How are end reactions calculated for a simply supported beam with multiple loads?

End reactions for a simply supported beam are calculated using static equilibrium: ΣM = 0 about one support to find the reaction at the other, then ΣFy = 0 for the remaining reaction. For a single point load P at distance a from the left support (L = total span): RA = P × b / L, RB = P × a / L where b = L − a. For a uniformly distributed load w over the full span: RA = RB = wL/2. For multiple loads, use superposition: calculate each load's contribution to RA and RB independently and sum them. Worked example — W18×35 beam, L = 30 ft with three loads: 5 kip at 8 ft, 8 kip at 15 ft, 3 kip at 22 ft. RA = (5×22 + 8×15 + 3×8)/30 = (110 + 120 + 24)/30 = 254/30 = 8.47 kips. RB = (5×8 + 8×15 + 3×22)/30 = (40 + 120 + 66)/30 = 226/30 = 7.53 kips. Check: RA + RB = 8.47 + 7.53 = 16.0 = 5 + 8 + 3. OK. For a trapezoidal load varying from w1 at left to w2 at right: total load = (w1 + w2) × L/2. Centroid from left = L × (2w2 + w1)/(3 × (w1 + w2)). RA = total × (L − centroid)/L, RB = total × centroid/L. For a beam with 2 k/ft at left increasing to 5 k/ft at right over 20 ft: total = (2+5)×20/2 = 70 kips, centroid = 20×(2×5+2)/(3×(2+5)) = 20×12/21 = 11.43 ft from left. RA = 70×(20−11.43)/20 = 30.0 kips, RB = 70×11.43/20 = 40.0 kips. The reaction at the heavier-loaded end is larger by the ratio of the centroid offsets.

What are the fixed-end moments and reactions for common load cases?

Fixed-end beams restrain rotation at supports, developing end moments that reduce the midspan positive moment but impose negative moments at the supports. For a uniformly distributed load w over full span L: end moments M_A = M_B = wL²/12 (fixed-end moment at each support), reactions R_A = R_B = wL/2. The midspan positive moment = wL²/24 (half of the simply supported moment wL²/8). For a center point load P: M_A = M_B = PL/8, R_A = R_B = P/2, midspan moment = PL/8 (the same magnitude as the fixed-end moment, but positive). For a point load at distance a from left (b = L − a): M_A = Pab²/L², M_B = Pa²b/L², R_A = Pb²(3a+b)/L³, R_B = Pa²(a+3b)/L³. For a beam fixed at both ends with UDL w and a W18×35 spanning 30 ft: M_A = M_B = 0.5×30²/12 = 37.5 kip-ft (assuming w = 1.0 k/ft). R_A = R_B = 1.0×30/2 = 15.0 kips. Midspan M = wL²/24 = 1.0×900/24 = 37.5 kip-ft. The maximum negative moment (-37.5 k-ft) at supports equals the positive moment at midspan — the beam is equally stressed positive and negative. For design, the fixed-end moments reduce the required section compared to a simply supported beam (M_max = wL²/8 = 112.5 k-ft vs 37.5 k-ft) but require moment-resisting connections at both ends to develop the full fixity.

How do cantilever beam reactions differ from simply supported beams?

A cantilever beam is fixed at one end and free at the other — the support must provide both vertical reaction and a moment reaction (fixity). For a uniformly distributed load w over the full cantilever length L: vertical reaction R = wL (acting upward at the fixed support to balance the total downward load), fixed-end moment M_fixed = wL²/2 (the maximum moment, occurring at the fixed support). For a concentrated load P at the free end: R = P, M_fixed = PL. For a linearly varying distributed load from zero at the free end to w_max at the fixed end: R = w_max × L/2, M_fixed = w_max × L²/6. For multiple loads on a cantilever, sum the reactions (R = ΣP_i for point loads) and sum the moments at the fixed end (M = Σ(P_i × a_i) for each point load at distance a_i from the fixed support). Worked example — 8 ft cantilever beam supporting: UDL w = 0.3 k/ft over full length, point load P1 = 4 kips at 3 ft from fixed end, P2 = 2 kips at 8 ft (free end). R = 0.3×8 + 4 + 2 = 2.4 + 4 + 2 = 8.4 kips upward. M_fixed = 0.3×8²/2 + 4×3 + 2×8 = 9.6 + 12 + 16 = 37.6 kip-ft. The fixed support must resist both — a base plate with anchor bolts in tension for the moment, combined with shear on the anchors/hole bearing for the vertical reaction. Deflection at the free end: Δ = wL⁴/(8EI) + Σ(P_i × a_i² × (3L − a_i)/(6EI)). For the example with W18×35 (I=510 in⁴): Δ_UDL = (0.3/12×96⁴)/(8×29000×510) + (4×36²×(3×96−36))/(6×29000×510) + (2×96²×96)/(3×29000×510) — calculate each term for total deflection.

