6.1 Determine internal normal force, shear force, and bending moment | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Internal Forces and Beam Diagrams

ENGG 130 Engineering Mechanics: Statics — Study topic 6.1

A beam can be in equilibrium as a whole and still have internal forces and moments at every section. To determine them at a location, imagine cutting the member there and isolating one side. The removed part’s action on the isolated part is represented by three resultants: a force along the member, a force perpendicular to it, and a couple moment. Equilibrium of the isolated segment lets us calculate these quantities. The method works with either side of the cut, provided the cut-face directions and signs are consistent.

What you will learn

Explain how a cut through a member reveals internal force and moment resultants.
Use a free-body diagram and planar equilibrium to determine internal normal force, shear force, and bending moment.
Apply a consistent sign convention and interpret negative results correctly.
Verify internal resultants using force and moment equilibrium.

1. Make a cut and define the resultants

Consider a straight, horizontal member in planar statics. At a chosen section, the material on one side exerts forces and a turning effect on the other side. We represent their combined effect by three internal resultants: the normal force

N

, parallel to the member; the shear force

V

, perpendicular to it; and the bending moment

M

, a couple that tends to turn the cut segment.

Begin by identifying the member and the section of interest. Draw the external forces and support reactions, then imagine a cut at the chosen location. Isolate one side and replace the removed part’s action at the cut with unknown

N

,

V

, and

M

. The isolated segment is now a free body: treat it as a separate object for equilibrium calculations.

For a horizontal member, choose

+x

to the right,

+y

upward, and counterclockwise moments as positive. On a left face of a cut, assume positive

N

points right, positive

V

points up, and positive

M

is counterclockwise. These are assumed directions, not guaranteed answers. A negative result means the actual direction is opposite. On opposite faces of the same cut, the internal actions are equal in magnitude and opposite in direction.

\sum F_x=0,\quad \sum F_y=0,\quad \sum M=0

Isolate one side of the cut and show all external loads and reactions acting on it.
For a left cut face, assume $N$ right, $V$ up, and $M$ counterclockwise.
A negative result indicates the direction or sense is opposite the assumed positive direction.

2. Use equilibrium to solve the cut

A segment at rest must have zero net force in each coordinate direction and zero net moment about any point. These three equations can determine the three cut resultants when the segment’s external forces and reactions are known. Taking moments about the cut is often convenient: the unknown forces

N

and

V

act at that point and therefore have zero moment arm, while the unknown couple

M

remains in the moment equation.

For a force acting away from the cut, calculate its moment using its perpendicular distance from the moment centre. Choose a sign from the force’s turning tendency. An applied couple is included directly in the moment equation with its stated sense; it does not need a lever arm. Keep units distinct: forces use newtons or kilonewtons, while moments use newton-metres or kilonewton-metres.

For equilibrium calculations, a distributed load over the isolated segment can be replaced by a single resultant at the load’s centroid. A uniform load of intensity

w

over length

L

has resultant magnitude

wL

, acting at the midpoint of that loaded length. Use only the part of the distributed load that lies on the chosen free body.

M_O=F d_{\perp}

Write separate horizontal-force, vertical-force, and moment equations.
A moment equation about the cut includes the cut couple but not the cut forces.
For a uniform load, use resultant magnitude $wL$ at the midpoint of the loaded length.

3. Read signs, units, and checks carefully

Solve with the directions you assumed at the cut. Do not reverse an arrow partway through because a result looks unexpected. Instead, report a negative value as a magnitude acting opposite the assumed direction. With the convention used here, a negative bending moment is clockwise on the isolated left segment.

Verify the result by substituting it into both force equations and the moment equation for the isolated segment. A force check alone cannot confirm a moment, and a moment check alone cannot confirm force balance. Also check dimensions:

N

and

V

have units of force, while

M

has units of force times length.

The loads included in the free body depend on the cut location. Moving the cut past a point load, or changing the length of a segment under distributed load, changes the free body. Mark the cut and include exactly the loads acting on the isolated side.

[N]=[V]=\mathrm{kN},\quad [M]=\mathrm{kN\cdot m}

Verify horizontal force, vertical force, and moment balance.
Internal normal and shear forces have force units; internal moment has force-times-length units.
Use only the loads acting on the isolated segment.

