Training Civil Engineering Beam Reactions & Equilibrium
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Beam Reactions & Equilibrium

24 min Civil Engineering

Beam Reactions & Equilibrium

Structural engineers analyze beams by applying static equilibrium: the sum of all forces and moments must equal zero.

Equilibrium Equations (2-D)

$$\sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum M = 0$$

These three equations let us find unknown support reactions on a statically determinate beam.

Types of Supports

SupportReactions Provided
Pin (hinge)Horizontal + Vertical ($R_{Ax}$, $R_{Ay}$)
RollerVertical only ($R_B$)
FixedHorizontal + Vertical + Moment ($R_x$, $R_y$, $M$)
Example 1

A simply supported beam of length $L = 10$ m carries a concentrated load $P = 20$ kN at 4 m from the left support A. Find the reactions at A and B.

Take moments about A:

$$\sum M_A = 0: \quad R_B(10) - 20(4) = 0 \implies R_B = 8 \text{ kN}$$

Vertical equilibrium:

$$\sum F_y = 0: \quad R_A + 8 - 20 = 0 \implies R_A = 12 \text{ kN}$$

Example 2

A cantilever beam of length 6 m has a uniform distributed load $w = 5$ kN/m over its entire span. Find the reactions at the fixed support.

Total load: $W = 5 \times 6 = 30$ kN acting at the midpoint (3 m from the fixed end).

$$R_y = 30 \text{ kN} \qquad M = 30 \times 3 = 90 \text{ kN·m (counterclockwise)}$$

Example 3

A simply supported beam carries loads of 10 kN at 2 m and 15 kN at 6 m from left support A. Span = 8 m. Find reactions.

$$\sum M_A = 0: \quad R_B(8) = 10(2) + 15(6) = 20 + 90 = 110$$

$$R_B = 13.75 \text{ kN}$$

$$R_A = 10 + 15 - 13.75 = 11.25 \text{ kN}$$

Practice Problems

1. Simply supported beam, $L = 12$ m, point load $P = 24$ kN at midpoint. Find $R_A$ and $R_B$.
2. Simply supported beam, $L = 10$ m, point load $P = 30$ kN at 3 m from A. Find reactions.
3. Cantilever beam, $L = 4$ m, point load $P = 10$ kN at the free end. Find the reaction and moment at the fixed end.
4. Simply supported beam, $L = 8$ m, UDL $w = 6$ kN/m over entire span. Find reactions.
5. Simply supported beam, two point loads: 20 kN at 2 m and 10 kN at 7 m from A. Span = 9 m. Find reactions.
6. Cantilever, $L = 5$ m, UDL $w = 8$ kN/m. Find fixed-end reactions.
7. Simply supported beam with an overhang: span AB = 6 m, overhang BC = 2 m. Load of 12 kN at C. Find $R_A$ and $R_B$.
8. Simply supported beam, $L = 10$ m, triangular load from 0 at A to $w = 12$ kN/m at B. Find reactions.
9. Simply supported beam, $L = 8$ m, moment of 40 kN·m applied at midpoint. Find reactions.
10. A beam rests on three supports (A, B at 5 m, C at 10 m). UDL of 4 kN/m over entire span. This is statically indeterminate — how many additional equations are needed?
11. Simply supported beam, $L = 6$ m, point loads of 8 kN at 1 m, 2 m, 3 m, 4 m, and 5 m from A. Find reactions.
12. Cantilever $L = 3$ m. Point load 15 kN at 1 m from fixed end. Find reactions.
Show Answer Key

1. $R_A = R_B = 12$ kN (symmetric)

2. $R_B = 9$ kN, $R_A = 21$ kN

3. $R = 10$ kN (up), $M = 40$ kN·m (CCW)

4. $R_A = R_B = 24$ kN (symmetric UDL)

5. $\sum M_A: R_B(9)=20(2)+10(7)=110$; $R_B=12.22$ kN, $R_A=17.78$ kN

6. $R=40$ kN, $M=100$ kN·m

7. $\sum M_A: R_B(6)=12(8)$; $R_B=16$ kN, $R_A=-4$ kN (downward)

8. Total load $= \frac{1}{2}(12)(10)=60$ kN at $\frac{2}{3}(10)=6.67$ m from A; $R_B=40$ kN, $R_A=20$ kN

9. $R_A = -5$ kN (down), $R_B = 5$ kN (up) — a couple

10. One additional equation (degree of indeterminacy = 1)

11. Total load $=40$ kN, $\sum M_A: R_B(6)=8(1+2+3+4+5)=120$; $R_B=20$ kN, $R_A=20$ kN

12. $R=15$ kN, $M=15$ kN·m

Bending Stress & the Flexure Formula