What are the ACI moment coefficients for continuous beams?

The ACI 318 moment coefficients (Section 6.5) provide approximate moments and shears for continuous beams without running a full structural analysis. They apply when: spans are approximately equal (longest span ≤ 1.2 × shortest span), loads are uniformly distributed, live load ≤ 3 × dead load, and there are at least two spans. For continuous beams: positive moment in end span = wL²/14, positive moment in interior spans = wL²/16, negative moment at exterior face of first interior support = wL²/10 (or wL²/9 if only two spans), negative moment at other interior supports = wL²/11, negative moment at exterior support (beam built integrally with support) = wL²/16, negative moment at exterior support (beam resting on support) = 0, shear in end member at first interior support = 1.15 × wL/2, shear at other supports = wL/2. For a 3-span continuous beam with L=25 ft, w = 1.5 k/ft: End span positive M = 1.5×25²/14 = 1.5×625/14 = 67.0 k-ft. Interior span positive M = 1.5×625/16 = 58.6 k-ft. First interior negative M = 1.5×625/10 = 93.8 k-ft. Interior negative M = 1.5×625/11 = 85.2 k-ft. Exterior negative M (beam integral with column) = 1.5×625/16 = 58.6 k-ft. These coefficients assume full moment continuity — partially restrained connections would produce values between simply supported and fully continuous.

How does load position affect the maximum shear and moment in a beam?

The position of a concentrated load dramatically changes the maximum shear and moment. For shear: the absolute maximum shear in a simply supported beam equals the larger of the two end reactions, which is maximized when the load is closest to that support. A point load P placed 1 ft from the left support of a 30 ft beam: RA = P × 29/30 = 0.967P (the maximum possible shear is ~P when the load is infinitesimally close to the support). For moment: the maximum moment under a point load P occurs when the load is at midspan — Mmax = PL/4 = 0.25 PL. If P moves to 1/4 point from left: Munder_load = P × b × a / L = P × 22.5 × 7.5 / 30 = P × 168.75/30 = 0.1875 PL, only 75% of the midspan value. For maximum absolute moment in a beam with two moving loads (truck/crane): the maximum moment occurs when one load is as far to one side of midspan as the center of gravity of the load group is to the other side. For two loads P₁ and P₂ separated by distance s: the critical position places the center of gravity of the loads halfway between midspan and P₁ (the heavier load). The maximum moment = (P₁ + P₂) × (L/2 − s × P₂/(P₁+P₂)/2)² / L. For a 50 ft beam with P₁ = 20 kips and P₂ = 10 kips at 12 ft spacing: CG from P₁ = 12×10/30 = 4 ft. Place CG at midspan + 2 ft (so P₁ is at midspan): RA = (20×(50/2+2) + 10×(50/2+2−12))/50 = (20×27 + 10×15)/50 = (540+150)/50 = 13.8 kips. M under P₁ = 13.8 × 27 = 372.6 k-ft. Compare to single 30-kip load at midspan: M = 30×50/4 = 375 k-ft — similar magnitude.

Beam End Reactions — Formulas for Common Conditions

Q: How do I calculate reactions for a beam with a partial uniform load?

For a partial UDL of intensity w from x = a to x = b on a simply supported beam of span L: RA = w x (b-a) x (L - (a+b)/2) / L. Example: 2 k/ft from 5 ft to 15 ft on 30 ft beam. RA = 2 x 10 x (30 - 10) / 30 = 13.33 kips. RB = 20 - 13.33 = 6.67 kips.

Q: What is the maximum reaction for a trapezoidal load?

For a trapezoidal load from w1 at left to w2 at right over span L: total = (w1+w2) x L/2. Centroid = L x (2w2+w1) / (3 x (w1+w2)). RA = total x (L - centroid) / L. Example: w1=3 k/ft, w2=6 k/ft, L=20 ft. RA = 40.0 kips, RB = 50.0 kips.

Beam end reactions are the forces that the beam transfers to its supports. For simply supported beams, these are vertical forces. For fixed-end beams, they include both forces and moments. This page provides reaction formulas for common loading conditions and a worked example showing how to calculate reactions for a steel beam with multiple loads.

