Statics: cram pack

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

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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 scalar projection compb a of a = (2, −3, −1) onto b = (2, −3, 0), that is, the component of a in the direction of b.

a = (2, −3, −1)b = (2, −3, 0)
compb a =
2.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−3, −1, 4), B(4, −1, 1) and C(−4, −1, −2). Coordinates are in metres.

A(−3, −1, 4)B(4, −1, 1)C(−4, −1, −2)
A = m²
3.Beam support reactionsMedium
P1​ = 25 kNP2​ = 40 kNABRA​RB​ = ?1.5 m2.5 m1 m

The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.

L = 5 mP1 = 25 kN at 1.5 mP2 = 40 kN at 4 m
RB = kN
4.Beam support reactionsHard
w0​ = 6 kN/mP = 60 kNABMA​ = ?5 m1 m

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 6 kN/m at the free end B, plus a point load P at 5 m from the wall. Find the magnitude of the reaction moment MA at the wall.

L = 6 mw0 = 6 kN/mP = 60 kN at 5 m
MA = kN·m
5.Truss: method of jointsMedium
ABCDEFGH35 kN15 kN25 kN4 × 3 m4 m

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

B: 35 kN ↓D: 15 kN ↓F: 25 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
6.Truss: method of jointsHard
ABCDEFGH35 kN35 kN4 × 4 m4 m

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

F: 35 kN →G: 35 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
20 kN3 m2.5 m0.45 mABCK

The frame shown has pins at A, B and C; A and C are fixed to the ground 3 m apart, and B is 2.5 m above the ground and 0.45 m to the right of A. A vertical load of 20 kN acts on member AB at K, 0.225 m (horizontally) from A. Find the force FBC in member BC.

P = 20 kNL = 3 mH = 2.5 mhorizontal distance A to B = 0.45 mhorizontal distance A to K = 0.225 m
FBC = kN
8.Frames & machinesHard
w = 1.5 kN/m5 kN1.2 m1.3 m1.5 mABCD

The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 1.5 m below A). The beam carries a uniform load of 1.5 kN/m over its full length and a point load of 5 kN at its free end B. Find the force FCD in the strut.

w = 1.5 kN/mP = 5 kNAD = 1.2 mAB = 2.5 mAC = 1.5 m
FCD = kN
9.Shear force & bending momentMedium
P1​ = 25 kNP2​ = 20 kNAB3 m6 m1 m

The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.

L = 10 mP1 = 25 kN at 3 mP2 = 20 kN at 9 m
Mmax = kN·m
10.Shear force & bending momentHard
w = 2 kN/mP = 35 kNABsection3 m1.5 m0.5 m

Find the bending moment Mx at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 35 kN at 3 mw = 2 kN/m from 0 to 5 mx = 4.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
14090t = 10ȳ = ?dimensions in mm

A channel section lies with its open side up: a base plate 140 mm wide with two upright legs, 90 mm high overall. Base and legs are all 10 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 140 mmh = 90 mmt = 10 mm
ȳ = mm
12.Centroid & second moment of areaHard
120 mm150 mm97.5 mm30 × 20xIx​ = ?

A 120 mm × 150 mm plate has a rectangular cut-out 30 mm wide and 20 mm high, centred on the vertical centre line with its centre 97.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.

b = 120 mmh = 150 mmcut-out: 30 × 20 mmhole centre: y = 97.5 mm
Ix = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: compb a = 3.606

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 = 2·2 + (−3)·(−3) + (−1)·0 = 13

Length of b

|b| = √(2² + (−3)² + 0²) = 3.606

Scalar projection

compb a = a·b/|b| = 13/3.606 = 3.606

Answer

compb a = 3.606
2.Vectors: dot & cross product

Answer: A = 22.5 m²

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.

Edge vectors

AB = B − A = (7, 0, −3)
AC = C − A = (−1, 0, −6)

Cross product

(AB × AC)x = AByACz − ABzACy = 0·(−6) − (−3)·0 = 0
(AB × AC)y = ABzACx − ABxACz = (−3)·(−1) − 7·(−6) = 45
(AB × AC)z = ABxACy − AByACx = 7·0 − 0·(−1) = 0
|AB × AC| = √(0² + 45² + 0²) = 45

Area of the triangle

A = ½·45 = 22.5 m²

Answer

A = 22.5 m²
3.Beam support reactions

Answer: RB = 39.5 kN

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.

