Frames & machines: practice problems with step-by-step solutions

Free Frames & machines practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: two-force members and simple machines, pin forces and linkages, multi-member frames.

Free practice problemEasy
15 kN4 m1.2 m0.8 m0.8 mABCDEK

The A-frame shown rests on a pin at A and a roller at B, 4 m apart. A vertical load of 15 kN acts on leg AC at the point K, 0.8 m (horizontally) from A. Find the vertical reaction By at the roller.

P = 15 kNAB = 4 mhorizontal distance A to K = 0.8 m
kN
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What you practise

  1. Easytwo-force members and simple machines
  2. Mediumpin forces and linkages
  3. Hardmulti-member frames

How to solve Frames & machines problems

  1. Draw the whole structure and find the support reactions if there are no more than three unknowns.
  2. Spot the two-force members: loaded only at two pins, their force acts along the line between the pins.
  3. Take the frame apart and draw every member on its own, with equal and opposite pin forces where members meet.
  4. For each member, take moments about a pin with unknown forces, then use ΣFx = 0 and ΣFy = 0; combine components into a magnitude at the end.

Common mistakes

  • Treating a member with a load between its pins as a two-force member.
  • Giving a pin force the same direction on both members it connects.
  • Taking the full inclined force as the vertical component in a moment equation.
  • Reporting only one component when the magnitude of a pin force is asked.

Key formulas

Equilibrium of each member
ΣFx = 0, ΣFy = 0, ΣM = 0
Two-force member
force along the line through its two pins
Pin force
equal and opposite on the two members it joins
Magnitude of a pin force
F = √(Fx² + Fy²)
Lever, pliers, bell crank
Fin·a = Fout·b
Toggle
Flink = F/(2 sin θ), Q = F/(2 tan θ)
Uniform load
resultant w·L at the middle of the load

Learn Frames & machines from scratch

Before this: Beam support reactions, Truss: method of joints

1What frames and machines are

In a truss every member is a straight bar with pins at its two ends and loads only at the joints, so each member is simply in tension or compression. A frame is different: at least one member carries forces at three or more points, or a load between its pins. Such a member also bends.

ABCDE8 kN

In this A-frame the legs AC and BC are multi-force members: each is loaded at the support, at the tie and at the pin C. The tie DE is pinned at both ends and carries nothing in between. A frame that moves and turns an input force into an output force, such as pliers, a lever or a clamp, is called a machine.

Self-check Which member of the A-frame is a two-force member?
  1. Every member, because each has two pins at its ends
  2. Leg AC, because it carries the load
  3. Leg BC, because it rests on a roller
  4. The tie DE

Answer: The tie DE. DE has forces only at its two pins, so its force acts along DE. Each leg also takes the tie force at D or E besides its support and pin C, so it carries three forces, not two.

2Two-force members and pin forces

A member loaded only at two pins is a two-force member. Its two forces are equal, opposite and lie on the line through the pins. So it brings one unknown, the size of that force, instead of two components.

Where two members meet at a pin, the pin pushes on one member with a force (Cx, Cy) and on the other with the opposite force (−Cx, −Cy). Draw them that way, with the same symbols, on both free-body diagrams.

Each separate member gives three equations: ΣFx = 0, ΣFy = 0 and ΣM = 0. Take moments about a pin where unknown forces act, so both of its components drop out.

3The method step by step

  1. Draw the whole frame and find the support reactions, if there are no more than three unknowns.
  2. Mark the two-force members: their forces act along the line between their pins.
  3. Take the frame apart and draw each multi-force member on its own, with equal and opposite pin forces.
  4. For each member, take moments about a pin, then use ΣFx = 0 and ΣFy = 0. Combine components into a magnitude at the end: F = √(Fx² + Fy²).

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

The A-frame stands on a pin at A and a roller at B, 3 m apart. The pin C is 2.4 m high and the tie DE is 0.8 m above the ground. A load of 8 kN acts at C. Find the tension T in the tie and the force of the pin C on leg BC.

1

Look at the frame

ABCDE8 kN3 m0.8 m1.6 m

The tie DE is a two-force member, so it pulls horizontally on both legs.

2

Whole frame

The load is in the middle, so each support carries half of it.

Ay = By = 8/2 = 4 kN, Ax = 0
3

Leg BC: moments about C

4 kNTCx​Cy​CEB

B is 1.5 m to the right of C, and the tie is 2.4 − 0.8 = 1.6 m below C.

4·1.5 = T·1.6 ⇒ T = 3.75 kN
4

Leg BC: force balance

Cx = T = 3.75 kN, Cy = −4 kN
C = √(3.75² + 4²) = 5.48 kN
5

Check the answer

On leg AC the pin force is opposite: (−3.75, +4) kN. With Ay = 4 kN, the tie +3.75 kN and the load −8 kN, both force sums are zero.

Answer

T = 3.75 kN (tension) and C = 5.48 kN on leg BC.

5Machines: pliers, levers and toggles

For a machine, draw one part and take moments about its pivot. The pin force drops out, and the input and output forces balance:

Fin·a = Fout·b

Pliers with the hand 100 mm from the pin and the jaws 25 mm from it multiply the force by 100/25 = 4. The pin then carries both forces: on one half they point the same way on opposite sides of the pin, so they add.

In a toggle, two links meet at a shallow angle θ. Each link is a two-force member with force F/(2 sin θ), and the clamping force is F/(2 tan θ). The flatter the links, the larger the force.

Self-check Pliers are squeezed with 50 N at 100 mm from the pin, and the jaws are 25 mm from the pin. What is the gripping force?
  1. 12.5 N
  2. 200 N
  3. 250 N
  4. 50 N

Answer: 200 N. Moments about the pin: 50·100 = Q·25, so Q = 200 N. Inverting the ratio gives 12.5 N, and 250 N is the pin force, the hand force plus the grip.