Worked example

Point load on a simply supported beam

A 6 m beam is supported by a pin at A and a roller at B. A 12 kN downward point load acts 2 m from A. Determine

N

,

V

, and

M

at a section 4 m from A, using the left portion as the free body.

Find the support reactions
Use the complete beam first. The pin and roller provide vertical reactions; there are no horizontal loads, so the horizontal reaction is zero. Taking moments about A gives the reaction at B, and vertical force balance gives the reaction at A.
$B_y(6)-12(2)=0,\quad B_y=4\,\mathrm{kN},\quad A_y=8\,\mathrm{kN}$
Isolate the left segment
At the cut, assume the left-face directions $N$ right, $V$ up, and $M$ counterclockwise. The segment contains the 8 kN reaction and the 12 kN load. Horizontal equilibrium shows the axial resultant is zero.
$\sum F_x=0:\quad N=0$
Solve for shear and moment
Vertical force balance gives the signed shear. Taking moments about the cut gives the bending moment. The negative moment means the actual cut moment is clockwise on this left segment.
$8-12+V=0,\quad V=4\,\mathrm{kN};\quad 8(4)-12(2)+M=0,\quad M=-8\,\mathrm{kN\cdot m}$

Answer:

N=0,\quad V=4\,\mathrm{kN}

upward, and

M=8\,\mathrm{kN\cdot m}

clockwise.

Check: Vertical forces balance:

8-12+4=0\,\mathrm{kN}

. Moments about the cut balance:

32-24-8=0\,\mathrm{kN\cdot m}

. Horizontal forces balance because

N=0

.

Worked example

Uniform load on a beam segment

A 4 m simply supported beam carries a uniform downward load of 3 kN/m over its full length. Determine

N

,

V

, and

M

at a section 1.5 m from the left support, using the left portion as the free body.

Find the support reaction
The full-length distributed load has total force 12 kN at the beam midpoint. Moment and vertical-force balance give equal vertical reactions.
$wL=3(4)=12\,\mathrm{kN},\quad A_y=B_y=6\,\mathrm{kN}$
Replace the load on the isolated segment
Only 1.5 m of the uniform load acts on the left segment. Its resultant is 4.5 kN downward, acting 0.75 m from A. There are no horizontal loads, so the normal force is zero.
$R=3(1.5)=4.5\,\mathrm{kN},\quad N=0$
Apply equilibrium at the cut
Vertical force balance determines the signed shear. Taking moments about the cut uses a 1.5 m arm for the reaction and a 0.75 m arm for the load resultant. Negative values indicate directions opposite the assumed upward shear and counterclockwise moment.
$6-4.5+V=0,\quad V=-1.5\,\mathrm{kN};\quad 6(1.5)-4.5(0.75)+M=0,\quad M=-5.625\,\mathrm{kN\cdot m}$

Answer:

N=0

,

V=1.5\,\mathrm{kN}

downward, and

M=5.625\,\mathrm{kN\cdot m}

clockwise.

Check: Vertical forces balance:

6-4.5-1.5=0\,\mathrm{kN}

. Moments about the cut balance:

9-3.375-5.625=0\,\mathrm{kN\cdot m}

. Horizontal forces balance because

N=0

.

Worked example

Angled force and applied couple on a fixed member

A horizontal member is fixed at A. A 10 kN force acts 1 m from A, directed down and to the right at 30° below the horizontal. A 2 kN·m counterclockwise couple acts at the same point. Determine the internal resultants at a cut 2 m from A, using the left segment.

Resolve the applied force
The force is 30° below the positive horizontal direction. Its horizontal component points right, and its vertical component points down. Resolve it before writing equilibrium equations.
$F_x=10\cos 30^\circ=8.66\,\mathrm{kN},\quad F_y=-10\sin 30^\circ=-5.00\,\mathrm{kN}$
Find the fixed-support reactions
For the complete member, horizontal and vertical force balance determine the reaction components. About A, the downward force component produces a clockwise moment of 5 kN·m. The applied couple is counterclockwise, so the support must provide a 3 kN·m counterclockwise reaction couple.
$A_x=-8.66\,\mathrm{kN},\quad A_y=5.00\,\mathrm{kN},\quad M_A+2-5(1)=0,\quad M_A=3\,\mathrm{kN\cdot m}$
Determine the cut resultants
For the left segment, the horizontal and vertical reaction components cancel the corresponding components of the applied force, so both cut forces are zero. Taking moments about the cut includes the support and applied couples, the vertical reaction at A, and the vertical component of the angled force. The horizontal forces have zero moment arm because they act along the member’s axis.
$N=0,\quad V=0,\quad -5(2)+5(1)+3+2+M=0,\quad M=0\,\mathrm{kN\cdot m}$

