Question 26.3.41: Determine the stiffness of the beam shown in Fig. 11 given b...
Determine the stiffness of the beam shown in Fig. 11 given below :
When : I = 375 × 10^{-4} m^4
L = 0.5 m
E = 200 GPa
The stiffness is given by
(a) 12 × 10^{10} N/m (b) 10 × 10^{10} N/m
(c) 4 × 10^{10} N/m (d) 8 × 10^{10} N/m.
(c)
EXPLANATION
\text { Stiffness }=\frac{\text { Load }}{\text { Deflection }}=\frac{P}{\text { Deflection under } P} …(i)
Let us find deflection under load P. This can be done by conjugate Beam Method. In this method, the beam carries the \frac{M}{E I} load corresponding to actual load. The deflection at any section will be equal to B. M. at that section due the load carried by conjugate beam. Refer to Fig. 20 (b).
Load for conjugate beam is \frac{M}{E I} .
B.M. at A = 0, hence value of \frac{M}{E I} at A = 0
B.M. at C = P × 2L, hence value of \frac{M}{E I} \text { at } C=\frac{P \times 2 L}{E \times(2 I)}=\frac{P L}{E I}
B.M. at B = P × L, hence value of \frac{M}{E I} \text { at } B \text { for } A B=\frac{P \times L}{E I}, \text { for } B C=\frac{P \times L}{E \times(2 I)}=\frac{P L}{2 E I}
Deflection (δ) at A = B.M. at A* due to load carried by conjugate beam
=\left(\frac{1}{2} \times \frac{P L}{E I} \times L\right) \times \frac{2 L}{3}+\left(\frac{P L}{2 E I} \times L\right) \times 1.5 L+\left(\frac{1}{2} \times \frac{P L}{2 E L} \times L\right) \times \frac{5 L}{3} \\ \space \\ =\frac{P L^3}{E I}\left(\frac{1}{3}+0.75+\frac{5}{12}\right)=\frac{P L^3}{E I} \times \frac{18}{12}=\frac{1.5 \times P L^3}{E I} \\ \space \\ \therefore \text { Stiffness }=\frac{P}{\delta}=\frac{P}{\left(\frac{1.5 \times P L^3}{E I}\right)}=\frac{E I}{1.5 \times L^3}=\frac{\left(200 \times 10^9\right) \times\left(375 \times 10^{-6}\right)}{1.5 \times(0.5)^3} \\ \space \\ =\frac{200 \times 10^9 \times 375 \times 10^{-4} \times 10^3}{1.5 \times 125}=\pmb{4 \times 10^{10} N / m.}