Answer:
The deflection at A is 1.93×10⁻⁵ in and
The deflection at the midspan is 2.24×10⁻⁵ in
The percentage difference is 15.11 %
Explanation:
Here we have
Taking moments about the right end
350 × 26 + 6.5 × 37²/2 = 13549.25 = R₁×37
R₁ = 13549.25 /37 = 366.2 lbf
Weight of steel shaft= 6.5 × 37 = 240.5 lbf
Total downward force = 350 + 240.5 = 590.5 lbf
∴ R₂ = 590.5 lbf - 366.2 lbf = 224.3 lbf
The bending moment equations are
![EI\frac{d^2y}{dx^2} = M = R_1[x] - 350[x-11]-\frac{wx^2}{2}](https://tex.z-dn.net/?f=EI%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20M%20%3D%20R_1%5Bx%5D%20-%20350%5Bx-11%5D-%5Cfrac%7Bwx%5E2%7D%7B2%7D)
![EI\frac{d^2y}{dx^2} = M = 366.2 \cdot [x] - 350[x-11]-\frac{6.5\cdot x^2}{2}](https://tex.z-dn.net/?f=EI%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20M%20%3D%20366.2%20%5Ccdot%20%5Bx%5D%20-%20350%5Bx-11%5D-%5Cfrac%7B6.5%5Ccdot%20x%5E2%7D%7B2%7D)
By integrating once we have
..................(1)
Second integration gives
......................(2)
The boundary conditions are, at x = 0, y = 0 and at x = 37. y = 0
From equation 2 we have at x = 0
![EI(0)= 366.2 \cdot \frac{ [0]^3}{6} - 350\cdot \frac{ [0-11]^3}{6}-\frac{6.5\cdot 0^4}{24} + C0 + B](https://tex.z-dn.net/?f=EI%280%29%3D%20%20366.2%20%5Ccdot%20%5Cfrac%7B%20%5B0%5D%5E3%7D%7B6%7D%20-%20350%5Ccdot%20%5Cfrac%7B%20%5B0-11%5D%5E3%7D%7B6%7D-%5Cfrac%7B6.5%5Ccdot%200%5E4%7D%7B24%7D%20%2B%20C0%20%2B%20B)
Which gives 0 - 0 - 0 + 0 + B Since we ignore the bracket with negative value
When x = 37, equation 2 becomes
![EI(0) = 366.2 \cdot \frac{ [37]^3}{6} - 350\cdot \frac{ [37-11]^3}{6}-\frac{6.5\cdot 37^4}{24} + C\cdot 37 + B](https://tex.z-dn.net/?f=EI%280%29%20%3D%20%20366.2%20%5Ccdot%20%5Cfrac%7B%20%5B37%5D%5E3%7D%7B6%7D%20-%20350%5Ccdot%20%5Cfrac%7B%20%5B37-11%5D%5E3%7D%7B6%7D-%5Cfrac%7B6.5%5Ccdot%2037%5E4%7D%7B24%7D%20%2B%20C%5Ccdot%2037%20%2B%20B)
![EI(0) = 3091521.433- 1025266.67-507585.27 + C\cdot 37](https://tex.z-dn.net/?f=EI%280%29%20%3D%20%203091521.433-%201025266.67-507585.27%20%2B%20C%5Ccdot%2037)
37·C = -1558669.5
C = -42126.2
Therefore we have
![EIy= 366.2 \cdot \frac{ [x]^3}{6} - 350\cdot \frac{ [x-11]^3}{6}-\frac{6.5\cdot x^4}{24} -42126.2\cdot x](https://tex.z-dn.net/?f=EIy%3D%20%20366.2%20%5Ccdot%20%5Cfrac%7B%20%5Bx%5D%5E3%7D%7B6%7D%20-%20350%5Ccdot%20%5Cfrac%7B%20%5Bx-11%5D%5E3%7D%7B6%7D-%5Cfrac%7B6.5%5Ccdot%20x%5E4%7D%7B24%7D%20-42126.2%5Ccdot%20x)
When x = A = 11 in, we have
= -386118.133
y
= -386118.133 /EI = -386118.133 /(200×10⁸ lbf·in²) = -1.93×10⁻⁵ in or 1.93×10⁻⁵ in
At the midspan, we have x = 37/2 in = 18.5
EIy = -449228.023 lbf·in⁴
= -449228.023 lbf·in⁴/(200×10⁸ lbf·in²) = -2.24×10⁻⁵ in or
2.24×10⁻⁵ in
The percentage difference is
= 0.1511×100 = 15.11 %.