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Beams: reactions, shear and moment

A beam is a long member that carries loads perpendicular to its axis to its supports, working in bending. Analyzing it means answering four questions: what forces the supports exert, which internal forces act at each section, how much it deflects and whether the material can take it. Here you will see the first three with examples that you then solve in the simulator.

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Supports and reactions

Each support prevents some movements and, to do so, exerts forces on the beam: the reactions. They are the unknowns of the problem.

Reactions provided by each type of support.

Support	Prevents	Reactions
Roller	displacement perpendicular to the surface	1: $A_{y}$
Pin	both displacements	2: $A_{x}, A_{y}$
Fixed	displacements and rotation	3: $A_{x}, A_{y}, M_{A}$

In the plane there are three equilibrium equations, so they give at most three reactions:

\sum F_{x} = 0 \sum F_{y} = 0 \sum M_{O} = 0

If the beam has exactly three independent reactions it is statically determinate and equilibrium is enough. If it has more (for example, a fixed support and a roller) it is statically indeterminate: extra deformation equations are needed. The simulator tells you which case you are in, on the reactions card and in the first step of the procedure.

When taking moments, pick a point that several unknowns pass through: they drop out of the equation and only one is left to solve for.

How to find the reactions, step by step

Draw the free-body diagram: the beam by itself, without its supports, with each support replaced by its reactions. If you do not know which way a reaction points, assume one: if it comes out negative, it was the other way.
Replace each distributed load by its resultant (see the table below), for this step only.
Take moments about one support: the equation directly gives the reaction at the other support.
Add up vertical and horizontal forces for the remaining ones. With no inclined loads, $A_{x} = 0$ .
Check with the equation you did not use, for example moments about the other support: it must give zero.

The simulator writes these same equations, with your numbers, in the «Step-by-step solution». Inclined loads are split into components before starting; the horizontal one only affects $A_{x}$ (and the axial force), not the $V$ and $M$ diagrams.

Types of load

To find the reactions, a distributed load is replaced by its resultant: a force equal to the area under the load diagram, applied at its centroid. Careful: this replacement only works for overall equilibrium, not for drawing the diagrams.

Point, uniform and triangular loads with their resultant.

Load over a length $a$	Resultant	Where it acts
uniform $w$	$w a$	at the middle, $a /2$
triangular, from 0 to $w$	$\frac{w a}{2}$	at $a /3$ from the large end ( $2 a /3$ from the tip)
trapezoidal, from $w_{1}$ to $w_{2}$	$\frac{( w _{1} + w _{2} ) a}{2}$	split it into a rectangle and a triangle, each with its own resultant
point moment $M_{0}$	not a force	it only enters the sum of moments, the same about any point

Units help you avoid mistakes: $w$ is in kN/m, so $w a$ is in kN; a moment is in kN·m. If a reaction calculation gives kN·m, a distance is missing or extra.

Shear force and bending moment

If the beam is cut at a section at distance $x$ , each part must stay in equilibrium. For that, two internal forces appear at the cut: the shear force $V$ (perpendicular to the axis) and the bending moment $M$ . With the usual convention, $M$ is positive when it compresses the top fiber (the beam «smiles»).

Positive senses of V and M on each side of the cut.

Method of sections

Split the beam into segments: a new segment starts at each support, point load, point moment and at each start or end of a distributed load.
In each segment cut at a distance $x$ and draw the left piece with $V$ and $M$ in their positive senses.
Equilibrium of that piece: $\sum F_{y} = 0$ gives $V (x)$ and $\sum M_{cut} = 0$ gives $M (x)$ . Distributed loads on the piece are replaced by their resultant, but only the part inside the piece.
If the right piece is simpler (an overhang, for example), use it: the result is the same, with the positive senses of the other side of the cut.

Not all textbooks use the same sign convention. Here and in the simulator,

M

is positive when it puts the bottom fiber in tension and is drawn on the tension side; if your course uses another one, the plot may come out mirrored, but the values are the same.

From the equilibrium of a small piece come two relations that let you draw the diagrams with hardly any calculation ( $w$ is the distributed load, positive downward):

\frac{d V}{d x} = - w (x) \frac{d M}{d x} = V (x)

Integrated between two sections $a$ and $b$ they say the same with areas, which is how they are used to draw by hand:

V_{b} - V_{a} = - \int_{a}^{b} w d x M_{b} - M_{a} = \int_{a}^{b} V d x

That is, the moment changes between two points by the area of the shear diagram between them. Walking along the beam from left to right and adding areas you reach the free end or the last support with $M = 0$ (unless there is a fixed end or an applied moment): if it does not give zero, the reactions are wrong.

The slope of the shear diagram is the load: where there is no load, $V$ is constant; with a uniform load, $V$ is a straight line.
The slope of the moment diagram is the shear: the moment is maximum or minimum where $V = 0$ (or where $V$ changes sign under a point load).
A point load produces a jump in $V$ equal to the load; a point moment, a jump in $M$ .