Simply Supported Beam Reactions

Point Load at Center

        P
        ↓
───────┬───────
  RA ↑         ↑ RB

RA = P/2, RB = P/2

Max moment: Mmax = PL/4

Point Load at Distance "a" from Left

           P
           ↓
─────┬───────────
 RA ↑    a    b    ↑ RB

RA = Pb/L, RB = Pa/L

where b = L - a. Max moment at load: Mmax = Pab/L

Uniform Distributed Load (UDL)

  w (per unit length)
  ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
─────────────────────
  RA ↑             ↑ RB

RA = wL/2, RB = wL/2

Max moment: Mmax = wL²/8

Triangular Load (Zero at Left, Max at Right)

        ↓ (increasing)
       ╱│
      ╱ │
     ╱  │
    ╱   │
─────────────
 RA ↑        ↑ RB

RA = wL/6, RB = wL/3

where w = maximum intensity at right end. Max moment at x = L/√3 from left.

Two Equal Point Loads at Third Points

    P         P
    ↓         ↓
────┬─────────┬────
 RA ↑             ↑ RB

RA = P, RB = P

Max moment: Mmax = PL/3

Moment at Left End

  M ↻
─────────────────────
  RA ↑             ↑ RB

RA = -M/L, RB = M/L

(Positive reaction = upward)

Fixed-End Beam Reactions

Point Load at Center

RA = RB = P/2 MA = MB = PL/8 (negative moment at supports)

Max positive moment: PL/8 at midspan

Uniform Distributed Load (UDL)

RA = RB = wL/2 MA = MB = wL²/12 (negative moment at supports)

Max positive moment: wL²/24 at midspan

Point Load at Distance "a" from Left

RA = Pb²(3a+b)/L³ RB = Pa²(a+3b)/L³ MA = Pab²/L² (negative) MB = Pa²b/L² (negative)

Cantilever Beam Reactions

Point Load at Free End

              P
              ↓
──────────────┤
  RA ↑     MA ↺

RA = P MA = PL (fixed-end moment, counterclockwise)

Max moment: PL at support

Uniform Distributed Load (UDL)

RA = wL MA = wL²/2 (fixed-end moment, counterclockwise)

Max moment: wL²/2 at support

Point Load at Distance "a" from Free End

RA = P MA = Pa (if a ≤ L)

Propped Cantilever (Fixed at Left, Pin at Right)

Uniform Distributed Load (UDL)

  ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
═════════════════════
  RA ↑, MA ↺      ↑ RB

RA = 5wL/8 RB = 3wL/8 MA = wL²/8 (at fixed end)

Max positive moment: 9wL²/128 at x = 3L/8 from left

Point Load at Center

RA = 11P/16 RB = 5P/16 MA = 3PL/16 (at fixed end)

Continuous Beam (Two Equal Spans)

UDL on Both Spans

Each span L, uniform load w:

RA = 5wL/8 RB (interior) = 5wL/4 (supports both spans) RC = 5wL/8

Interior support moment: MB = wL²/8

Quick Reference Table

Load Case	RA	RB	Max M
Central point load P	P/2	P/2	PL/4
UDL w	wL/2	wL/2	wL²/8
Eccentric load P at a	Pb/L	Pa/L	Pab/L
Two loads P at 1/3 pts	P	P	PL/3
Triangular (max at right)	wL/6	wL/3	0.064wL²
End moment M	-M/L	M/L	M
Fixed-fixed, central P	P/2	P/2	PL/8
Fixed-fixed, UDL w	wL/2	wL/2	wL²/12
Cantilever, tip load P	P	—	PL
Cantilever, UDL w	wL	—	wL²/2

Worked Example: Steel Beam with Multiple Loads

Problem

A W18x55 beam spans 30 ft. It carries a uniform dead load of 0.4 klf and a uniform live load of 0.8 klf, plus a concentrated live load of 15 kips at 10 ft from the left support.

