Engineer's Notebook §2.1

Reactions and shear

A simply supported beam carries a point load P at distance a from support A. Taking moments about each support gives the reactions:

RA = P·bL,RB = P·aL
P
point load
N
L
span
m
a
load distance from A
m
b
load distance from B, L − a
m
RA
reaction at A
N
RB
reaction at B
N
V
shear force
N
w
load per length
N/m
x
distance from A
m
A beam on a pin at A and a roller at B, with a point load that can move along it. Arrows show the two support reactions, and below the beam the shear force diagram steps from plus R A to minus R B under the load. A B P = 10 kN at a = 2 m R = 6.67 kN R = 3.33 kN SHEAR FORCE V +6.67 kN −3.33 kN
Fig. 2.1-1Reaction arrows and shear diagram are drawn to scale with the load.
RA = Pb/L6.67 kN
RB = Pa/L3.33 kN
|V|max6.67 kN

Worked example

Uses the values in Fig. 2.1-1. Change them and the working updates.

Given
  • Ppoint load10 kN
  • Lspan between supports6 m
  • aload distance from A2 m
Find

The support reactions RA and RB, and the largest shear force.

Solution
b = L − a = 6 m − 2 m = 4 m ΣMA = 0: RBL = Pa RB = Pa/L = 10 kN × 2 m / 6 m = 3.33 kN ΣF = 0: RA = P − RB = 10 kN − 3.33 kN = 6.67 kN
Answer
RA = 6.67 kN, RB = 3.33 kN

Shear equals the reaction at A left of the load and minus the reaction at B right of it, so the largest is 6.67 kN.

Related cases

Reactions and largest shear force V for other common beams.

  • Simply supported, central load

    R = P2V = P2

  • Simply supported, uniform load

    R = wL2Vmax = wL2

  • Cantilever, end load

    R = PV = PMwall = PL

Assumptions
  • The supports are a pin and a roller, so with only vertical loads there is no horizontal reaction.
  • The beam is statically determinate: its reactions come from equilibrium alone and do not depend on its stiffness.
  • The beam's own weight is ignored. Add it as a uniform load if it matters.
  • Shear is positive when the part of the beam left of a cut is pushed up.

Derivation

Take moments about support A, with the load at distance a and support B at L.

  1. ΣMA = RBL − Pa = 0, so RB = Pa/L.
  2. ΣF = RA + RB − P = 0, so RA = P(1 − a/L) = Pb/L.
  3. Cut the beam at x < a. The only force left of the cut is RA, so V = RA.
  4. Cut at x > a: V = RA − P = −RB.

The shear jumps by P under the load. With the load at mid-span, a = b = L/2 and each support carries P/2.