Please see attached sheet for the help required.

I have an answer of 15KN and the length as 2.667. Please can anyone help and show workings out so I can use it as a check,

Many Thanks.

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- Mar 28th 2014, 02:55 AMCivilTraineeHelp Needed Please. Beam Calculations
Please see attached sheet for the help required.

I have an answer of 15KN and the length as 2.667. Please can anyone help and show workings out so I can use it as a check,

Many Thanks. - Mar 28th 2014, 05:17 AMebainesRe: Help Needed Please. Beam Calculations
Yes, those answers are correct. Start with $\displaystyle \normalsiize \sum F_y = 0$ to find $\displaystyle R_A$. Then use $\displaystyle \normalsize \sum T_A = 0 $ to solve for d.

- Mar 28th 2014, 06:04 AMCivilTraineeRe: Help Needed Please. Beam Calculations
Thank you for the quick response,

However, I am unsure of what your mean by Fy and Ta.

Please could you expand on this in more detail to help me understand how you worked this out.

Thanks - Mar 28th 2014, 07:32 AMebainesRe: Help Needed Please. Beam Calculations
For the beam to be in equilibrium the sum of all forces acting in the vertical direction (or the y-direction in an x-y coordinate system) must equal 0. So:

$\displaystyle \sum F_y = 0 = R_A + R_B - 10KN - 15KN - 30KN$

You know that R_B = 40KN, so you can solve for R_A.

Also to be in equilibrium the sum of moments (torques) about any point must equal 0. If you take sum of moments about support A you get:

$\displaystyle \sum T_A = 0 = 10KN(0.5m) -15KN(1m) + R_B(1.5m) -30KN(d-1)$

Again, given that R_B = 40KN, you can now solve for d.