Solution to Problem 830 at Strength of Materials Forum

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November 2, 2017 - 4:10am

#1

griflyn

Solution to Problem 830

I'm learning how to do reactions with the Three Moment Theorem and in the solution for problem 830, it has 1077.27/12 = 538.64.....this clearly isn't correct.

Am I missing something? I tried searching to see if this has been addressed elsewhere, but some guidance would be very helpful and many thanks in advance.

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work problem

November 2, 2017 - 7:05am

#2

Jhun Vert

Yes, it is clearly not correct. Thank you for pointing this out. We flag the page for correction and will update it the soonest possible time. Again, thank you.

November 2, 2017 - 7:50am

#3

Jhun Vert

Correction has been applied to the page. For this update, the maximum positive moment occurred under the 1st 300 lb concentrated load, it is equivalent to 840.92 lb·ft. Link to page: https://www.mathalino.com/node/2322

We retained the erroneous solution below for record purposes.

Click here to show or hide the solution

Problem 830
Refer to Problem 815.

Solution 830
From Solution 815
$M_2 = -1077.27 ~ \text{lb}\cdot\text{ft}$

$M_3 = -640 ~ \text{lb}\cdot\text{ft}$

$y = 60 ~ \text{lb/ft}$

From the first span

Reactions due to loadings
$V_{1L} = V_{1R} = 300 ~ \text{lb}$

Reactions due to end moments
${R_1}' = \dfrac{M_2 - M_1}{L_1} = \dfrac{1077.27 - 0}{12} = 538.64 ~ \text{lb}$

Thus,
$R_{1L} = V_{1L} - {R_1}' = 300 - 538.64$

$R_{1L} = -238.64 ~ \text{lb}$

$R_{1R} = V_{1R} + {R_1}' = 300 + 538.64$

$R_{1R} = 838.64 ~ \text{lb}$

From the second span

Reactions due to loadings
$\Sigma M_{V_{2R}} = 0$

$10V_{2L} = \frac{1}{2}(10)(135)[ \, \frac{2}{3}(10) \, ] + \frac{1}{2}(10)(60)[ \, \frac{1}{3}(10) \,]$

$V_{2L} = 550 ~ \text{lb}$

$\Sigma M_{V_{2L}} = 0$

$10V_{2R} = \frac{1}{2}(10)(135)[ \, \frac{1}{3}(10) \, ] + \frac{1}{2}(10)(60)[ \, \frac{2}{3}(10) \,]$

$V_{2R} = 425 ~ \text{lb}$

Reactions due to end moments
${R_2}' = \dfrac{M_2 - M_3}{L_2} = \dfrac{1077.27 - 640}{10}$

${R_2}' = 43.73 ~ \text{lb}$

Thus,
$R_{2L} = V_{2L} + {R_2}' = 550 + 43.73$

$R_{2L} = 593.73 ~ \text{lb}$

$R_{2R} = V_{2R} - {R_2}' = 425 - 43.73$

$R_{2R} = 381.27 ~ \text{lb}$

From the third span

$\Sigma F_V = 0$

$R_{3L} = \frac{1}{2}(8)(60) = 240 ~ \text{lb}$

From the figure above
$R_1 = 238.64 ~ \text{lb downward}$ answer

$R_2 = 838.64 + 593.73 = 1432.37 ~ \text{lb}$ answer

$R_3 = 381.27 + 240 = 621.27 ~ \text{lb}$ answer

From the triangular load
$\dfrac{y}{x} = \dfrac{135}{18}$

$y = 7.5x$

Solve for x
$\Sigma F_{\text{to the right of }A} = 0$

$621.27 - \frac{1}{2}xy = 0$

$621.27 - \frac{1}{2}x(7.5x) = 0$

$3.75x^2 = 621.27$

$x = 12.8714 ~ \text{ft}$

Another way to solve for x

Another way to solve x is by squared property of parabola
$\dfrac{x^2}{240 + 381.27} = \dfrac{18^2}{240 + 381.27 + 593.73}$

$x = 12.8714 ~ \text{ft}$ okay

$M_{max(+)} = \Sigma M_{\text{to the right of }A}$

$M_{max(+)} = 621.27(x - 8) - \frac{1}{2}xy(\frac{1}{3}x) = 621.27(x - 8) - \frac{1}{6}x^2(7.5x)$

$M_{max(+)} = 621.27(x - 8) - 1.25x^3 = 621.27(12.8714 - 8) - 1.25(12.8714^3)$

$M_{max(+)} = 360.90 ~\text{lb}\cdot\text{ft}$ answer

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