Statics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Vectors: dot & cross product

Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|

Method

  1. Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
  2. For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
  3. Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
  4. Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.

Beam support reactions

Equilibrium
ΣFx = 0, ΣFy = 0, ΣM = 0
Uniform load
resultant w·L at the middle of the load
Moment
M = F·d

Method

  1. Draw a free-body diagram and replace each support by its reactions.
  2. Replace every distributed load by a single force at its centroid.
  3. Take moments about one support to find the reaction at the other.
  4. Use ΣFy = 0 to find the remaining reaction, and check with a second moment equation.

Truss: method of joints

Method of joints
ΣFx = 0, ΣFy = 0 at each joint
Method of sections
ΣM = 0 about a cut point
Member force
positive = tension, negative = compression

Method

  1. Find the support reactions for the whole truss.
  2. Method of joints: start at a joint with at most two unknown member forces and use ΣFx = 0 and ΣFy = 0.
  3. Method of sections: cut through the member you need and take moments about a point where the other unknown members meet.
  4. Assume tension; a negative answer means compression.

Frames & machines

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

Method

  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.

Shear force & bending moment

Shear and moment
V = ΣF on one side, M = ΣF·d on one side
Bending stress
σ = M·y/I
Rectangle
I = bh³/12
Section modulus
W = I/ymax
Simply supported, uniform load
Mmax = wL²/8

Method

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid
ȳ = ΣAiyi/ΣAi
Rectangle
I = bh³/12
Parallel-axis theorem
I = Ic + A·d²
Composite section
I = Σ(Ic,i + Aidi²)

2Practice problems

1.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−2, −5, −1) and b = (3, 1, −2), in degrees.

a = (−2, −5, −1)b = (3, 1, −2)
θ = °
2.Vectors: dot & cross productHard
xyzab

For a = (5, 5, 4) and b = (−5, 2, −1), find the y-component (a × b)y of the cross product a × b.

a = (5, 5, 4)b = (−5, 2, −1)
(a × b)y =
3.Beam support reactionsMedium
w = 5 kN/mP = 50 kNABMA​ = ?4 m2 m4 m

The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 6 m and a point load P at 4 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 10 mP = 50 kN at 4 mw = 5 kN/m from 0 to 6 m
MA = kN·m
4.Beam support reactionsHard
P = 65 kNADCMA​ = ?5 m2 m3 m

A compound beam is fixed to the wall at A, has an internal hinge at C (5 m from A) and rests on a roller at D (10 m from A). It carries a point load P at 7 m from A. Find the magnitude of the reaction moment MA at the wall.

AC = 5 mAD = 10 mP = 65 kN at 7 m
MA = kN·m
5.Truss: method of jointsMedium
ABBCACACB40 kN4 m1 m3 m

Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.

span = 4 mheight = 3 mB is 1 m from Avertical load = 40 kN
FAC = kN
6.Truss: method of jointsHard
ABCDEFGH15 kN4 × 4 m3 m

Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.

H: 15 kN ←
  1. 1member AB
  2. 2member BC
  3. 3member CD
  4. 4member DE
  5. 5member FG
  6. 6member GH
  7. 7member AF
  8. 8member HE
  9. 9member BF
  10. 10member CG
  11. 11member DH
  12. 12member BG
  13. 13member DG
7.Frames & machinesMedium
4 kN3.2 m0.8 m2.8 mABCDE

The A-frame shown stands on a pin at A and a roller at B, 3.2 m apart. The legs meet at the pin C, 3.6 m high, and the horizontal tie DE is 0.8 m above the ground. A load of 4 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.

P = 4 kNAB = 3.2 mheight of C = 3.6 mheight of DE = 0.8 m
C = kN
8.Frames & machinesHard
6 kN4 m0.8 m2.2 mABCDE

The A-frame shown stands on a pin at A and a roller at B, 4 m apart. The legs meet at the pin C, 3 m high, and are tied by the horizontal bar DE at 0.8 m. A horizontal force of 6 kN to the right acts at C. Find the tension T in the tie.

P = 6 kNAB = 4 mheight of C = 3 mheight of DE = 0.8 m
T = kN
9.Shear force & bending momentMedium
10 kN25 kNABCDEFG2 m2 m2 m2 m2 m2 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

10 kN @ 10 m25 kN @ 8 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
10.Shear force & bending momentHard
4 kN/m15 kN20 kN25 kNABCDEFG1 m1 m1 m1 m1 m1 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

15 kN @ 0 m20 kN @ 6 m25 kN @ 2 m4 kN/m, 1–4 m
  1. 1A (x = 0 m)
  2. 2B (x = 1 m)
  3. 3C (x = 2 m)
  4. 4D (x = 3 m)
  5. 5E (x = 4 m)
  6. 6F (x = 5 m)
  7. 7G (x = 6 m)
11.Centroid & second moment of areaMedium
120 mm200 mm60 mmØ40xIx​ = ?

A 120 mm × 200 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.

b = 120 mmh = 200 mmd = 40 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
120230101501410xdimensions in mm

A welded I-section has a top flange 120 × 10 mm, a web 150 mm high and 10 mm thick, and a bottom flange 230 × 14 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.

top flange: 120 × 10 mmweb: 10 × 150 mmbottom flange: 230 × 14 mm
Ix = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: θ = 116 °

Dot product: a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.