Answer:

N=0

,

V=0

, and

M=0

at the cut.

Check: For the complete member, horizontal forces balance:

-8.66+8.66=0\,\mathrm{kN}

. Vertical forces balance:

5-5=0\,\mathrm{kN}

. Moments about A balance:

3+2-5=0\,\mathrm{kN\cdot m}

. For the left segment, moments about the cut balance:

-10+5+3+2+0=0\,\mathrm{kN\cdot m}

.

Common mistakes and how to avoid them

Using the same cut-face arrows on both sides of a cut.

Correction: The internal actions on opposite faces are equal and opposite. State which side is isolated and apply the chosen convention to that face.

Forgetting the moment of a force when taking moments about the cut.

Correction: Include each force’s perpendicular distance from the moment centre, and include an applied couple directly with its sign.

Treating a negative shear or moment as impossible.

Correction: A negative answer means the actual direction or turning sense opposes the assumed positive direction.

Using the full-beam distributed load on a partial segment.

Correction: Use only the load on the isolated segment, with its resultant located at that segment’s load centroid.

Assuming a pin can balance a net couple.

Correction: A pin provides force reactions but no reaction couple. Check whole-body moment equilibrium before using its reactions to analyze a cut.

Lesson summary

Cut the member at the position of interest and isolate one side.
Replace the removed side’s action with the internal resultants $N$ , $V$ , and $M$ .
Choose a sign convention and apply planar force and moment equilibrium.
Interpret negative results as opposite to the assumed directions and verify force and moment balance.

Check your understanding

Question 1

On a left-side free body, assume positive shear is upward. Equilibrium gives

V=-3\,\mathrm{kN}

. What does this mean?

The shear is 3 kN upward.
The shear is 3 kN downward.
The shear is a 3 kN·m moment.
The member is not in equilibrium.

Show answer and explanation

The shear is 3 kN downward.

The negative sign means the actual shear acts opposite to the assumed upward direction: 3 kN downward.

Question 2

A uniform load of 2 kN/m acts over the 1.5 m segment to the left of a cut. What is the magnitude and location of its equivalent resultant on that segment?

3 kN, 0.75 m from the start of the loaded segment.
2 kN, 0.75 m from the start of the loaded segment.
3 kN, 1.5 m from the start of the loaded segment.
0.75 kN, 1.5 m from the start of the loaded segment.

Show answer and explanation

3 kN, 0.75 m from the start of the loaded segment.

The resultant is the intensity times the loaded length,

2(1.5)=3\,\mathrm{kN}

, and a uniform load acts at the midpoint, 0.75 m from the segment’s start.

Question 3

Which equilibrium equation is most directly useful for finding the cut moment when taking moments about the cut?

$\sum F_x=0$
$\sum F_y=0$
$\sum M_{\mathrm{cut}}=0$
$N=V$

Show answer and explanation

\sum M_{\mathrm{cut}}=0

The moment equation about the cut includes the internal couple directly, while the cut forces have zero moment arm about that point.

Key terms

Normal force: The internal force component parallel to the member at a cut.
Shear force: The internal force component perpendicular to the member at a cut.
Bending moment: The internal couple moment acting at a cut.
Resultant: A single force or couple used to represent the combined effect of a load for equilibrium calculations.

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6.1 · Determine internal normal force, shear force, and bending moment

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 Engineering Mechanics: Statics — Study topic 6.1

What you will learn

1. Make a cut and define the resultants

2. Use equilibrium to solve the cut

3. Read signs, units, and checks carefully

Point load on a simply supported beam

Uniform load on a beam segment

Angled force and applied couple on a fixed member

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through ENGG 130

About this lesson and its review