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 = (0, −5, 2) onto b = (5, −1, −4), that is, the component of a in the direction of b.

a = (0, −5, 2)b = (5, −1, −4)
compb a =
2.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (1, 4, 1) and b = (3, 3, 2) and has a positive z-component. Find its y-component n̂y.

a = (1, 4, 1)b = (3, 3, 2)n̂z > 0
n̂y =
3.Beam support reactionsMedium
P1​ = 45 kNP2​ = 10 kNABRA​ = ?RB​1 m1 m2 m

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

L = 4 mP1 = 45 kN at 1 mP2 = 10 kN at 2 m
RA = kN
4.Beam support reactionsHard
w = 12 kN/mP = 45 kNABRA​RB​ = ?1 m1 m3 m

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

L = 5 mP = 45 kN at 1 mw = 12 kN/m from 0 to 2 m
RB = kN
5.Truss: method of jointsMedium
ABCDEFGH40 kN10 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: 40 kN ↓D: 10 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 FC
  13. 13member HC
6.Truss: method of jointsHard
ABCDEFGH40 kN40 kN30 kN25 kN40 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.

B: 40 kN ↓D: 40 kN ↓F: 30 kN ↓F: 25 kN →H: 40 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 FC
  13. 13member HC
7.Frames & machinesMedium
OAB250 NQ150 mm120 mmFO​ = ?

The bell crank shown turns about the pin O. A vertical force of 250 N at A (150 mm from O) is balanced by a horizontal force at B (120 mm above O). Find the magnitude of the pin reaction FO.

F = 250 NOA = 150 mmOB = 120 mm
FO = N
8.Frames & machinesHard
6 kN2 m1.8 m2.7 m0.25 mABCDEK

The A-frame shown stands on a pin at A and a roller at B, 2 m apart, with its apex pin C 4.5 m high and the horizontal tie DE 1.8 m above the ground. A vertical load of 6 kN acts on leg AC at K, 0.25 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.

P = 6 kNAB = 2 mheight of C = 4.5 mheight of DE = 1.8 mhorizontal distance A to K = 0.25 m
C = kN
9.Shear force & bending momentMedium
50 kN30 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.

50 kN @ 8 m30 kN @ 10 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
w = 20 kN/mAB7 m3 m

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.

L = 10 mw = 20 kN/m from 0 to 7 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
1202020020xdimensions in mm

Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.

bf = 120 mmtf = 20 mmtw = 20 mmhw = 200 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
150280183001410xdimensions in mm

A welded I-section has a top flange 150 × 18 mm, a web 300 mm high and 10 mm thick, and a bottom flange 280 × 14 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.

top flange: 150 × 18 mmweb: 10 × 300 mmbottom flange: 280 × 14 mm
Ix = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: compb a = −0.4629

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 = 0·5 + (−5)·(−1) + 2·(−4) = −3

Length of b

|b| = √(5² + (−1)² + (−4)²) = 6.481

Scalar projection

compb a = a·b/|b| = −3/6.481 = −0.4629

Answer

compb a = −0.4629

Negative: a has a component pointing against b.

2.Vectors: dot & cross product

Answer: n̂y = −0.09667

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

(a × b)x = aybz − azby = 4·2 − 1·3 = 5
(a × b)y = azbx − axbz = 1·3 − 1·2 = 1
(a × b)z = axby − aybx = 1·3 − 4·3 = −9

Normalise

The z-component of a × b is negative, so use −(a × b).

|a × b| = √(5² + 1² + (−9)²) = 10.34
n̂ = −(5, 1, −9)/10.34 = (−0.4834, −0.09667, 0.8701)
n̂y = −0.09667

Answer

n̂y = −0.09667
3.Beam support reactions

Answer: RA = 38.75 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​ = 45 kNP2​ = 10 kNABRA​RB​1 m1 m2 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·4 = 45·1 + 10·2
RB = 65/4 = 16.25 kN
ΣFy = 0: RA = 55 − 16.25 = 38.75 kN

Answer

RA = 38.75 kN
4.Beam support reactions

Answer: RB = 13.8 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.

w = 12 kN/mP = 45 kNABRA​RB​1 m1 m3 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.

