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Analysis and Design of Structures (Solved Numerical Problems)

Structural Analysis

Analyse the continuous beam shown in figure by slope deflection method and draw bending moment, shear force diagram and elastic curve. (AMIE Summer 2023).

Solution

Fixed end moments

\(\begin{array}{l}{M_{FAB}} =  - \frac{{60(4){{(2)}^2}\,}}{{{6^2}}} =  - 26.67\,kNm\\{M_{FBA}} = \frac{{60(2){{(4)}^2}}}{{{6^2}}} = 53.33\,kNm\\{M_{FBC}} =  - \frac{{30{{(6)}^2}}}{{12}} =  - 90\,kNm\\{M_{FCB}} = 90\,kNm\end{array}\)

Slope deflection equations

\(\begin{array}{l}{M_{AB}} =  - 26.67 + \frac{{2EI}}{6}(2{\theta _A} + {\theta _B} - 0)\\ - 26.67 + \frac{1}{3}EI{\theta _B}\\{M_{BA}} = 53.33 + \frac{{2EI}}{6}({\theta _A} + 2{\theta _B} - 0)\\ = 53.33 + \frac{2}{3}EI{\theta _B}\\{M_{BC}} =  - 90 + \frac{{2EI}}{6}(2{\theta _B} + {\theta _C} - 0)\\ =  - 90 + \frac{2}{3}EI{\theta _B} + \frac{1}{3}EI{\theta _C}\\{M_{CB}} = 90 + \frac{{2EI}}{6}({\theta _B} + 2{\theta _C} - 0)\\ = 90 + \frac{1}{3}EI{\theta _B} + \frac{2}{3}EI{\theta _C}\end{array}\)

Equilibrium equations

\(\begin{array}{l}\sum {M_B} = 0\\{M_{BA}} + {M_{BC}} = 0\\\therefore \,53.33 + \frac{2}{3}EI{\theta _B} - 90 + \frac{2}{3}EI{\theta _B} + \frac{1}{3}EI{\theta _C} = 0\\4EI{\theta _B} + EI{\theta _C} = 110\,.....(1)\\\sum {M_C} = 0 \Rightarrow {M_{CB}} = 0\\90 + \frac{1}{3}EI{\theta _B} + \frac{2}{3}EI{\theta _C} = 0\\EI{\theta _B} + 2EI{\theta _C} =  - 270.....(2)\end{array}\)

Solving (1) and (2)

\(\begin{array}{l}EI{\theta _B} = 70\\EI{\theta _C} =  - 170\end{array}\)

End moments

Substituting these values in slope deflection equation,

\(\begin{array}{l}{M_{AB}} =  - 26.67 + \frac{1}{3}(70) =  - 3.33\,kNm\\{M_{BA}} = 53.33 + \frac{2}{3}(70) = 100\,kNm\\{M_{BC}} =  - 90 + \frac{2}{3}(70) + \frac{1}{3}( - 170) =  - 100\,kNm\\{M_{CB}} = 90 + \frac{1}{3}(70) + \frac{2}{3}( - 170) = 0\end{array}\)

The sign conventions for these moments are clockwise direction positive and anticlockwise direction negative. Hence, the moments are as shown in the following figure. 

But while drawing bending moment diagrams, sign convention followed is tension at bottom positive and tension at top negative. Free moments (moment in a simply supported beam for given loading) give tension at bottom and hence are positive. End moments give tension at top and hence they are negative moments. Since, we need difference of these diagrams, to get the final diagram, we draw them on the same side and mark the difference diagram as the final diagram.

The free moment diagram in AB is a triangle with maximum ordinate under load, its magnitude being 60 x 4 x 2/6 = 80 kNm.

Free moment diagram in BC is a symmetric parabola with maximum ordinate,

\( = \frac{{30{{(6)}^2}}}{8} = 135\,kNm\)

BMD is shown below.

