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UPPSC AE Mechanical 2013 Official Paper I

Option 2 : Shear force at the section

CT 1: Current Affairs (Government Policies and Schemes)

54560

10 Questions
10 Marks
10 Mins

**Concept:**

**Shear force at any section of the beam is equal to the slope of the Bending Moment Diagram**

\(S.F = \left( {\frac{{dM}}{{dx}}} \right)\)

It is given that \(M = EI\frac{{{d^2}y}}{{d{x^2}}}\)

∴ \(\left( {\frac{{dM}}{{dx}}} \right) = \frac{d}{{dx}}\left( {EI\frac{{{d^2}y}}{{d{x^2}}}} \right) = \;EI\frac{{{d^3}y}}{{d{x^3}}}\;\)

∴ \(EI\frac{{{d^3}y}}{{d{x^3}}}\) **represents the shear force at the section**

Note that slope of Shear force diagram gives the load acting at that section

\(W = \left( {\frac{{dF}}{{dx}}} \right)\)

\(EI\frac{{{d^2}y}}{{d{x^2}}} = M\) , in this equation, y is the deflection

Deflection y is obtained by double integration method

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

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

here, \(\frac{{dy}}{{dx}} = {\rm{slope\;or\;angular\;deviation\;of\;the\;axis\;of\;the\;beam}}\)

and the constant C_{1} is determined by using boundary conditions for the slope.

\(EI.\;y = \int\!\!\!\int M + {C_1}x + {C_2}\)

C