6Common mistakes

  • Treating a member with a load between its pins as a two-force member.
  • Drawing a pin force in the same direction on both members it connects.
  • Using the whole force of an inclined strut as its vertical component.
  • Giving one component when the size of a pin force is asked.
Self-check A strut pushes on a beam with 10 kN at 30° to the beam. What is the component perpendicular to the beam?
  1. 10 kN
  2. 20 kN
  3. 5 kN
  4. 8.66 kN

Answer: 5 kN. The perpendicular component is 10·sin 30° = 5 kN. 8.66 kN is the component along the beam (cos 30°), and 10 kN is the whole force.

7How exam questions look

  • Easy: two-force members and simple machines. You find the grip of pliers or the pin force, the strut force of a wall bracket, the tie force or a support reaction of an A-frame, or the output force of a bell crank.
  • Medium: pin forces and linkages. You find the pin reaction of a bracket or a bell crank, the tie force for an off-centre load, the force of a pin on a leg, the force in a member of a three-pinned frame, or the clamping force of a toggle.
  • Hard: multi-member frames. You add a uniform load to a bracket, find the cable force or pin reaction of a boom with its own weight, the pin force for an off-centre load, a support reaction with a horizontal load, or the force a toggle press needs.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
12 kN0.5 m1 m1.2 mABCD

A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 1.2 m below A. A load of 12 kN hangs at the free end B. Neglect the weights. Find the force FCD in the strut.

P = 12 kNAD = 0.5 mAB = 1.5 mAC = 1.2 m

Find FCD (kN)

Show hints
  1. CD is loaded only at its two pins, so it is a two-force member: its force acts along CD.
  2. Take moments about A for the beam: only the strut and the load have a moment.
  3. sin α = 1.2/√(0.5² + 1.2²) = 0.9231.
Show the answer and the step-by-step solution

Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.

Two-force member

The strut has pins at both ends and no load between them.

tan α = 1.2/0.5, sin α = 0.9231

Moments about A for the beam

FCD·sin α·0.5 = 12·1.5
FCD = 18/(0.9231·0.5) = 39 kN

Answer

FCD = 39 kN

The strut pushes the beam up: it is in compression.

Example 2Medium
8 kN0.75 m1.25 m0.75 mABCD

A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 0.75 m below A. A load of 8 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.

P = 8 kNAD = 0.75 mAB = 2 mAC = 0.75 m

Find A (kN)

Show hints
  1. Find the strut force first: moments about A for the beam.
  2. FCD = 30.17 kN, sin α = 0.7071, cos α = 0.7071.
  3. Then ΣFx = 0 and ΣFy = 0 for the beam give the components at A.
Show the answer and the step-by-step solution

Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.

Strut force

FCD = 8·2/(0.7071·0.75) = 30.17 kN

Components at A

Ax = FCD·cos α = 21.33 kN (towards the wall)
Ay = 8 − FCD·sin α = −13.33 kN

Magnitude

A = √(21.33² + −13.33²) = 25.16 kN

Answer

A = 25.16 kN

A negative Ay means the wall pulls the beam down at A.

Example 3Hard
8 kN1 kN2 m1 mABC

A uniform horizontal boom AB of weight 1 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 1 m above A. A load of 8 kN hangs at B. Find the tension T in the cable.

W = 8 kNWb = 1 kNAB = 2 mAC = 1 m

Find T (kN)

Show hints
  1. The cable pulls along BC. Take moments about A for the boom.
  2. The cable's vertical component T·sin β has the arm AB; its horizontal component passes through A.
  3. sin β = 1/√(2² + 1²) = 0.4472.
Show the answer and the step-by-step solution

Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.

Cable angle

BC = √(2² + 1²) = 2.236 m, sin β = 0.4472

Moments about A

T·0.4472·2 = 8·2 + 1·1
T = 19.01 kN

Answer

T = 19.01 kN

Questions and answers

What is Frames & machines?

In a truss every member is a straight bar with pins at its two ends and loads only at the joints, so each member is simply in tension or compression. A frame is different: at least one member carries forces at three or more points, or a load between its pins. Such a member also bends.

In this A-frame the legs AC and BC are multi-force members: each is loaded at the support, at the tie and at the pin C. The tie DE is pinned at both ends and carries nothing in between. A frame that moves and turns an input force into an output force, such as pliers, a lever or a clamp, is called a machine.

How do I solve Frames & machines problems step by step?
  1. Draw the whole structure and find the support reactions if there are no more than three unknowns.
  2. Spot the two-force members: loaded only at two pins, their force acts along the line between the pins.
  3. Take the frame apart and draw every member on its own, with equal and opposite pin forces where members meet.
  4. For each member, take moments about a pin with unknown forces, then use ΣFx = 0 and ΣFy = 0; combine components into a magnitude at the end.
What are common mistakes in Frames & machines?
  • Treating a member with a load between its pins as a two-force member.
  • Giving a pin force the same direction on both members it connects.
  • Taking the full inclined force as the vertical component in a moment equation.
  • Reporting only one component when the magnitude of a pin force is asked.
What do the Frames & machines exercises cover?
  • Easy: two-force members and simple machines
  • Medium: pin forces and linkages
  • Hard: multi-member frames

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Beam support reactionsTruss: method of joints

Courses with this topic

Statics

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Beam support reactionsAxial stress & strainTorsion of shaftsMohr’s circle & stress transformationProjectile motionWork & energyRigid-body rotationMechanical vibrationsHeat engines & refrigeratorsGas power cyclesContinuity & BernoulliPipe friction & head lossPumps & pump powerHeat conductionGears & gear trains

Printable worksheet

Frames & machines worksheet with answers
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