Beam Reaction Calculator
Calculate support reactions, maximum moment, and shear for simply supported beams with uniform loads, point loads, or combined loading.
Beam Reaction Calculator
Calculator
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Formula: Ra + Rb = Total Load; Sum of Moments about A = 0 to find Rb
Worked example โ Ra = 5,000 lbs, Rb = 5,000 lbs, M_max = 25.00 kip-ft
Formula
Ra + Rb = Total Load; Sum of Moments about A = 0 to find Rb
Apply static equilibrium: the sum of all vertical forces equals zero (Ra + Rb = total applied load), and the sum of moments about any point equals zero. Taking moments about the left support eliminates Ra and allows solving for Rb directly. For uniform load: Ra = Rb = wL/2. For point load at distance a: Ra = P(L-a)/L, Rb = Pa/L.
Worked Examples
Example 1: Uniform Load on Simply Supported Beam
Problem:Find reactions for a 20-ft beam with 500 plf uniform load.
Solution:Total load = 500 x 20 = 10,000 lbs Ra = Rb = 10,000 / 2 = 5,000 lbs each Max moment = 500 x 20^2 / 8 = 25,000 lb-ft Max shear = 5,000 lbs at supports
Result:Ra = 5,000 lbs, Rb = 5,000 lbs, M_max = 25.00 kip-ft
Example 2: Off-Center Point Load
Problem:Find reactions for a 16-ft beam with 8,000 lb point load at 6 ft from left support.
Solution:a = 6 ft, b = 10 ft, P = 8,000 lbs Ra = P x b/L = 8,000 x 10/16 = 5,000 lbs Rb = P x a/L = 8,000 x 6/16 = 3,000 lbs Moment at load = 5,000 x 6 = 30,000 lb-ft
Result:Ra = 5,000 lbs, Rb = 3,000 lbs, M = 30.00 kip-ft
Frequently Asked Questions
How do I calculate beam reactions for a simply supported beam?
For a simply supported beam, use static equilibrium equations. Sum of vertical forces equals zero: Ra + Rb = total load. Sum of moments about one support equals zero to solve for the other reaction. For a uniform load w over span L, both reactions equal wL/2. For a point load P at distance a from the left support, Ra = P(L-a)/L and Rb = Pa/L. These principles apply to any combination of loads using superposition for linear elastic analysis.
What is the difference between a reaction force and an internal force?
Reaction forces are the external forces that the supports exert on the beam to maintain equilibrium. They act at the support points and represent the load transferred to the supporting structure below. Internal forces (shear and moment) exist within the beam at every cross section and represent the forces the beam material must resist. A reaction force at a wall column or footing determines the required capacity of that supporting element. Internal forces determine the required size and strength of the beam itself.
How does the location of a point load affect the reactions?
Moving a point load closer to one support increases the reaction at the closer support and decreases the reaction at the far support. When the load is at midspan, both reactions are equal at P/2. When the load is directly over one support, that support carries the entire load and the other reaction is zero. The relationship is linear: the reaction at a support equals the point load multiplied by the distance from the load to the opposite support, divided by the span length.
Can I add multiple load types using superposition?
Yes, for linear elastic structures you can analyze each load separately and add the results. This is called the principle of superposition. Calculate reactions for the uniform load alone, then for each point load alone, then sum all the individual reactions at each support. This works for reactions, shear forces, bending moments, and deflections. Superposition is valid as long as the material remains in the elastic range and the deformations are small relative to the beam dimensions.
What is static determinacy and why does it matter for hand calculations?
A simply supported beam has one pin and one roller, giving two unknown vertical reactions that two equilibrium equations, sum of vertical forces and sum of moments, can solve directly from geometry alone. Add a third support or fix an end and the beam becomes statically indeterminate: the reactions then depend on member stiffness, and you need a compatibility or stiffness method rather than statics alone.
What is the difference between a pin support and a roller support?
A pin restrains translation in both directions and provides vertical and horizontal reactions while allowing rotation. A roller restrains only translation perpendicular to its surface, providing a single vertical reaction and permitting the beam to move along its axis. Pairing one of each lets the beam expand and contract with temperature without generating axial force, which is why that combination is the standard idealisation.
Where does the maximum moment occur under a uniform load compared with an off-centre point load?
Under a symmetric uniform load the peak moment is at midspan and equals w L squared over eight. Under a single point load the peak sits directly beneath the load and equals P times a times b divided by L, where a and b are the distances from the load to each support. With the load at midspan those distances are both L over two, and the expression reduces to P L over four.
How do I locate the point of zero shear on a beam?
The maximum moment occurs where the shear diagram crosses zero, because shear is the rate of change of moment along the beam. Under a uniform load, start from the left reaction and travel a distance equal to that reaction divided by the load per unit length. With point loads the shear diagram steps rather than slopes, so the crossing usually falls exactly at a load location.
What happens to the reactions when part of the load overhangs a support?
Load beyond a support acts as a cantilever and rotates the beam about that support, which reduces the reaction at the far end. If the overhanging load is large enough, the far reaction goes negative, meaning the beam tries to lift off its support. That uplift must be resisted by a hold-down connection or by sufficient counteracting dead load, otherwise the beam is unstable regardless of its bending capacity.
How do reactions change when a beam runs continuously over three supports?
A continuous beam is statically indeterminate, so the equilibrium equations alone cannot give the reactions; compatibility of slopes and deflections is needed too, through methods such as the three-moment equation or moment distribution. The interior support attracts a disproportionate share of the load and also carries a hogging moment over it. Treating each span as separately simply supported understates that interior reaction and misses the negative moment entirely.
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Reviewed for accuracy by Abdullah, Technical Content Specialist ยท Editorial policy
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