P1​ = 25 kNP2​ = 40 kNABRA​RB​1.5 m2.5 m1 m

Free-body diagram

Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·5 = 25·1.5 + 40·4
RB = 197.5/5 = 39.5 kN
ΣFy = 0: RA = 65 − 39.5 = 25.5 kN

Answer

RB = 39.5 kN
4.Beam support reactions

Answer: MA = 372 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 triangular load

W = ½·6·6 = 18 kN
x̄ = 2L/3 = 4 m

Moment about the wall

MA = W·x̄ + P·a = 18·4 + 60·5 = 372 kN·m

Answer

MA = 372 kN·m
5.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: 36.56 kN (T), BC: 28.13 kN (T), CD: 28.13 kN (T), DE: 19.69 kN (T), FG: 36.56 kN (C), GH: 19.69 kN (C), AF: 60.94 kN (C), HE: 32.81 kN (C), BF: 23.75 kN (T), CG: 0, DH: 26.25 kN (T), BG: 14.06 kN (T), DG: 14.06 kN (C)

Answer

Zero-force member: CG
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: 43.75 kN (T), BC: 52.5 kN (T), CD: 52.5 kN (T), DE: 26.25 kN (T), FG: 43.75 kN (C), GH: 26.25 kN (C), AF: 12.37 kN (C), HE: 37.12 kN (C), BF: 8.75 kN (T), CG: 0, DH: 26.25 kN (T), BG: 12.37 kN (C), DG: 37.12 kN (C)

Answer

Zero-force member: CG
7.Frames & machines

Answer: FBC = 2.143 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.

Two-force member BC

BC = √(2.55² + 2.5²) = 3.571 m

Whole frame: moments about A

The ground pushes on BC at C along CB; its horizontal component passes through A, only the vertical one has a moment.

FBC·(2.5/3.571)·3 = 20·0.225
FBC = 2.143 kN (compression)

Answer

FBC = 2.143 kN
8.Frames & machines

Answer: FCD = 18.34 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.

Resultant of the uniform load

w·L = 1.5·2.5 = 3.75 kN at 1.25 m

Moments about A for the beam

FCD·0.7809·1.2 = 3.75·1.25 + 5·2.5 = 17.19 kN·m
FCD = 18.34 kN

Answer

FCD = 18.34 kN
9.Shear force & bending moment

Answer: Mmax = 58.5 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

ΣMA = 0: RB·10 = 25·3 + 20·9
RB = 255/10 = 25.5 kN
ΣFy = 0: RA = 45 − 25.5 = 19.5 kN

Moment under each load

Keep the part left of each load.

under P1: M = RA·3 = 19.5·3 = 58.5 kN·m
under P2: M = RB·1 = 25.5·1 = 25.5 kN·m

Largest value

Mmax = 58.5 kN·m (under P1)

Answer

Mmax = 58.5 kN·m
10.Shear force & bending moment

Answer: Mx = 12.75 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 2·5 = 10 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 35·3 + 2·5·2.5
RB = 130/5 = 26 kN
ΣFy = 0: RA = 45 − 26 = 19 kN
w = 2 kN/mP = 35 kNABRA​ = 19RB​ = 26section3 m1.5 m0.5 m

Cut at x = 4.5 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 19·4.5 − 35·(4.5 − 3) − 2·4.5·2.25 = 12.75 kN·m

Answer

Mx = 12.75 kN·m
11.Centroid & second moment of area

Answer: ȳ = 29 mm

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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A1 = 140·10 = 1400 mm², y1 = 5 mm
two legs: A2 = 2·10·80 = 1600 mm², y2 = 10 + 80/2 = 50 mm

Centroid

ȳ = (1400·5 + 1600·50)/3000 = 29 mm

Answer

ȳ = 29 mm
12.Centroid & second moment of area

Answer: Ix = 33.42 × 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 = (18000·75 − 600·97.5)/(18000 − 600) = 74.22 mm

Full plate about ȳ

120·150³/12 + 18000·0.7759² = 33.76 × 10⁶ mm⁴

Hole about ȳ

30·20³/12 + 600·23.28² = 0.3451 × 10⁶ mm⁴

Subtract

Ix = 33.42 × 10⁶ mm⁴

Answer

Ix = 33.42 × 10⁶ mm⁴

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