Step 1: Factored Loads

wD = 0.4 klf, wL = 0.8 klf, PL = 15 kips

wu = 1.2 × 0.4 + 1.6 × 0.8 = 0.48 + 1.28 = 1.76 klf

Pu = 1.6 × 15 = 24 kips (live load only, no dead component)

Step 2: Reactions from UDL

RA1 = RB1 = wu × L / 2 = 1.76 × 30 / 2 = 26.4 kips

Step 3: Reactions from Point Load

Load at a = 10 ft from left, b = 20 ft

RA2 = Pu × b / L = 24 × 20 / 30 = 16.0 kips RB2 = Pu × a / L = 24 × 10 / 30 = 8.0 kips

Step 4: Total Reactions

RA = 26.4 + 16.0 = 42.4 kips RB = 26.4 + 8.0 = 34.4 kips

Check: RA + RB = 42.4 + 34.4 = 76.8 kips Total load = 1.76 × 30 + 24 = 52.8 + 24 = 76.8 kips ✓

Step 5: Select Connection Type

Reaction (kips)	Recommended Connection
≤ 30	Single plate (shear tab)
30-50	Single plate (thicker) or double angle
50-100	Double angle or stiffened seat
> 100	End plate or stiffened single plate

Left reaction RA = 42.4 kips → Use 1/2 in single plate shear tab with (4) 3/4 in A325-N bolts.

Right reaction RB = 34.4 kips → Use 3/8 in single plate shear tab with (3) 3/4 in A325-N bolts.

ACI Moment Coefficients for Continuous Beams

For continuous beams with approximately equal spans (longest <= 1.2 x shortest), uniform loading, and LL <= 3 x DL, the ACI 318 moment coefficients provide a quick approximate analysis without running a full structural analysis.

Positive Moments

Location	Coefficient	Moment = Coefficient x w x L²
End span, discontinuous end	wL² / 11	Positive moment
End span, continuous end	wL² / 14	Positive moment
Interior spans	wL² / 16	Positive moment

Negative Moments at Supports

Location	Coefficient	Moment = Coefficient x w x L²
Exterior support, beam integral	wL² / 16	Negative moment
Exterior support, beam resting	0	No continuity
First interior support (2 spans)	wL² / 9	Negative moment
First interior support (3+ spans)	wL² / 10	Negative moment
Other interior supports	wL² / 11	Negative moment

Shear Coefficients

Location	Coefficient	Shear = Coefficient x w x L
End reaction at exterior support	0.40
End reaction at first interior support	0.60
Interior support reactions (both sides)	0.50

Worked Example — 3-Span Continuous Beam

Problem: A three-span continuous steel beam (W21x44, Ix = 843 in^4), each span = 30 ft, uniform load w = 2.5 k/ft (dead + live, factored). Use ACI coefficients to find reactions and moments.

End span positive moment (continuous end): M = wL²/14 = 2.5 x 900 / 14 = 160.7 kip-ft.

Interior span positive moment: M = wL²/16 = 2.5 x 900 / 16 = 140.6 kip-ft.

First interior negative moment (3 spans): M = wL²/10 = 2.5 x 900 / 10 = 225.0 kip-ft.

Interior support negative moment: M = wL²/11 = 2.5 x 900 / 11 = 204.5 kip-ft.

Exterior negative moment (beam built into column): M = wL²/16 = 140.6 kip-ft.

End reactions: Rexterior = 0.40 x 2.5 x 30 = 30.0 kips. Rfirst_interior = 0.60 x 2.5 x 30 = 45.0 kips.

Comparison to exact analysis (three-moment equation): ACI coefficients provide results within 5-10% of exact elastic analysis for uniform loading with equal spans. The difference comes from the simplified LL pattern (ACI assumes all spans loaded, which is conservative for reactions but slightly unconservative for some positive moment locations).

Reaction Coefficient Table for Common Loading Patterns

The following table provides multiplication factors for calculating reactions for various load cases on simply supported beams. Multiply the total load by the coefficient to get the reaction at the left support (RA). RB = total load - RA.

Load Pattern	RA Coefficient	RB Coefficient	Notes
UDL over full span	0.500	0.500	Symmetric
Point load at midspan	0.500	0.500	Symmetric
Point load at quarter span (a = L/4)	0.750	0.250	Load closer to left support
Triangular load (zero at left, max at right)	0.333	0.667	Centroid at 2L/3 from left
Triangular load (max at left, zero at right)	0.667	0.333	Centroid at L/3 from left
Two equal loads at third points	0.500	0.500	P each, total = 2P
Three equal loads at quarter points	0.500	0.500	P each, total = 3P
Trapezoidal load (w1 at left, w2 at right)	(2w2+w1)/(3(w1+w2))	(2w1+w2)/(3(w1+w2))	w1 and w2 are end intensities
Partial UDL (length a from left, length b from right)	(2L-a)/(2L) x wa/L	see notes	For partial UDL, RA = w x a x (2L-a) / (2L)
Moment at right end	1.0	-1.0	RA = M/L, RB = -M/L

Frequently Asked Questions

How do I calculate beam reactions? For statically determinate beams, use equilibrium: sum of vertical forces = 0, and sum of moments about any point = 0. For a simply supported beam with a UDL: RA = RB = wL/2. For a point load at distance a: RA = Pb/L, RB = Pa/L.

