4.4 Analyze selected truss members with the method of sections | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Truss Analysis

ENGG 130 Engineering Mechanics: Statics — Study topic 4.4

The method of sections finds forces in selected truss members without first solving every member. Imagine cutting through the truss and isolating one side. Forces in members severed by the cut become external forces on the isolated portion. Since that portion is in static equilibrium, its force and moment sums are zero.

This lesson uses the ideal planar truss model: straight members are connected at their ends by pins, loads and support reactions act at joints, and each member carries force along its own axis. The method works most directly when a cut crosses no more than three unknown member forces, because a planar free body provides three independent equilibrium equations.

What you will learn

Explain why the method of sections can find selected truss-member forces without solving every member.
Draw a free-body diagram of one cut truss portion with consistent cut-member force directions.
Use planar equilibrium equations and member geometry to solve selected forces.
Identify tension or compression and verify force and moment balance.

1. Set up a method-of-sections analysis

Start by identifying the member forces you need. Choose a cut through those members, then retain the side with the simpler known loads and reactions. If a reaction on that side is unknown, find it first from equilibrium of the whole truss. A useful cut often crosses no more than three members with unknown forces.

Draw the retained portion as a free-body diagram. Include every load and support reaction on that portion. For each severed member, draw one axial cut force along the member. A convenient convention is to assume every unknown cut force is tension: its arrow points away from the retained portion. Keep this convention in all equations. A negative result means the actual force is opposite the assumed direction, so the member is in compression.

Choose axes, commonly positive

x

to the right and positive

y

upward. Take counterclockwise moments as positive. Use the three planar equilibrium equations. To make the moment equation simpler, take moments about the intersection of two unknown cut-force lines when possible; those two forces then have zero moment about that point.

A member force is a vector along the member. If its horizontal and vertical direction differences are

b

and

h

, its length is

L=\sqrt{b^2+h^2}

. For a member rising to the right, an assumed tension force of magnitude

F

has components

F(b/L)

to the right and

F(h/L)

upward. Reverse signs when the member slopes the other way or its assumed arrow points in the opposite direction.

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

The cut exposes member forces as external forces on the isolated portion.
Assume tension consistently; a negative solution indicates compression.
Choose the retained portion and moment centre to simplify the equations.

2. Solve and verify

Write the moment equation using signed moments. A force contributes its magnitude times its perpendicular distance from the moment centre, with the sign set by its turning direction. Equivalently, resolve the force into components and use their signed lever arms. A force whose line of action passes through the moment centre has zero moment there.

Use the moment equation and force equations together to solve the unknown cut forces. Keep exact values or extra digits during the calculation, then round the reported answers sensibly. Use one consistent unit system: for example, forces in kilonewtons, distances in metres, and moments in kilonewton-metres.

Check the result on the same isolated portion. Substitute the solved forces, with their actual directions, into both force-component sums and a moment sum. All should equal zero within rounding. A force balance by itself is not enough: an incorrect line of action or lever arm can still produce a failed moment check.

L=\sqrt{b^2+h^2},\quad F_x=F\frac{b}{L},\quad F_y=F\frac{h}{L}

Use perpendicular distances or signed component lever arms for moments.
After solving, use the actual directions implied by the signs in the force and moment checks.
A complete section solution must satisfy both force equations and moment equilibrium.

3. Common mistakes

A section free-body diagram represents only the retained portion. A member wholly within that portion is not an external force on it. A severed member contributes one force along its axis, not a force in an arbitrary direction.

Do not decide whether a member is in tension or compression just by looking at the truss. Solve using the stated assumed directions. A negative answer is useful information: it means the actual force points opposite the assumed tension arrow.

Do not use a full distance as a moment arm unless it is perpendicular to the force. A force can be left out of a moment equation only if its line of action passes through the moment centre. Finally, do not stop after finding a plausible number from one equation: check that the complete isolated portion balances.

Draw forces only on the isolated body, including all forces exposed by the cut.
Use the member direction and the assumed tension convention consistently.
Check line of action, moment arm, units, and all equilibrium equations.

Worked example

Finding one diagonal force from a section

A planar truss has joints

A=(0,0)

,

B=(4,0)

, and

C=(4,3)

, with coordinates in metres. Its members are

AB

,

BC

, and

AC

. Joint

A

is pinned and joint

B

is a roller on a horizontal surface. A

10\ \mathrm{kN}

downward load acts at

C

. Find the force in

AC

by cutting members

AB

,

BC

, and

AC

, and retaining the triangular portion containing

B

and

C

.

