Finding Reactions In Simply Supported Beams

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The discussion focuses on calculating reactions in simply supported beams, specifically at points Ra and Rb. The initial calculations led to Ra being 3.33 kN and Rb 11.67 kN, but there were errors in summing moments and assigning signs to forces. Participants emphasized the importance of correctly measuring perpendicular distances and using appropriate signs for forces when calculating moments about a point. A corrected formula was suggested to properly account for the forces and distances involved. Accurate moment summation is crucial for determining the correct reaction forces in beam analysis.
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Homework Statement


Hope you can access this link for question http://s353.photobucket.com/albums/r387/james_nikko/?action=view&current=img002.jpg

Supports are at 1.5m and 4.5m

Homework Equations


The Attempt at a Solution


Ok so i have attempted the problem using the following
(5x1.5) + (5x3) + (5x6) = (Rb x 4.5)

Therefore giving:
(Rb x 4.5) = 52.5
Rb= 11.67kN

Therefore:
Ra = (5+5+5) - Rb
Ra = 3.33kN
 
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When the beam is in equilibrium, you can sum moments of a force about any point and set the sum equal to zero, but you must compute the moments of each force about that same chosen point, and then watch your plus and minus signs (clockwise vs. counterclockwise). Although you can choose any point, it is convenient to choose a point on the beam at one of the reaction supports. Try again to sum moments about R_a, and watch your plus and minus signs. Note from the symmetry of the loading and beam, the values of the end reactions, as you gain more experrence, should pop right out at you.
 
ok so my first force which is before point Ra on an overhang should be negative whereas the other two forces are positive? assuming i use a negative reaction at b of course. therefore formula would be, -(5x1.5) + (5x3) + (5x6) - (Rbx4.5)?
 
Jamesnikko said:
ok so my first force which is before point Ra on an overhang should be negative whereas the other two forces are positive? assuming i use a negative reaction at b of course. therefore formula would be, -(5x1.5) + (5x3) + (5x6) - (Rbx4.5)?
You are not summing moments correctly. All perpendicular distances must be measured between the force and the point (Ra) about which you are summing moments.

So I'll start you off, it's -(5 x 1.5) + (5 x 1.5) + (5 x ___) -(Rb x ___) = 0. Fill in the blanks and solve for Rb.
 
sorry i was working from memory :) formula should be
-(1.5 x 5) + (5 x 1.5) + (5 x 4.5) - (Rb x 3)
My major problem was that i wasn't using the force to the left of reaction a as a - number.
Thanks.
 
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