{w{L^3}}}{{24EI}}\)
	\({y_c} = \frac{{P{L^3}}}{{48EI}}\)	\({\ta _B} = \frac{{w{L^2}}}{{16EI\;}}\)
	\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\)	\({\ta _B} = \frac{{w{L^3}}}{{24EI}}\)
	\({y_c} = 0\)	\({\ta _B} = \frac{{ML}}{{24EI}}\)
	\({y_c} = \frac{{M{L^2}}}{{8EI}}\)	\({\ta _B} = \frac{{ML}}{{2EI}}\)
	\({y_c} = \frac{{P{L^3}}}{{192EI}}\)	\({\ta _A} = {\ta _B} = {\ta _C} = 0\)
	\({y_c} = \frac{{w{L^4}}}{{384EI}}\)	\({\ta _A} = {\ta _B} = {\ta _C} = 0\)

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

Hint Answer

5. The deflection of a simply supported beam with an uniformly distributed load w per unit length, having flexural rigidity EI and of length L is

\(\frac{5}{{384}}.\frac{{w{L^3}}}{{EI}}\) \(\frac{3}{{584}}.\frac{{w{L^4}}}{{EI}}\) \(\frac{3}{{584}}.\frac{{w{L^3}}}{{EI}}\) \(\frac{5}{{384}}.\frac{{w{L^4}}}{{EI}}\)

\(\frac{5}{{384}}.\frac{{w{L^4}}}{{EI}}\)

Concept:

The maximum deflection (at centre) of a simply supported beam with uniformly distributed load (UDL) is given by

\(\delta = {5\over 384}\frac{{w{L^4}}}{{EI}}\)

where w is weight per unit length,

Additional Information

Deflection and slope of various beams is given by:

	\({y_B} = \frac{{P{L^3}}}{{3EI}}\)	\({\ta _B} = \frac{{P{L^2}}}{{2EI}}\)
	\({y_B} = \frac{{w{L^4}}}{{8EI}}\)	\({\ta _B} = \frac{{w{L^3}}}{{6EI}}\)
	\({y_B} = \frac{{M{L^2}}}{{2EI}}\)	\({\ta _B} = \frac{{ML}}{{EI}}\)
	\({y_B} = \frac{{w{L^4}}}{{30EI}}\)	\({\ta _B} = \frac{{w{L^3}}}{{24EI}}\)
	\({y_c} = \frac{{P{L^3}}}{{48EI}}\)	\({\ta _B} = \frac{{w{L^2}}}{{16EI\;}}\)
	\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\)	\({\ta _B} = \frac{{w{L^3}}}{{24EI}}\)
	\({y_c} = 0\)	\({\ta _B} = \frac{{ML}}{{24EI}}\)
	\({y_c} = \frac{{M{L^2}}}{{8EI}}\)	\({\ta _B} = \frac{{ML}}{{2EI}}\)
	\({y_c} = \frac{{P{L^3}}}{{192EI}}\)	\({\ta _A} = {\ta _B} = {\ta _C} = 0\)
	\({y_c} = \frac{{w{L^4}}}{{384EI}}\)	\({\ta _A} = {\ta _B} = {\ta _C} = 0\)

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

Hint Answer

6. Bending moment at any section in a conjugate beam gives _______ in actual beam.

Slope Curvature Deflection Bending moment

Deflection

Explanation:

Real Beam	Conjugate beam
Free end	Fixed end
Internal hinge	Internal pin or roller support
End pin or roller connection	Remains same
M/EI diagram of real beam due to top applied load	Loading on Conjugate beam
Slope at any point in real beam	Shear force at that point or section in Conjugate beam
Deflection at any point in real beam	Bending moment at that point or section in Conjugate beam

Hint Answer

7. Reaction at level of prop, at free end of a cantilever of span I carrying a u.d.I. 'W' per unit length is

\(\frac{WI}{8}\) \(\frac{WI}{4}\) \(\frac{3WI}{8}\) \(\frac{5WI}{8}\)