Find the support reaction
Use the whole truss to determine the vertical reaction at $B$ . Taking moments about $A$ removes both reaction components at the pin. The $10\ \mathrm{kN}$ load is $4\ \mathrm{m}$ horizontally from $A$ .
$\sum M_A=0:\quad 4B_y-10(4)=0$
Set the cut-force directions
The reaction is $10\ \mathrm{kN}$ upward. On the retained portion, assume tension in all three cut members. Member $AC$ runs from $A$ to $C$ with direction ratios $4$ horizontal and $3$ vertical, so its length is $5\ \mathrm{m}$ . At the cut, the tension force on the retained portion points from $C$ toward $A$ , down and left. \hat{u}_{C\to A}=(-\frac45,-\frac35)
Use moments about B
The forces in $AB$ and $BC$ have lines of action through $B$ , so they create no moment about $B$ . The vertical reaction at $B$ also has zero moment there. The downward load at $C$ creates a clockwise moment. The assumed tension force in $AC$ has an upward component at $C$ equal to $3F_{AC}/5$ , creating a counterclockwise moment with a $4\ \mathrm{m}$ arm.
$\sum M_B=0:\quad -10(4)+\frac35F_{AC}(4)=0$
Solve and check
The moment equation gives $F_{AC}=16.7\ \mathrm{kN}$ in tension. Horizontal balance then gives $F_{AB}=13.3\ \mathrm{kN}$ in tension. Vertical balance gives $F_{BC}=0$ . Check the retained portion: horizontal forces are $10-16.7(4/5)+13.3=0\ \mathrm{kN}$ , and vertical forces are $10-10+16.7(3/5)-0=0\ \mathrm{kN}$ . Moments about $B$ are $-40+16.7(3/5)(4)=0\ \mathrm{kN\cdot m}$ , allowing for rounding.
$F_{AC}=\frac{50}{3}\ \mathrm{kN},\quad F_{AB}=\frac{40}{3}\ \mathrm{kN},\quad F_{BC}=0$

Answer:

F_{AC}=16.7\ \mathrm{kN}

in tension.

Check: With exact values, horizontal forces are

\frac{40}{3}-\frac45\left(\frac{50}{3}\right)=0\ \mathrm{kN}

. Vertical forces are

10-10+\frac35\left(\frac{50}{3}\right)=0\ \mathrm{kN}

. Moments about

B

are

-40+4\left[\frac35\left(\frac{50}{3}\right)\right]=0\ \mathrm{kN\cdot m}

.

Worked example

Finding a horizontal member force by moments

A cut isolates a portion of a truss with joint

P

at

(0,2)

and joints

Q=(3,0)

and

R=(-3,0)

, coordinates in metres. A

6\ \mathrm{kN}

downward load acts at

P

. The cut exposes members

PQ

,

PR

, and horizontal

QR

. Assume all three cut forces are tension on the retained portion, which contains joint

P

. Find the forces in the three cut members.

Take moments about P
Both sloping-member forces and the applied load act through $P$ , so they have zero moment about $P$ . The horizontal force in $QR$ acts along a line $2\ \mathrm{m}$ below $P$ and therefore creates a moment. Its assumed tension direction on this portion is to the right, producing a counterclockwise moment.
$\sum M_P=0:\quad 2F_{QR}=0$
Resolve the sloping forces
The direction from $P$ to either lower joint has horizontal magnitude $3\ \mathrm{m}$ , vertical change $2\ \mathrm{m}$ , and length $\sqrt{13}\ \mathrm{m}$ . The assumed tension forces point down and outward. Horizontal balance requires equal magnitudes; vertical balance requires their downward components to be balanced by the applied load, which is also downward. This indicates the assumed tension signs cannot balance the load, so the solved sloping forces must be negative: both members are in compression.
$\sum F_x=0:\quad \frac{3}{\sqrt{13}}(F_{PQ}-F_{PR})=0,\quad \sum F_y=0:\quad -\frac{2}{\sqrt{13}}(F_{PQ}+F_{PR})-6=0$
Solve and verify
The equations give equal forces of $-3\sqrt{13}/2\ \mathrm{kN}$ under the tension-positive convention. Thus both sloping members are in compression with magnitude $3\sqrt{13}/2\ \mathrm{kN}$ , while $F_{QR}=0$ . Their actual upward components each equal $3\ \mathrm{kN}$ , balancing the downward load. Their horizontal components cancel, and the zero force in $QR$ creates no moment.
$F_{PQ}=F_{PR}=-\frac{3\sqrt{13}}{2}\ \mathrm{kN},\quad F_{QR}=0$

Answer:

F_{PQ}=F_{PR}=5.41\ \mathrm{kN}

in compression;

F_{QR}=0\ \mathrm{kN}

.