 

Analyse the continuous beam shown in the figure and draw a bending moment diagram. (AMIE Summer 2023)

Fixed end moments

\(\begin{array}{l}{M_{FAB}} =  - \frac{{60(4){{(2)}^2}}}{{{6^2}}} =  - 26.67\,kNm\\{M_{FBA}} = \frac{{60{{(4)}^2}(2)}}{{{6^2}}} = 53.33\,kNm\\{M_{FBC}} =  - \frac{{20{{(3)}^2}}}{{12}} =  - 15\,kNm\\{M_{FCD}} =  - \frac{{30x8}}{8} - 30\,kNm\\{M_{FDC}} = 30\,kNm\end{array}\)

Distribution tables

Moment distribution

Free moment diagram for AB is a triangle with maximum ordinate under the load

= 60 x 4 x 2/6 = 80 kNm

Free moment diagram for BC is a symmetric parabola with maximum ordinate

\( = \frac{{20x{3^2}}}{4} = 22.5\,kNm\)

Free moment diagram for CD is a triangle with maximum ordinate under the load

\( = \frac{{30x8}}{4} = 60\,kNm\)

A bending moment diagram is given below.

A three-hinged parabolic arch having supports at different levels shown in the figure carriers a uniformly distributed load of intensity 30 kN/m over the portion left of the crown. Determine the horizontal thrust developed. Find also the bending moment, normal thrust and radial shear force developed at a section 15 m from the left support. (AMIE Summer 2023)

 

Solution

Taking C as origin.

 \(\frac{{{x^2}}}{y} = a\)

The horizontal distance between L and C is L1 and that of C and D is L2 .

 \(\begin{array}{l}\frac{{{L_1}^2}}{5} = a = \frac{{{L_2}^2}}{3}\\\frac{{{L_1}}}{{\sqrt 5 }} = \frac{{{L_2}}}{{\sqrt 3 }} = \frac{{{L_1} + {L_2}}}{{\sqrt 5  + \sqrt 3 }} = \frac{L}{{\sqrt 5  + \sqrt 3 }}\\\therefore \,{L_1} = \frac{{L\sqrt 5 }}{{\sqrt 5  + \sqrt 3 }} = 22.54\,m\\{L_2} = 40 - 22.54 = 17.46\,m\\\sum {M_C} = 0\\{V_B}x17.46 = Hx3\\H = 5.82{V_B}\\\sum {M_A} = 0\\{V_B}x40 + Hx2 = 30x22.54x\frac{{22.54}}{2}\\ = 7620.774\end{array}\)

From above two equations

\(\begin{array}{l}{V_B}x40 - 5.82{V_B}x2 = 7620.774\\{V_B} = 147.58\\H = 5.82x147.58 = 858.92\\{V_A} = 30{L_1} - {V_B} = 30x22.5 - 147.58\\ = 528.02\,kN\end{array}\)

The portion left of C may be treated as a parabola of span 

L' = 2 x 22.54 = 45.08 m 

Hence, equation of parabola is

\(\begin{array}{l}y = \frac{{4{h_1}x(L' - x)}}{{L{'^2}}}\\At = 15\,m;y\, = 4.44\,m\\\therefore \,M = {V_A}x15 - Hx4.44 - 30x15x\frac{{15}}{2}\\ = 740\,kNm\\\frac{{dy}}{{dx}} = \tan \theta  = \frac{{4h(L' - 2x)}}{{L{'^2}}}\\At\,x = 15\,m\\\tan \theta  = \frac{{4x5x(45.08 - 2x15)}}{{{{45.08}^2}}}\\\therefore \theta  = {8.44^0}\end{array}\)

V = 528.62 - 30 x 15 = 76.62 kN
N = Vsinθ + Hcosθ
=78.62sin8.440 + 858.92cos8.440
= 861.25 kN
Q = Vcosθ - Hsinθ
= 78.62cos8.440 - 858.92sin8.440
= -48.30 kN

A circular arch to span 25 m with a central rise 5 m is hinged at the crown and springing. It carries a point load of 100 kN at 6 m from the left support. 