Dot product

a·b = (−2)·3 + (−5)·1 + (−1)·(−2) = −9

Magnitudes

|a| = √((−2)² + (−5)² + (−1)²) = 5.477
|b| = √(3² + 1² + (−2)²) = 3.742

Angle

cos θ = −9/(5.477·3.742) = −0.4392
θ = arccos(−0.4392) = 116°

Answer

θ = 116°
2.Vectors: dot & cross product

Answer: (a × b)y = −15

Cross product: a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.

Cross product, component by component

(a × b)x = aybz − azby = 5·(−1) − 4·2 = −13
(a × b)y = azbx − axbz = 4·(−5) − 5·(−1) = −15
(a × b)z = axby − aybx = 5·2 − 5·(−5) = 35

Requested component

(a × b)y = −15

Answer

(a × b)y = −15

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

3.Beam support reactions

Answer: MA = 290 kN·m

Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction RA and a reaction moment MA. ΣFy = 0 gives RA, and moments about the wall give MA = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.

Resultant of the distributed load

W = 5·6 = 30 kN, x̄ = 3 m

Moment about the wall

MA = P·a + W·x̄ = 50·4 + 30·3 = 290 kN·m

Answer

MA = 290 kN·m
4.Beam support reactions

Answer: MA = 195 kN·m

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.

Part CD

The hinge has no moment, so moments about C on CD alone must balance.

ΣMC = 0: RD·5 = 65·2
RD = 26 kN
VC = 65 − 26 = 39 kN

Part AC

The hinge force VC acts downward on AC (equal and opposite to its action on CD).

RA = 39 = 39 kN
MA = 39·5 = 195 kN·m

Answer

MA = 195 kN·m
5.Truss: method of joints

Answer: FAC = 10 kN

Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.

ABBCACACB40 kNAy​Cy​

Support reactions

Equilibrium of the whole truss. Moments about A, counter-clockwise positive.

ΣMA = 0: Cy·4 = 40·1 → Cy = 10 kN
ΣFy = 0: Ay = 40 − 10 = 30 kN

Joint C

The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.

sin γ = 3/4.243 = 0.7071, cos γ = 3/4.243 = 0.7071
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −10/0.7071 = −14.14 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−14.14)·0.7071 = 10 kN

Answer

FAC = 10 kN (tension)
6.Truss: method of joints

Answer: member CG

Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.

Joint C

BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.

ΣF⊥ = 0 → FCG = 0

Check with the method of joints

Solving all joints gives these member forces (T = tension, C = compression).

AB: 11.25 kN (C), BC: 7.5 kN (C), CD: 7.5 kN (C), DE: 3.75 kN (C), FG: 3.75 kN (C), GH: 11.25 kN (C), AF: 4.688 kN (C), HE: 4.687 kN (T), BF: 2.812 kN (T), CG: 0, DH: 2.813 kN (C), BG: 4.687 kN (C), DG: 4.687 kN (T)

Answer

Zero-force member: CG
7.Frames & machines

Answer: C = 2.304 kN

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.

Tie force from leg BC

By = 2 kN
T = 2·1.6/2.8 = 1.143 kN

Components of the pin force on BC

Cx = T = 1.143 kN, Cy = −By = −2 kN

Magnitude

C = √(1.143² + 2²) = 2.304 kN

Answer

C = 2.304 kN
8.Frames & machines

Answer: T = 4.091 kN

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.

Whole frame: moments about A

By·4 = 6·3
By = 4.5 kN

Leg BC: moments about C

4.5·2 = T·2.2
T = 4.091 kN

Answer

T = 4.091 kN

A positive T means the tie is in tension, as assumed.

9.Shear force & bending moment

Answer: E (x = 8 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 10 kN, R2 = 25 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = 20, MC = 40, MD = 60, ME = 80 kN·m, MF = 50, MG = 0

Answer

Mark E: ME = 80 kN·m
10.Shear force & bending moment

Answer: E (x = 4 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 29.33 kN, R2 = 42.67 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = −15, MC = −2.667, MD = −19.33, ME = −40 kN·m, MF = −20, MG = 0

Answer

Mark E: ME = −40 kN·m
11.Centroid & second moment of area

Answer: Ix = 77.75 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.

Centroid

ȳ = (24000·100 − 1257·60)/(24000 − 1257) = 102.2 mm

Full plate about ȳ

120·200³/12 + 24000·−2.21² = 80.12 × 10⁶ mm⁴

Hole about ȳ

π·40⁴/64 + 1257·−42.21² = 2.365 × 10⁶ mm⁴

Subtract

Ix = 77.75 × 10⁶ mm⁴

Answer

Ix = 77.75 × 10⁶ mm⁴
12.Centroid & second moment of area

Answer: Ix = 27.44 × 10⁶ mm⁴

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into three rectangles

Measure every y from the bottom edge.

top flange: A = 1200 mm², y = 169 mm
web: A = 1500 mm², y = 89 mm
bottom flange: A = 3220 mm², y = 7 mm

Centroid

ȳ = (1200·169 + 1500·89 + 3220·7)/5920 = 60.61 mm

Parallel-axis theorem

top flange: 120·10³/12 + 1200·108.4² = 1.411 × 10⁷ mm⁴
web: 10·150³/12 + 1500·28.39² = 4.021 × 10⁶ mm⁴
bottom flange: 230·14³/12 + 3220·53.61² = 9.309 × 10⁶ mm⁴
Ix = 1.411 × 10⁷ + 4.021 × 10⁶ + 9.309 × 10⁶
Ix = 2.744 × 10⁷ mm⁴ = 27.44 × 10⁶ mm⁴

Answer

Ix = 27.44 × 10⁶ mm⁴

ExamProblems · examproblems.com/courses/statics