Segment with…	V diagram	M diagram
no load	constant	straight line
uniform load	straight line	parabola (2nd degree)
triangular load	parabola	cubic (3rd degree)
point load	jump equal to P	change of slope (kink)

Bending and shear stresses

The bending moment stretches the fibers on one side of the neutral axis and shortens those on the other. The normal stress grows linearly with the distance $y$ from the neutral axis and is largest at the farthest fiber, at a distance $c$ :

σ = \frac{M y}{I} σ_{m \overset{a}{ˊ} x} = \frac{∣ M ∣ _{m \overset{a}{ˊ} x} c}{I} = \frac{∣ M ∣ _{m \overset{a}{ˊ} x}}{S}

$I$ is the second moment of area of the section about the neutral axis and $S = I / c$ the section modulus. That is why design looks at the moment diagram: the critical section is the one with the largest $∣ M ∣$ . Shear produces a stress $τ = \frac{V Q}{I t}$ , largest at the neutral axis; in a rectangular section it is $τ_{max} = \frac{3 V}{2 A}$ , 50 % more than the average.

In a long beam bending almost always governs; shear matters in short beams with large loads near the supports and in thin webs. In the simulator, the Design tab computes these stresses for the section and material you choose.

Beam stresses and failure criteriaNormal and shear stress in the section, principal stresses, von Mises and Tresca, criteria for brittle materials and sizing the section.

Deflection

The beam bends according to its flexural stiffness $E I$ (modulus of elasticity times the second moment of area of the section). The elastic curve $y (x)$ satisfies, for small deflections,

E I \frac{d ^{2} y}{d x ^{2}} = M (x)

Integrating twice and using the support conditions gives the deflection. For the classic cases it is worth remembering these results:

Case	Maximum moment	Maximum deflection
Simply supported, uniform load $w$	$\frac{w L ^{2}}{8}$	$\frac{5 w L ^{4}}{384 E I}$
Simply supported, $P$ at midspan	$\frac{P L}{4}$	$\frac{P L ^{3}}{48 E I}$
Cantilever, $P$ at the free end	$P L$	$\frac{P L ^{3}}{3 E I}$
Cantilever, uniform load $w$	$\frac{w L ^{2}}{2}$	$\frac{w L ^{4}}{8 E I}$

The two integration constants come from the boundary conditions, which depend on each support:

Point	Conditions
Pin or roller	$y = 0$ (it can rotate)
Fixed end	$y = 0$ and $y^{'} = 0$ (it neither drops nor rotates)
Free end	$M = 0$ and $V = 0$ (with no loads applied there)

Since the equation is linear, superposition holds: the deflection under several loads is the sum of the deflections of each load alone. That is how the cases in the table are combined, and how example 4 is solved. The maximum deflection is usually compared with a serviceability limit such as $L /360$ or $L /240$ , depending on what the beam is for.

Watch the units of

E I

: with

E

in GPa and

I

in cm⁴ or mm⁴ it is easy to be off by powers of ten. The simulator converts them for you and gives the deflection in mm (or in).

Common mistakes

Using the resultant of a distributed load to draw $V$ and $M$ : it is valid for the reactions, not inside the diagrams.
Forgetting the fixed-end moment $M_{A}$ when taking moments, or giving it the wrong sign.
Looking for the maximum moment only under the loads: with a distributed load it is where $V = 0$ , which can be between two loads.
Keeping only the positive maximum: in overhanging beams the negative moment over the support can be the largest in magnitude.
Solving an indeterminate beam with equilibrium alone: with more than three reactions the system has no unique solution without the deformation.

Worked examples

Example 1Simply supported with a point load

A 6 m beam is pinned at A (

x = 0

) and rests on a roller at B (

x = 6

m). It carries a 12 kN load 2 m from A. Find the reactions and the maximum bending moment.

Solution

Moments about A (counterclockwise positive): $6 B_{y} - 12 \cdot 2 = 0 \Rightarrow B_{y} = 4$ kN.
Vertical forces: $A_{y} + B_{y} - 12 = 0 \Rightarrow A_{y} = 8$ kN.
Shear: $V = 8$ kN between A and the load, and $V = 8 - 12 = - 4$ kN after it. It changes sign under the load, so that is where the maximum moment is.
$M_{max} = A_{y} \cdot 2 = 16$ kN·m (the area of the shear diagram between A and the load).

A_{y} = 8

kN,

B_{y} = 4

kN and

M_{max} = 16

kN·m at

x = 2

Solve it in Simuveliq

In Engineering → Beams, type the length 6 m.
In Supports, keep a Pin at 0 and set the other one as a Roller at 6.
In Loads, press + Point, set 12 kN at $x = 2$ and delete the other loads.
Read the reactions in the Reactions card and the moment in Max moment; the diagrams and the «Step-by-step solution» show the same reasoning.