Calculate
1. The reactions at the supports
2. The reactions at crown
3. Moment at 5 m from the left support. (AMIE Winter 2022)

Solution 

VA x 25 = 100 x (25 - 6)

Giving VA = 76 kN

VB = 100 - 76 = 24 kN

Moment about C

0 = 24 x 12.5 - H x 5

H = 60 kN

Considering the equilibrium of the left-half of the arch, the reactions at crown are

Moment at 5 m from the left support:

From the property of circles, we get

\(\begin{array}{l}h(2R - h) = \frac{L}{2}x\frac{L}{2}\\5(2R - 5) = \frac{{25}}{2}x\frac{{25}}{2}\\\therefore \,R\, = \,18.125\,m\end{array}\)

\(\begin{array}{l}R\sin \theta  = 12.5 - 5 = 6.5\\\therefore \,\,\theta  = {24.443^0}\\{y_D} = h - R(1 - \cos \theta )\\ = 5 - 18.125(1 - \cos {24.443^0})\\ = \,3.375\,\end{array}\)

\(\begin{array}{l}{M_D} = {V_A}x5 - H{y_D}\\ = 76x5 - 60x3.376\\ = 177.5\,kNm\end{array}\)

Using influence line diagrams, determine the shear force and bending moment at section C in the simply supported beam shown in figure. (AMIE Summer 2023)

Solution

Shear force at C

ILD for shear force at C is given below.

\(\begin{array}{l}{F_C} =  - 40\,\left( {\frac{1}{7}} \right) - 10\left( {\frac{1}{2}} \right)\left[ {\frac{1}{7} + \frac{2}{7}} \right](2)\\ + \frac{{10(1)}}{2}\left[ {\frac{5}{7} + \frac{4}{7}} \right](2)\\ + 60\left( {\frac{3}{7}} \right) + 80\left( {\frac{2}{7}} \right)\\ = 51.43\,kN\end{array}\)

Bending moment at C

ILD for bending moment at C is as shown below

Maximum ordinate

\(\begin{array}{l}{y_c} = \frac{{4(14 - 4)}}{{14}} = \frac{{20}}{7}\\{y_1} = \frac{{10}}{7}\\{y_2} = \frac{8}{{10}}\,x\,\frac{{20}}{7} = \frac{{16}}{7}\\{y_3} = \frac{6}{{10}}\,x\,\frac{{20}}{7} = \frac{{12}}{7}\\{y_4} = \frac{4}{{10}}\,x\,\frac{{20}}{7} = \frac{8}{7}\end{array}\)

\(\begin{array}{l}{M_C} = 40\left( {\frac{{10}}{7}} \right) + \frac{{10(1)}}{2}\left[ {\frac{{10}}{7} + \frac{{20}}{7} + \frac{{20}}{7} + \frac{{16}}{7}} \right](2)\\ + 60\left( {\frac{{12}}{7}} \right) + 80\left( {\frac{8}{7}} \right)\\ = 345.71\,kNm\end{array}\)

A train of concentrated loads shown in figure moves from left to right on a simply supported girder of span 16 m. Determine the absolute maximum shear force and bending moment developed in the beam. (AMIE Summer 2023).

Solution

ILD for shear force at A is as shown below.

When 20 kN load is just on A,

Positive shear force at A

\(\begin{array}{l} = 20(1) + 60\left( {\frac{{13}}{{16}}} \right) + 80\left( {\frac{{11}}{{16}}} \right) + 40\left( {\frac{9}{{16}}} \right)\\ = 146.25\,kNm\end{array}\)

Since. 20 kN load is lighter, one more trial with 60 kN load at A is made. 

For this position, SF at A

\( = 60(1) + 80\left( {\frac{{14}}{{16}}} \right) + 40\left( {\frac{{12}}{{16}}} \right) = 160\,kNm\)

Hence, maximum positive S.F. occurs at A and is equal to 160 kN.
Maximum negative shear force occurs at B when leading load is on B (refer to following figure).