What is the difference between reaction force and reaction moment? Reaction forces are the vertical (and sometimes horizontal) forces at supports. Reaction moments occur at fixed supports where rotation is prevented. Simple supports have only reaction forces. Fixed supports have both forces and moments.

How do I find reactions for a beam with multiple loads? Use superposition: calculate reactions for each load case separately, then add them together. For example, a beam with a UDL and a point load has total reactions equal to the sum of the UDL reactions and the point load reactions.

What connection type should I use for a 40 kip reaction? For 40 kips, a single plate shear tab (1/2 inch thick) with 4 bolts typically works. Alternatively, a double clip angle (2L4×3.5×3/8) provides more ductility. Check bolt shear, plate shear, and weld capacity.

Do I need to consider beam self-weight in reactions? Yes. Include the beam self-weight as part of the dead load. For a W18x55, add 55 lb/ft = 0.055 klf to the distributed dead load. In the example above, if beam weight was not already included in the 0.4 klf dead load, the total wD would be 0.455 klf.

What are the maximum reactions for common connection types? Single plate shear tab: 50-60 kips (typical limit). Double angle: 70-100 kips. End plate: 100+ kips. Stiffened seat: 150+ kips. These are practical limits, not code limits. Actual capacity depends on the specific connection design.

How do I calculate reactions for a beam with a partial uniform load? For a partial UDL of intensity w extending from x = a to x = b on a simply supported beam of span L: RA = w x (b-a) x (L - (a+b)/2) / L. For example, a 2 k/ft load from 5 ft to 15 ft on a 30 ft beam: RA = 2 x 10 x (30 - 10) / 30 = 20 x 20 / 30 = 13.33 kips. RB = total load - RA = 20 - 13.33 = 6.67 kips. The max shear is RA = 13.33 kips (at left). The shear changes by w x (x - a) across the loaded region.

What is the maximum reaction for a trapezoidal load? For a trapezoidal load varying from w1 at the left to w2 at the right over span L: total load = (w1 + w2) x L / 2. Centroid from left = L x (2w2 + w1) / (3 x (w1 + w2)). RA = total x (L - centroid) / L. RB = total x centroid / L. Example: w1 = 3 k/ft, w2 = 6 k/ft, L = 20 ft. Total = (3+6) x 20/2 = 90 kips. Centroid = 20 x (12+3) / (3x9) = 20 x 15/27 = 11.11 ft. RA = 90 x (20-11.11)/20 = 40.0 kips. RB = 90 x 11.11/20 = 50.0 kips. Note the reaction is larger at the heavier-loaded end.

Beam Calculator — Full beam analysis tool
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SFD & BMD Calculator — Shear and moment diagram tool

Disclaimer

This is a calculation tool, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in construction, fabrication, or permit documents. The user is responsible for the accuracy of all inputs and the verification of all outputs.

Beam End Reactions — Formulas for Common Conditions

Simply Supported Beam Reactions

Point Load at Center

Point Load at Distance "a" from Left

Uniform Distributed Load (UDL)

Triangular Load (Zero at Left, Max at Right)

Two Equal Point Loads at Third Points

Moment at Left End

Fixed-End Beam Reactions

Point Load at Center

Uniform Distributed Load (UDL)

Point Load at Distance "a" from Left

Cantilever Beam Reactions

Point Load at Free End

Uniform Distributed Load (UDL)

Point Load at Distance "a" from Free End

Propped Cantilever (Fixed at Left, Pin at Right)

Uniform Distributed Load (UDL)

Point Load at Center

Continuous Beam (Two Equal Spans)

UDL on Both Spans

Quick Reference Table

Worked Example: Steel Beam with Multiple Loads

Problem

Step 1: Factored Loads

Step 2: Reactions from UDL

Step 3: Reactions from Point Load

Step 4: Total Reactions

Step 5: Select Connection Type

ACI Moment Coefficients for Continuous Beams

Positive Moments

Negative Moments at Supports

Shear Coefficients

Worked Example — 3-Span Continuous Beam

Reaction Coefficient Table for Common Loading Patterns

Frequently Asked Questions

Related Pages

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