Open in the simulator →

Diagrams of example 1: the moment peaks where the shear changes sign.

Example 2Cantilever with a uniform load

A 3 m cantilever is fixed at A and carries a uniform load of 4 kN/m along its whole length. Find the reactions at the fixed support and the moment diagram.

Solution

Resultant of the load: $R = w L = 4 \cdot 3 = 12$ kN, applied at the middle (1.5 m).
Vertical forces: $A_{y} = 12$ kN.
Moments at A: $M_{A} = 12 \cdot 1.5 = 18$ kN·m (the fixed support must balance the rotation).
Measuring from the free end, $M (s) = - \frac{w s ^{2}}{2}$ : a parabola, zero at the tip and $- 18$ kN·m at the support (negative: the top fiber is in tension).

A_{y} = 12

kN and

M_{A} = 18

kN·m; the largest moment in magnitude is at the support:

\frac{w L ^{2}}{2}

Solve it in Simuveliq

Length 3 m; in Supports keep a single Fixed support at 0.
In Loads, press + Distributed, shape Uniform, from 0 to 3 with $w = 4$ kN/m.
The Reactions card shows $A_{y}$ and $M_{A}$ ; in Diagrams you will see the straight shear line and the moment parabola.

Open in the simulator →

Example 3Overhanging beam: positive and negative moment

An 8 m beam is pinned at A (

x = 0

) and supported at B (

x = 6

m), with a 2 m overhang. It carries 3 kN/m along its whole length. Find the reactions and the maximum positive and negative moments.

Solution

Resultant: $3 \cdot 8 = 24$ kN at $x = 4$ m. Moments about A: $6 B_{y} - 24 \cdot 4 = 0 \Rightarrow B_{y} = 16$ kN; then $A_{y} = 24 - 16 = 8$ kN.
Between A and B: $V (x) = 8 - 3 x$ , which is zero at $x = 8/3 \approx 2.667$ m.
There the moment is maximum: $M = 8 x - 1.5 x^{2} = 10.67$ kN·m.
Over support B the overhang hangs: $M_{B} = - \frac{3 \cdot 2 ^{2}}{2} = - 6$ kN·m.

A_{y} = 8

kN,

B_{y} = 16

kN;

M_{max}^{+} = 10.67

kN·m at

x = 2.667

m and

M_{max}^{-} = - 6

kN·m over B.

Solve it in Simuveliq

Length 8 m; Pin at 0 and Roller at 6.
A uniform Distributed load from 0 to 8 with 3 kN/m.
In Diagrams, move the cursor over the plot: at $x = 2.667$ m the shear is 0 and the moment 10.67 kN·m. The «Critical values» table of the solution lists both extremes.

Open in the simulator →

Diagrams of example 3: the positive maximum is where V = 0 and the negative one over the overhang support.

Example 4Propped cantilever (statically indeterminate)

A 6 m beam is fixed at A and rests on a roller at B, with a uniform load of 4 kN/m. Find the reactions.

Solution

There are four reactions ( $A_{x}, A_{y}, M_{A}, B_{y}$ ) and three equations: the beam is indeterminate to the first degree and equilibrium is not enough.
Extra condition (compatibility): the deflection at B is zero. Removing the roller, the load lowers the tip by $\frac{w L ^{4}}{8 E I}$ and the reaction $B_{y}$ raises it by $\frac{B _{y} L ^{3}}{3 E I}$ ; equating them, $B_{y} = \frac{3 w L}{8} = 9$ kN.
Equilibrium: $A_{y} = 24 - 9 = 15$ kN and $M_{A} = 24 \cdot 3 - 9 \cdot 6 = 18$ kN·m $= \frac{w L ^{2}}{8}$ .

A_{y} = 15

kN,

B_{y} = 9

kN and

M_{A} = 18

kN·m. They do not depend on

E I

because the beam has a single material and section.

Solve it in Simuveliq

Length 6 m; Fixed support at 0 and Roller at 6.
A uniform Distributed load from 0 to 6 with 4 kN/m.
The Reactions card says «Indeterminate · degree 1» and gives the values; the step-by-step solution shows the compatibility equation the simulator used.

Open in the simulator →

Try changing the data of any example in the simulator (move a support, add a load) and watch how the diagrams change: it is the quickest way to understand the rules above.

Beams: reactions, shear and moment

Supports and reactions

How to find the reactions, step by step

Types of load

Shear force and bending moment

Method of sections

Bending and shear stresses

Deflection

Common mistakes

Worked examples

Example 1Simply supported with a point load

Solution

Solve it in Simuveliq

Example 2Cantilever with a uniform load

Solution

Solve it in Simuveliq

Example 3Overhanging beam: positive and negative moment

Solution

Solve it in Simuveliq

Example 4Propped cantilever (statically indeterminate)

Solution

Solve it in Simuveliq

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