Hence, maximum negative shear force

\(\begin{array}{l} = 40(1) + 80\left( {\frac{{14}}{{16}}} \right) + 60\left( {\frac{{12}}{{16}}} \right) + 20\left( {\frac{9}{{16}}} \right)\\ = 166.25\,kNm\end{array}\)

Hence, absolute maximum shear force = 166.25 kNm

For finding position for absolute maximum moment, position of C.G. of load system is to be located.

Distance of C.G. for the leading load of 40 kN

\(a = \frac{{80(2) + 60(4) + 20(7)}}{{40 + 80 + 60 + 20}} = 2.7\,m\)

It is nearer to 80 kN load and this load is heavier than another nearer load of 60 kN. Hence, maximum moment will occur under 80 kN load. Distance between this load and the resultant,

d = 2.7 - 2 = 0.7

Position of 80 kN load for maximum moment

\( = \frac{{16}}{2} + \frac{{0.7}}{2} = 8.35\,m\,from\,A.\)

For the section at 8.35 m from A, ILD
Ordinate for moment

\(\begin{array}{l}{y_c} = \frac{{z(L - z)}}{L} = \frac{{8.35(16 - 8.35)}}{{16}}\\ = 3.9923\end{array}\)

Absolute maximum moment (Refer following figure)


 \(\begin{array}{l} = 20{y_1} + 60{y_2} + 80{y_c} + 40{y_3}\\ = \left[ {20\left( {\frac{{3.35}}{{8.35}}} \right) + 60\left( {\frac{{6.35}}{{8.35}}} \right) + 80 + 40\left( {\frac{{5.65}}{{7.65}}} \right)} \right]\\ = 651.524\,kNm\end{array}\)

Problem

Determine the vertical deflection of point D in the truss shown in figure. The cross-sectional areas of members AD and DE arc 1500 mm2 while those of the other members arc 1000 mm2. Take E - 200 kNmm2. (AMIE Winter 2024, 20 marks)

Solution

P forces

From the equilibrium of joint C

PCB x 0.8 = 60 

PCB = 75 kN (tensile)

PCB x 0.6 = PCD

PCD = 45 kN (compressive)

At joint B

PBD = PCB x 0.8 = 75 x 0.8 = 60 kN (Compressive)

PBA = PCB x 0.6 = 45 kN (tensile)

At joint D

PDA x (1/√2) = PBD = 60 kN

PDA = 60√2 kN (tensile)

PDE = PDA x (1/√2) + PDC

= 60 + 45 = 105 kN (compressive)

k forces due to unit load at D

Referring to following figure.

From joint equilibrium C and B

PCB = PCD = PBA = PBD = 0

From joint D

PDA x (1/√2) = 1

PDA = √2 kN (tensile)

and PDE = PDA x (1/√2) = 1 kN (compressive)


Σ(PkL/A) = 732.55

Deflection at D = Σ(PkL/AE)

= (1/E) Σ(PkL/A)

= (1/200)(732.55)

= 3.663 mm

Problem

A simply supported beam has a span of 16 m is subjected to a UDL (live load) of 13 kN/m (longer than the span) traveling from left to right. Draw the ILD for shear force and bending moment at a section 4 m from the left end. Use these diagrams to determine the maximum shear force and bending moment at this section. (AMIE Winter 2024)

Solution

Maximum shear force

ILD ordinate for positive

= (L - x)/L

= (16 - 4)/16 = 0.75

Maximum positive shear force

= load x area

= 13 [0.5 x 12 x 0.75)

= 58.5 kN

Maximum negative shear force

ILD for negative ordinate = x/L = 4/16 = 0.25

Maximum negative shear force

= 13 (0.5 x 4 x 0.25)

= 6.5 kn

Maximum bending moment

= L/4 = 16/4 = 4 m

Since the UDL is acting longer than span.

Maximum bending moment will occur right of D.

Maximum ordinate at D

= x(L - x)/L

= 4(12 - 4)/12 = 2.67

Maximum BM = load x area

= 13(0.5 x 8 x 2.67)

= 138.84 kNm

 

---

The study material for AMIE/BTech/Junior Engineer exams is available at https://amiestudycircle.com

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