5.2 Create connected free-body diagrams for a frame | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Frames and Machines

Isolate each member, show forces at shared pins, and use equilibrium to connect the diagrams.

A frame is an assembly of connected members that supports applied loads. A free-body diagram (FBD) shows the forces and couples acting on an isolated body after its supports and connections have been replaced by their effects on that body. A connected set of FBDs contains one diagram for each isolated member. At a shared pin, the force on one member has an equal-magnitude, opposite-direction partner on the other member’s diagram. These diagrams let you track how the members interact while applying equilibrium to the whole frame and to individual members. In this lesson, members are treated as rigid for equilibrium calculations, pins transmit forces but no couples, and a roller supplies one reaction normal to its supporting surface.

What you will learn

Identify the members, supports, loads, and joints in a planar frame.
Draw whole-frame and member free-body diagrams with complete support reactions and pin forces.
Use equal-and-opposite forces at shared pins to connect member diagrams.
Apply planar equilibrium equations and verify force and moment balance.

1. Choose the body and draw complete diagrams

Begin by identifying the frame’s members, supports, applied loads, and joints. For the whole-frame FBD, show external loads and support reactions. Forces exchanged between members are internal to the complete frame, so omit them from that diagram. To find a force transmitted at a pin, isolate one member; the force exerted by the pin on that member is now external to the isolated body.

A member FBD must show every force and couple acting on that member. Include support reactions at supports attached to it, applied loads, and forces exerted at its joints. A ground pin can exert two in-plane reaction components, commonly labelled

A_x

and

A_y

. A pin joining two members also transmits two in-plane force components, such as

B_x

and

B_y

.

For a pin joining two members, draw the force components acting on the first member. On the second member’s FBD, draw the corresponding forces in the opposite directions. These are equal-and-opposite forces acting on different bodies; do not draw both members’ pin forces on one isolated-member diagram. If a direction is uncertain, assume one and retain its sign in the calculation. A negative answer means the actual direction is opposite to the assumed direction.

\mathbf{F}_{B\text{ on }1}=-\mathbf{F}_{B\text{ on }2}

Whole-frame FBD: include external loads and support reactions; omit internal member-to-member forces.
Member FBD: include that member’s support reactions, applied loads, and joint forces.
At a shared pin, show equal-and-opposite forces on the two separate member diagrams.

2. Choose signs and apply equilibrium

Use one coordinate convention throughout a connected set of diagrams. In these examples, positive

x

is right, positive

y

is up, and positive moment is counterclockwise. For a planar rigid body in static equilibrium, the horizontal force sum, vertical force sum, and moment sum about any point are each zero.

A force creates a moment about a point when its line of action does not pass through that point. Its moment magnitude is the force times the perpendicular distance to the line of action. Taking moments about a pin removes the moment contribution of forces applied at that pin because their moment arms are zero. A pure couple contributes a moment independent of where it is drawn on the FBD.

Apply equilibrium to the whole frame when that makes the external reactions easy to find. Then isolate members and apply equilibrium again to determine connection forces. Whole-frame equations alone generally do not give the internal pin forces; the separate member diagrams add the needed equations.

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

A pin support can exert horizontal and vertical reaction components; a roller supplies one normal reaction.
Use the same positive directions on every connected diagram.
Choose a moment centre that removes unknown forces from the moment equation when possible.

3. Solve connected diagrams and verify

A useful sequence is to draw the whole-frame FBD, find the external reactions, then draw each member FBD with all of its reactions and connection forces. Use matching labels for the shared pin components, reversing their directions from one member diagram to the other. Apply equilibrium to each isolated member and interpret negative answers using the assumed directions shown on the FBD.

Keep units consistent: for example, use kilonewtons for forces, metres for lengths, and kilonewton-metres for moments. Check the force and moment sums for the whole frame. Also check member equilibrium and confirm that each shared-pin pair is equal and opposite. When the members are considered together, their internal pin forces cancel.

1\ \mathrm{kN}\cdot1\ \mathrm{m}=1\ \mathrm{kN\cdot m}

Solve external reactions first when this simplifies the member calculations.
A complete member FBD includes its support reactions as well as its pin forces and applied loads.
Check both force balance and moment balance; a single satisfied equation is not enough.

Worked example

A right-angle frame with a centred load

A horizontal member

AB

is 4 m long and meets a vertical member

BC

at pin

B

. Pin

A

is attached to the ground. The vertical member ends at a ground roller at

C

, which supplies a vertical reaction. A 5 kN downward load acts at the midpoint of

AB

. Find the support reactions and pin forces. Assume static equilibrium.

