Respuesta :
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
[tex]EI\frac{d^2y}{dx^2} = M = R_1[x] - 350[x-11]-\frac{wx^2}{2}[/tex]
[tex]EI\frac{d^2y}{dx^2} = M = 366.2 \cdot [x] - 350[x-11]-\frac{6.5\cdot x^2}{2}[/tex]
By integrating once we have
[tex]EI\frac{dy}{dx} = 366.2 \cdot \frac{ [x]^2}{2} - 350\cdot \frac{ [x-11]^2}{2}-\frac{6.5\cdot x^3}{6} + C[/tex]..................(1)
Second integration gives
[tex]EIy= 366.2 \cdot \frac{ [x]^3}{6} - 350\cdot \frac{ [x-11]^3}{6}-\frac{6.5\cdot x^4}{24} + Cx + B[/tex]......................(2)
The boundary conditions are, at x = 0, y = 0 and at x = 37. y = 0
From equation 2 we have at x = 0
[tex]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[/tex]
Which gives 0 - 0 - 0 + 0 + B Since we ignore the bracket with negative value
When x = 37, equation 2 becomes
[tex]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[/tex]
[tex]EI(0) = 3091521.433- 1025266.67-507585.27 + C\cdot 37[/tex]
37·C = -1558669.5
C = -42126.2
Therefore we have
[tex]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[/tex]
When x = A = 11 in, we have
[tex]EIy= 366.2 \cdot \frac{ [11]^3}{6} - 350\cdot \frac{ [11-11]^3}{6}-\frac{6.5\cdot 11^4}{24} -42126.2\cdot 11[/tex]
= -386118.133
y[tex]_{11}[/tex] = -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
[tex]EIy= 366.2 \cdot \frac{ [18.5]^3}{6} - 350\cdot \frac{ [18.5-11]^3}{6}-\frac{6.5\cdot 18.5^4}{24} -42126.2\cdot 18.5[/tex]
EIy = -449228.023 lbf·in⁴
[tex]y_{(mid \, span)}[/tex] = -449228.023 lbf·in⁴/(200×10⁸ lbf·in²) = -2.24×10⁻⁵ in or
2.24×10⁻⁵ in
The percentage difference is
[tex]\% \,Difference = \frac{Difference}{\frac{New\,Value+Initial \,Value}{2} } \times 100 =\frac{2.24-1.93}{\frac{1.93+2.24}{2} } \times 100[/tex] = 0.1511×100 = 15.11 %.