UDL resultant: W = w·ℓ = 12·2 = 24 kN, acting at x̄ = 1 m
ΣMA = 0: RB·5 = 45·1 + 12·2·1
RB = 69/5 = 13.8 kN
ΣFy = 0: RA = 69 − 13.8 = 55.2 kN

Answer

RB = 13.8 kN
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 G

FG and GH 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: 24.38 kN (T), BC: 24.38 kN (T), CD: 13.12 kN (T), DE: 13.12 kN (T), FG: 18.75 kN (C), GH: 18.75 kN (C), AF: 40.63 kN (C), HE: 21.87 kN (C), BF: 40 kN (T), CG: 0, DH: 10 kN (T), FC: 9.375 kN (C), HC: 9.375 kN (T)

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 G

FG and GH 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: 115.4 kN (T), BC: 115.4 kN (T), CD: 109.6 kN (T), DE: 109.6 kN (T), FG: 112.5 kN (C), GH: 112.5 kN (C), AF: 113 kN (C), HE: 137 kN (C), BF: 40 kN (T), CG: 0, DH: 40 kN (T), FC: 3.646 kN (C), HC: 3.646 kN (T)

Answer

Zero-force member: CG
7.Frames & machines

Answer: FO = 400.2 N

Machines: pliers, levers, linkages and clamps turn an input force into a larger or a differently directed output force. Draw one part as a free body and take moments about its pivot: the pin force drops out, and Fin·a = Fout·b gives the mechanical advantage a/b. The pin force then follows from ΣF = 0 on the same part. In a toggle, two links meet at a shallow angle θ; each link is a two-force member, so a small force F at the joint creates large link forces F/(2 sin θ) and a clamping force F/(2 tan θ).

Force at B

Q = 250·150/120 = 312.5 N

Pin reaction

Ox = 312.5 N, Oy = 250 N
FO = √(312.5² + 250²) = 400.2 N

Answer

FO = 400.2 N
8.Frames & machines

Answer: C = 0.7998 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.

Support reaction at B

By = 6·0.25/2 = 0.75 kN

Tie force

T = 0.75·1/2.7 = 0.2778 kN

Pin force on BC

Cx = 0.2778 kN, Cy = −0.75 kN
C = √(0.2778² + 0.75²) = 0.7998 kN

Answer

C = 0.7998 kN
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 = 21.67 kN, R2 = 58.33 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 = 43.33, MC = 86.67, MD = 130, ME = 173.3 kN·m, MF = 116.7, MG = 0

Answer

Mark E: ME = 173.3 kN·m
10.Shear force & bending moment

Answer: Mmax = 207 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.

UDL resultant: W = w·ℓ = 20·7 = 140 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·10 = 20·7·3.5
RB = 490/10 = 49 kN
ΣFy = 0: RA = 140 − 49 = 91 kN

Zero shear

The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.

RA − w·(x0 − 0) = 0
x0 = 0 + 91/20 = 4.55 m

Maximum moment

Mmax = RA·x0 − w·(x0 − 0)²/2 = 91·4.55 − 20·4.55²/2 = 207 kN·m

Answer

Mmax = 207 kN·m
11.Centroid & second moment of area

Answer: Ix = 31.56 × 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 two rectangles

Measure every y from the bottom edge.

flange: Af = 120·20 = 2400 mm², yf = 200 + 20/2 = 210 mm
web: Aw = 20·200 = 4000 mm², yw = 200/2 = 100 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (2400·210 + 4000·100)/6400 = 141.3 mm

Parallel-axis theorem

flange: 120·20³/12 + 2400·(210 − 141.3)² = 1.142 × 10⁷ mm⁴
web: 20·200³/12 + 4000·(100 − 141.3)² = 2.014 × 10⁷ mm⁴
Ix = 1.142 × 10⁷ + 2.014 × 10⁷ = 3.156 × 10⁷ mm⁴ = 31.56 × 10⁶ mm⁴

Answer

Ix = 31.56 × 10⁶ mm⁴
12.Centroid & second moment of area

Answer: Ix = 183.9 × 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 = 2700 mm², y = 323 mm
web: A = 3000 mm², y = 164 mm
bottom flange: A = 3920 mm², y = 7 mm

Centroid

ȳ = (2700·323 + 3000·164 + 3920·7)/9620 = 144.7 mm

Parallel-axis theorem

top flange: 150·18³/12 + 2700·178.3² = 8.596 × 10⁷ mm⁴
web: 10·300³/12 + 3000·19.35² = 2.362 × 10⁷ mm⁴
bottom flange: 280·14³/12 + 3920·137.7² = 7.434 × 10⁷ mm⁴
Ix = 8.596 × 10⁷ + 2.362 × 10⁷ + 7.434 × 10⁷
Ix = 1.839 × 10⁸ mm⁴ = 183.9 × 10⁶ mm⁴

Answer

Ix = 183.9 × 10⁶ mm⁴

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