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.
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)
\(\begin{array}{l}{M_{FAB}} = - \frac{{60(4){{(2)}^2}}}{{{6^2}}} = - 26.67\,kNm\\{M_{FBA}} = \frac{{60{{(4)}^2}(2)}}{{{6^2}}} = 53.33\,kNm\\{M_{FBC}} = - \frac{{20{{(3)}^2}}}{{12}} = - 15\,kNm\\{M_{FCD}} = - \frac{{30x8}}{8} - 30\,kNm\\{M_{FDC}} = 30\,kNm\end{array}\)
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)
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
\(\begin{array}{l}y = \frac{{4{h_1}x(L' - x)}}{{L{'^2}}}\\At = 15\,m;y\, = 4.44\,m\\\therefore \,M = {V_A}x15 - Hx4.44 - 30x15x\frac{{15}}{2}\\ = 740\,kNm\\\frac{{dy}}{{dx}} = \tan \theta = \frac{{4h(L' - 2x)}}{{L{'^2}}}\\At\,x = 15\,m\\\tan \theta = \frac{{4x5x(45.08 - 2x15)}}{{{{45.08}^2}}}\\\therefore \theta = {8.44^0}\end{array}\)
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.
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
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.
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)
\(\begin{array}{l} = 20{y_1} + 60{y_2} + 80{y_c} + 40{y_3}\\ = \left[ {20\left( {\frac{{3.35}}{{8.35}}} \right) + 60\left( {\frac{{6.35}}{{8.35}}} \right) + 80 + 40\left( {\frac{{5.65}}{{7.65}}} \right)} \right]\\ = 651.524\,kNm\end{array}\)
---
The study material for AMIE/BTech/Junior Engineer exams is available at https://amiestudycircle.com
Comments