\(\frac{3WI}{8}\)

Explanation:

Structure SSC JE CE 9th SEPT images Q12

Deflection of end B = 0

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

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

\(\begin{array}{l} {R_B} = \frac{3}{8}wl\\ {R_A} = wl - \frac{3}{8}wl = \frac{5}{8}wl \end{array}\)

As asked at propped end i.e. at B

∴ RB = 3wl/8

Important Points Deflection of various beams is given by:

RRB JE CE R45 10Q Strength Of Materials(Hindi) 1

Hint Answer

8. A cantilever beam of length, L, with uniform cross-section and flexural rigidity, EI, is loaded uniformly by a vertical load, w per unit length. The maximum vertical deflection of beam is given by

\(\frac{{w{L^4}}}{{8EI}}\) \(\frac{{w{L^4}}}{{16EI}}\) \(\frac{{w{L^4}}}{{4EI}}\) \(\frac{{w{L^4}}}{{24EI}}\)

\(\frac{{w{L^4}}}{{8EI}}\)

Concept:

Area moment method

Theorem 1: The difference of slope of any two points of a beam is equal to 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 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 = (Ax?)

Calculation:

GATE ME 2014 B Images-Q41

δ_max = Ax?

\({δ _{max}} = \frac{1}{3} \times L \times \frac{{W{L^2}}}{{2EI}} \times \frac{3}{4}L\)

\(\Rightarrow {δ _{max}} = \frac{{W{L^4}}}{{8EI}}\)

Hint Answer

9. The ratio of deflections of free end of a cantilever due to an isolated load at 1/3rd and 2/3rd of span is

1/7 2/7 3/7 2/5

2/7

I^stcondition:

For a cantilever beam subjected to load W at distance of L/3 from free end, deflection is given by:

F1 N.M Madhu 25.02.20 D4

\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)

II^ndcondition:

For a cantilever beam subjected to load W at distance of 2L/3 from free end:

F1 N.M Madhu 25.02.20 D5

\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)

\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)

\(\refore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)

Hint Answer

10. In case of a beam simply supported at both ends, if same load instead of being concentrated at centre is distributed uniformly throughout length, n deflection at centre will get reduced by

1/2 times 1/4 times 5/8 times 3/8 times

3/8 times

Concept:

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

\(δ_1=\frac{WL^3}{48EI}\)

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

\(δ_2=\frac{5wL^4}{384EI}=\frac{5WL^3}{384EI}\;\;\;(\because w=\frac{W}{L})\)

Calculation:

Given:

\(\frac{\delta_2}{\delta_1}=\frac{5WL^3}{384EI}\times \frac{48EI}{WL^3}=\frac{5}{8}\)

Hence,

The deflection will be reduced by

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

Hint Answer

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Topics

This Chapter Cables-and-Arches consists of the following topics

Deflections

A free end of a cantilever beam rotates by 0.001 radians under a point load 10 kN. Then deflection at free end due to a moment of 100 KN - m is:

For a simply supported beam of given material and cross section, if udl throughout is replaced by a central point load equal to total load due to udl, central deflection of beam would

Which of following represents bending at a section of beam?

The ratio of deflection at center of a fixed beam and simply supported beam under a concentrated load up at centre of space is.

The deflection of a simply supported beam with an uniformly distributed load w per unit length, having flexural rigidity EI and of length L is

Bending moment at any section in a conjugate beam gives _______ in actual beam.

Reaction at level of prop, at free end of a cantilever of span I carrying a u.d.I. 'W' per unit length is

A cantilever beam of length, L, with uniform cross-section and flexural rigidity, EI, is loaded uniformly by a vertical load, w per unit length. The maximum vertical deflection of beam is given by

The ratio of deflections of free end of a cantilever due to an isolated load at 1/3rd and 2/3rd of span is

In case of a beam simply supported at both ends, if same load instead of being concentrated at centre is distributed uniformly throughout length, n deflection at centre will get reduced by

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