Find whole-frame reactions
For the complete frame, forces exchanged at B are internal, so omit them. Take counterclockwise moments about A to find the vertical roller reaction. The horizontal distance from A to C is 4 m.
$4C_y-5(2)=0$
Solve support forces
The moment equation gives the roller reaction. Whole-frame vertical force balance then gives $A_y$ . There is no external horizontal load, so horizontal force balance gives $A_x=0$ .
$C_y=2.5\ \mathrm{kN},\quad A_y=2.5\ \mathrm{kN},\quad A_x=0$
Isolate member AB
The FBD of AB includes both ground-pin reactions at A and the two pin-force components at B. Taking moments about A removes the moments of $A_x$ and $A_y$ . The 5 kN load acts 2 m from A, and $B_y$ acts 4 m from A. Horizontal force balance gives $B_x=0$ .
$4B_y-5(2)=0,\quad B_y=2.5\ \mathrm{kN},\quad B_x=0$
Connect and verify member BC
The force of BC on AB at B is upward, so the force of AB on BC is downward with the same magnitude. Member BC also has the 2.5 kN upward roller reaction at C. Its vertical forces balance, and these collinear forces create no net moment. For the whole frame, the upward reactions total 5 kN and their moments about A balance the applied load.
$\mathbf{B}_{\text{on BC}}=(0,-2.5)\ \mathrm{kN}$

Answer:

A_x=0

,

A_y=2.5\ \mathrm{kN}

, and

C_y=2.5\ \mathrm{kN}

. At B, BC exerts

(0,2.5)\ \mathrm{kN}

on AB, while AB exerts

(0,-2.5)\ \mathrm{kN}

on BC.

Check: Whole-frame vertical force:

2.5+2.5-5=0\ \mathrm{kN}

. Whole-frame moment about A:

2.5(4)-5(2)=0\ \mathrm{kN\cdot m}

. The pin forces at B are equal and opposite.

Worked example

A load closer to the pin

Use the same 4 m by 3 m frame, but place a 6 kN downward load on AB, 1 m from A. Find the reactions and pin forces.

Find whole-frame reactions
Take moments about A for the complete frame. The load is 1 m from A, and the roller reaction has a 4 m horizontal moment arm. Since all external forces are vertical, horizontal force balance gives $A_x=0$ .
$4C_y-6(1)=0,\quad C_y=1.5\ \mathrm{kN},\quad A_y=4.5\ \mathrm{kN},\quad A_x=0$
Find the pin force on AB
On member AB, take moments about A. The load’s moment arm is 1 m and the vertical pin force’s moment arm is 4 m. The resulting upward pin force also satisfies vertical force balance with $A_y$ .
$4B_y-6(1)=0,\quad B_y=1.5\ \mathrm{kN},\quad A_y+B_y-6=0$
Find the horizontal pin component
The member FBD shows $A_x$ and $B_x$ . Horizontal force balance, together with the whole-frame result $A_x=0$ , gives $B_x=0$ . On BC, the pin force at B is opposite to the force on AB; the vertical pin force and roller reaction balance.
$A_x+B_x=0,\quad B_x=0$
Verify
The whole-frame vertical reactions sum to the applied load. Their moments about A balance the load moment, and the horizontal force sum is zero. Member AB also balances vertically and in moment about A.
$4.5+1.5-6=0\ \mathrm{kN},\quad 1.5(4)-6(1)=0\ \mathrm{kN\cdot m}$

Answer:

A_x=0

,

A_y=4.5\ \mathrm{kN}

, and

C_y=1.5\ \mathrm{kN}

. On AB,

B_x=0

and

B_y=1.5\ \mathrm{kN}

upward. On BC, the force at B is 1.5 kN downward.

Check: Whole-frame force and moment sums are zero. Member AB’s vertical force sum is

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

, and its moment about A is zero.

Worked example

A couple applied to the beam

Use the same 4 m by 3 m right-angle frame, with no point load. Apply a 4 kN·m counterclockwise couple to AB. Find the support reactions and pin forces. A pure couple has a moment but no net force.

