Question
Download Solution PDFEIy \(=-\frac{5}{6} x^3+30 * x+\frac{5}{2}(x-6)^3\) is the equation of deflection obtained by using Macaulay's method for the beam shown in the following figure. Find the slope at A. Given EI = 10 × 1013 Nmm2.
Where x = horizontal distance measured from support A
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFNOTE: The equation given in the question is not correct, so the correct equation as per the correct option will be
EIy \(=-\frac{5}{6} x^3+30 * x+\frac{5}{2}(x-6)^3\)
Concept: Use of Discontinuity function: Macaulay’s Method
It is a method in which a single equation is formed for all loads on the beam and the equation is constructed in such a way that the integral constants apply to all the sections of the beam.
Single Moment Equation valid for all values of x (all the sections of the beam) would be:
\(M = EI × \frac{{\left( {{d^2}y} \right)}}{{d{x^2}}} = \left. {{R_A} × \frac{x}{1}\;} \right|_0^a - \left. {W × \frac{{\left( {x - a} \right)}}{1}\;} \right|_0^b - \left. {w × \frac{{{{\left( {x - b} \right)}^2}}}{2}\;} \right|_0^L\)
The oblique sign indicates that the remaining terms may be or may not be valid depending upon the range of x
To determine the deflection by this method, the differential equation of flexure is integrated twice.
\({\rm{M}} = {\rm{EI}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}{\rm{\;}}\)
\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \smallint \frac{{\rm{M}}}{{{\rm{EI}}}}\)
\(\therefore {\rm{y}} = \smallint \left( {\smallint \frac{{\rm{M}}}{{{\rm{EI}}}} + {\rm{C}}} \right)\)
where M = Moment produced at the section, E = Modulus of elasticity of cross-sectional material’s
\({\rm{θ }} = \frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) = Slope and y = Deflection.
Calculation:
EIy \(=-\frac{5}{6} x^3+30 * x+\frac{5}{2}(x-6)^3\)
Differentiating the given equation we get
\({d(EIy)\over dx} = =-\frac{5}{6} x^3+30 * x+\frac{5}{2}(x-6)^3 \ \\\ EI (dy/dx) ={ -\frac{5}{2} x^2+30 +\frac{15}{2}(x-6)^2} \)
Slope over the support A, we have to put x = 0 in the above equation and ignore the terms which comes negative after putting the value of x, like in this case (x - 6) comes negative (-6) after putting the value of x, so we have to ignore this term and solve the remaining terms
EI θA = -5/2 × 0 + 30
θA = 3 × 103 / EI = \({ 3 × 10^3 \over{ 10 × 10^{13}}} = {0.0003 \ radians}\)
Last updated on May 28, 2025
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