Support reactions for a beam at an incline

May 2019
7
0

img: https://imgur.com/a/IqYT8tq
Hello

I have a problem with calculating the support reactions for a beam. Lefts side of beam has a pin connection so it takes both \(\displaystyle Fx, Fy\). Right side of the beam has a horizontal roller and it takes only \(\displaystyle Fy\) in the direction of the wall. Therefore at the pin support \(\displaystyle Fy=9kN\), but how do i figure out the reactions for X-direction? i Thought about the X-component as a tangent to the 2 point loads so \(\displaystyle 5kN/tan(21.8); 4kN/tan(21.8)\) and they give me 12,5 and 10kN respectively. I don't Know if they are correct and if they are how do they distribute to each support?

I can also think of another way of doing this and that is by making the beam horizontal and turning the roller support 68.2 degrees, then calculating the reactions which come out to 3,8kN for Ay 9,5kN to Ax
5,2kN for By and 13kN for Bx. Is either of these methods correct?
Thanks
 
Last edited:
Jan 2019
44
33
Translational equilibrium

$\displaystyle \sum F_x = 0 \implies F_{x_1} = F_{x_2}$

$\displaystyle \sum F_y = 0 \implies F_{y_1} = 4kN + 5kN$


Rotational equilibrium about the lower left pivot point

$\displaystyle \sum \tau = 0 \implies F_{x_2} \cdot L \cdot \sin{\phi} = 4kN \cdot \dfrac{3L}{10} \cdot \sin{\theta} + 5kN \cdot \dfrac{4L}{5} \cdot \sin{\theta}$

$F_{x_2} \cdot L \cdot \dfrac{2}{L} = 4kN \cdot \dfrac{3L}{10} \cdot \dfrac{5}{L} + 5kN \cdot \dfrac{4L}{5} \cdot \dfrac{5}{L}$

$F_{x_2} = 13kN$
 

Attachments

  • Like
Reactions: 1 person
May 2019
7
0
Translational equilibrium

$\displaystyle \sum F_x = 0 \implies F_{x_1} = F_{x_2}$

$\displaystyle \sum F_y = 0 \implies F_{y_1} = 4kN + 5kN$


Rotational equilibrium about the lower left pivot point

$\displaystyle \sum \tau = 0 \implies F_{x_2} \cdot L \cdot \sin{\phi} = 4kN \cdot \dfrac{3L}{10} \cdot \sin{\theta} + 5kN \cdot \dfrac{4L}{5} \cdot \sin{\theta}$

$F_{x_2} \cdot L \cdot \dfrac{2}{L} = 4kN \cdot \dfrac{3L}{10} \cdot \dfrac{5}{L} + 5kN \cdot \dfrac{4L}{5} \cdot \dfrac{5}{L}$

$F_{x_2} = 13kN$
Thank you very much. My mistake was thinking the distance between those supports is 5m but instead it is 2m.