Find support reactions
For the whole frame, the applied couple contributes positive 4 kN·m. Taking counterclockwise as positive, the roller reaction at C has a 4 m horizontal moment arm. Force balance then gives the ground-pin reactions.
$4C_y+4=0,\quad C_y=-1.0\ \mathrm{kN},\quad A_y=1.0\ \mathrm{kN},\quad A_x=0$
Solve member AB
The negative value of $C_y$ means the actual roller force is downward. For AB, moment balance about A includes the applied couple and the moment of $B_y$ . Horizontal force balance gives $B_x=0$ ; vertical force balance is consistent with the calculated $B_y$ .
$4B_y+4=0,\quad B_y=-1.0\ \mathrm{kN},\quad B_x=0$
Connect and check
The force on BC at B is opposite to the downward force on AB, so it acts upward. It balances the downward roller force at C. For the whole frame, the vertical reactions cancel and the clockwise moment of the roller reaction balances the applied counterclockwise couple.
$4(-1)+4=0\ \mathrm{kN\cdot m}$

Answer:

A_x=0

,

A_y=1.0\ \mathrm{kN}

upward, and

C_y=1.0\ \mathrm{kN}

downward. At B, BC exerts 1.0 kN downward on AB, while AB exerts 1.0 kN upward on BC; the horizontal pin component is zero.

Check: The whole-frame horizontal and vertical force sums are zero. About A, the downward roller reaction gives a clockwise moment of 4 kN·m, balancing the applied counterclockwise couple.

Common mistakes and how to avoid them

Leaving the support reactions off a member FBD because the support is already drawn.

Correction: Show every reaction component acting on the isolated member. At a pin, include both in-plane components; at a roller, include its one normal reaction.

Drawing both members’ pin forces on one member FBD.

Correction: Draw one force on each isolated-member FBD. Use equal magnitudes and opposite directions across the connected diagrams.

Including internal pin forces on the whole-frame FBD.

Correction: For the complete frame, omit forces exchanged between its members; include only external loads and support reactions.

Treating a roller like a pin.

Correction: A roller supplies one reaction normal to its surface, while a pin can supply horizontal and vertical reaction components.

Assuming a negative answer means the equations failed.

Correction: A negative signed force means the actual direction is opposite to the direction assumed in the FBD.

Lesson summary

Isolate the whole frame to find external reactions; isolate individual members to find connection forces.
Include every force acting on each isolated member, including all support-reaction components.
Show each shared-pin force as an equal-and-opposite pair on the connected member diagrams.
Use consistent axes, signs, units, and planar equilibrium equations; verify force and moment balance.

Check your understanding

Question 1

At a pin connecting members 1 and 2, the force of member 1 on member 2 is 3 kN to the right. What force should appear on member 1’s FBD?

3 kN to the right
3 kN to the left
3 kN upward
No force, because the pin is internal

Show answer and explanation

3 kN to the left

The connected-member forces at a pin are equal in magnitude and opposite in direction. If member 1 exerts 3 kN right on member 2, member 2 exerts 3 kN left on member 1.

Question 2

When drawing an FBD of the complete frame, which forces should normally be omitted?

External support reactions
Applied loads
Forces exchanged between members at their internal pins
An applied couple

Show answer and explanation

Forces exchanged between members at their internal pins

Forces exchanged between members are internal to the complete frame and cancel as a pair. They are included when the members are isolated separately.

Question 3

A whole-frame moment equation gives a reaction of

-2\ \mathrm{kN}

when you assumed it acted upward. What does the sign mean?

The reaction is actually 2 kN downward
The reaction is zero
The frame cannot be in equilibrium
The reaction is a 2 kN·m couple

Show answer and explanation

The reaction is actually 2 kN downward

The negative sign indicates the actual force direction is opposite to the assumed upward direction. It remains a force measured in kN.

Key terms

Frame: An assembly of connected members that supports applied loads.
Free-body diagram: A sketch of an isolated body showing the external forces and couples acting on it.
Pin force: The force transmitted between connected members at a pin; it appears as an equal-and-opposite pair on their separate FBDs.
Reaction: A support force exerted on a body to represent the support’s effect.
Couple: A pure turning effect represented by a moment, with no net force.

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5.2 · Create connected free-body diagrams for a frame

University of Alberta ENGG 130: Engineering Mechanics: Statics

Isolate each member, show forces at shared pins, and use equilibrium to connect the diagrams.

What you will learn

1. Choose the body and draw complete diagrams

2. Choose signs and apply equilibrium

3. Solve connected diagrams and verify

A right-angle frame with a centred load

A load closer to the pin

A couple applied to the beam

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