Check: Each compressed member acts up and inward on the retained portion. Their horizontal components cancel; their vertical components total

2(3\ \mathrm{kN})=6\ \mathrm{kN}

upward, balancing the load. The horizontal-member force is zero, so moment balance about

P

is also satisfied.

Worked example

A section with three cut forces

A planar truss section has an applied

5\ \mathrm{kN}

downward load at point

L

, located

2\ \mathrm{m}

to the right of point

O

. A cut-member force

F

acts at point

K

,

3\ \mathrm{m}

to the right of

O

, along a member directed up and right with direction ratios

3

horizontal and

4

vertical. Two other cut-member forces have lines of action through

O

. Find

F

from moment equilibrium about

O

. Assume tension acts up and right on the retained portion.

Resolve the useful component
Only the vertical component of $F$ creates a moment about $O$ , since the force is applied directly to the right of $O$ . The direction ratios give a vertical component equal to $4F/5$ .
$F_y=\frac45F$
Write moment equilibrium
Take counterclockwise moments as positive. The upward component of $F$ produces a counterclockwise moment with a $3\ \mathrm{m}$ arm. The downward load produces a clockwise moment with a $2\ \mathrm{m}$ arm. The other cut-member forces contribute zero moment because their lines of action pass through $O$ .
$\sum M_O=0:\quad 3\left(\frac45F\right)-5(2)=0$
Solve and state the limit of the result
The moment equation gives $F=25/6\ \mathrm{kN}$ in the assumed tension direction. This determines the selected force from moments, but it is not by itself a complete section solution: the remaining cut forces must still be found or otherwise known so that horizontal and vertical balance can be checked.
$F=\frac{25}{6}\ \mathrm{kN}$

Answer: The moment equation gives

F=4.17\ \mathrm{kN}

in the assumed tension direction.

Check: The moment balance is

3(4/5)(25/6)-5(2)=0\ \mathrm{kN\cdot m}

. A complete analysis also requires horizontal and vertical force balance for the isolated portion.

Common mistakes and how to avoid them

Drawing severed-member forces in arbitrary directions or changing their directions midway through the solution.

Correction: Assume tension consistently, with arrows away from the retained portion along each member. A negative answer indicates compression.

Taking moments without checking which lines of action pass through the chosen moment centre.

Correction: A force has zero moment about a point only when its line of action passes through that point.

Using a member's horizontal or vertical distance directly as its force component fraction.

Correction: Divide each direction difference by the member length to obtain the component fraction.

Stopping after one equilibrium equation and treating its answer as a complete section solution.

Correction: Use the remaining independent force and moment equations to check equilibrium of the isolated portion.

Lesson summary

Choose a useful cut and isolate one truss portion.
Draw its external loads, reactions, and one axial force for each severed member.
Assume tension, choose axes and a moment sign, and apply planar equilibrium.
Use member geometry to resolve forces and select a moment centre that eliminates unknowns.
Interpret negative results as compression and verify both force and moment balance.

Check your understanding

Question 1

A cut-member force assumed to be tension solves to

-7\ \mathrm{kN}

. What does this indicate?

The member is in compression with magnitude $7\ \mathrm{kN}$ .
The member is in tension with magnitude $7\ \mathrm{kN}$ .
The member carries no force.
The section is automatically indeterminate.

Show answer and explanation

The member is in compression with magnitude

7\ \mathrm{kN}

.

A negative result means the actual force is opposite the assumed tension direction, so the member is in compression.

Question 2

Two unknown cut-member force lines pass through point

O

. Why can moments about

O

be useful?

Those forces have zero moment about $O$ , which may leave an equation for another cut force.
Every force through $O$ must be zero.
Moment equilibrium replaces both force equations.
It turns every force into a vertical component.

Show answer and explanation

Those forces have zero moment about

O

, which may leave an equation for another cut force.

A force whose line of action passes through the moment centre has zero moment there, which can simplify the equation for the remaining unknown force.

Key terms

Method of sections: A truss method that cuts selected members and applies equilibrium to one isolated portion to determine their forces.
Cut-member force: The axial force exerted on the retained portion by a member severed by the imagined cut.
Tension: A member force that pulls away from the joint or retained portion along the member.
Compression: A member force that pushes toward the joint or retained portion along the member.
Line of action: The straight line along which a force acts.

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4.4 · Analyze selected truss members with the method of sections

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 Engineering Mechanics: Statics — Study topic 4.4

What you will learn

1. Set up a method-of-sections analysis

2. Solve and verify

3. Common mistakes

Finding one diagonal force from a section

Finding a horizontal member force by moments

A section with three cut forces

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through ENGG 130

About this lesson and its review