50KN20tw/mE=2006 PaII Jo EB200 mmcross-sectionof the beamzm5mImusing ony methad, solvefur6) Domax A

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Answered: Using ony method, solvefor @tc max A

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

50KN
20tw/m
E=2006 Pa
I
I Jo E
B
200 mm
cross-section
of the beam
zm
5m
Im
using ony methad, solvefur
6) Do
max A

Expert Solution

Step 1 : Diagram

Step 2 : Equations

$\begin{array}{rcl} B y m a c a u l a y' s m e t h o d, \\ E I \frac{d^{2} y}{d x^{2}} & = & M_{x} = 75 x - 50 (x - 2) - 20 \times 5 \times (x - 6.5) \\ I n t e g r a t e a b o v e e q n w i t h r e s p e c t t o x, \\ E I \frac{d y}{d x} & = & \frac{75 x^{2}}{2} - \frac{50 {(x - 2)}^{2}}{2} - 100 \times \frac{{(x - 6.5)}^{2}}{2} + c_{1} . . (1) \\ A g a i n, i n t e g r a t i o n \\ E I y & = & 37.5 \frac{x^{3}}{3} - 25 \times \frac{{(x - 2)}^{3}}{3} - \frac{50}{3} \times {(x - 6.5)}^{3} + c_{1} x + c_{2} . . (2) \end{array}$

Step 3 : Boundary condition

$\begin{array}{rcl} A t x & = & 0 m, y = 0 \Rightarrow c_{2} = 0 (f r o m e q n 2) \\ A t x & = & 10 m, y = 0 \\ F r o m e q n 2, \\ 0 & = & \frac{37.5 \times 10^{3}}{3} - \frac{25 \times 8^{3}}{3} - \frac{50 \times 3 . 5^{3}}{3} + 10 c_{1} \\ c_{1} & = & - 751.875 \\ E q n 1 b e c o m e s, \\ E I \frac{d y}{d x} & = & 37.5 x^{2} - 25 {(x - 2)}^{2} - 50 {(x - 6.5)}^{2} - 751.875 . . (3) \\ E q n 2 b e c o m e s, \\ E I y & = & \frac{37.5 x^{3}}{3} - \frac{25 {(x - 2)}^{3}}{3} - \frac{50 {(x - 6.5)}^{3}}{3} - 751.875 x . . . (4) \end{array}$

Step 4 : Terms

$\begin{array}{rcl} M o m e n t o f i n e r t i a (I) & = & \frac{π}{64} D^{4} = \frac{π}{64} \times 0 . 2^{4} \\ = & 7.85398 \times 10^{- 5} m^{4} \\ M o d u l u s o f e l a s t i c i t y (E) & = & 200 G P a = 200 \times 10^{6} K N / m^{2} \\ F l e x u r a l R i g i d i t y (E I) & = & 200 \times 10^{6} \times 7.85398 \times 10^{- 5} \\ = & 15707.96 \approx 15708 K N . m^{2} \end{array}$

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