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
xyzdFAB

A constant force F = (25, −25, 30) N acts on an object while it moves in a straight line from point A(0, 0, 0) to point B(1, 1, −2). Coordinates are in metres. Find the work W done by the force.

F = (25, −25, 30) NA(0, 0, 0) mB(1, 1, −2) m
W = J
2.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (5, 0, 1) and b = (−1, −5, −3) and has a positive z-component. Find its x-component n̂x.

a = (5, 0, 1)b = (−1, −5, −3)n̂z > 0
n̂x =
3.Beam support reactionsMedium
P1​ = 50 kNP2​ = 60 kNABRA​RB​ = ?3 m5 m2 m

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

L = 10 mP1 = 50 kN at 3 mP2 = 60 kN at 8 m
RB = kN
4.Beam support reactionsHard
w0​ = 12 kN/mP = 35 kNABMA​ = ?5 m1 m

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is 12 kN/m at the wall and falls to zero 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 = 12 kN/mP = 35 kN at 5 m
MA = kN·m
5.Truss: method of jointsMedium
ABCDEFGH25 kN40 kN15 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: 25 kN ↓D: 40 kN ↓G: 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
6.Truss: method of jointsHard
ABBCACACB15 kN30 kN8 m5 m4 m

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

span = 8 mheight = 4 mB is 5 m from Avertical load = 15 kNhorizontal load = 30 kN →
FAC = kN
7.Frames & machinesMedium
4 kN2 m1.5 m2.5 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 m high and the horizontal tie DE 1.5 m above the ground. A vertical load of 4 kN acts on leg AC at K, 0.25 m (horizontally) from A. Find the tension T in the tie.

P = 4 kNAB = 2 mheight of C = 4 mheight of DE = 1.5 mhorizontal distance A to K = 0.25 m
T = kN
8.Frames & machinesHard
8 kN0.5 kN5 m1 mABC

A uniform horizontal boom AB of weight 0.5 kN is pinned to a wall at A and held by a cable from B to the point C on the wall, 1 m above A. A load of 8 kN hangs at B. Find the magnitude of the pin reaction A.

W = 8 kNWb = 0.5 kNAB = 5 mAC = 1 m
A = kN
9.Shear force & bending momentMedium
w = 4 kN/mABsection2 m6 m

Find the shear force Vx at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 4 kN/m from 0 to 8 mx = 2 m
Vx = kN
10.Shear force & bending momentHard
w = 6 kN/mP = 45 kNABsection1 m1 m2 m

Find the bending moment Mx at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 45 kN at 1 mw = 6 kN/m from 0 to 4 mx = 2 m
Mx = kN·m
11.Centroid & second moment of areaMedium
140152008xdimensions in mm

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

bf = 140 mmtf = 15 mmtw = 8 mmhw = 200 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
190160181901212xdimensions in mm

A welded I-section has a top flange 190 × 18 mm, a web 190 mm high and 12 mm thick, and a bottom flange 160 × 12 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.

top flange: 190 × 18 mmweb: 12 × 190 mmbottom flange: 160 × 12 mm
Ix = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: W = −60 J

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.

Displacement

d = B − A = (1, 1, −2) − (0, 0, 0) = (1, 1, −2) m

Work

W = F·d = 25·1 + (−25)·1 + 30·(−2) = −60 J

Answer

W = −60 J

Negative work: the force acts mostly against the motion.

2.Vectors: dot & cross product

Answer: n̂x = −0.1719

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 = 0·(−3) − 1·(−5) = 5
(a × b)y = azbx − axbz = 1·(−1) − 5·(−3) = 14
(a × b)z = axby − aybx = 5·(−5) − 0·(−1) = −25

Normalise

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

|a × b| = √(5² + 14² + (−25)²) = 29.09
n̂ = −(5, 14, −25)/29.09 = (−0.1719, −0.4813, 0.8595)
n̂x = −0.1719

Answer

n̂x = −0.1719
3.Beam support reactions

Answer: RB = 63 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​ = 50 kNP2​ = 60 kNABRA​RB​3 m5 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·10 = 50·3 + 60·8
RB = 630/10 = 63 kN
ΣFy = 0: RA = 110 − 63 = 47 kN

Answer

RB = 63 kN
4.Beam support reactions

Answer: MA = 247 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 = ½·12·6 = 36 kN
x̄ = L/3 = 2 m

Moment about the wall

MA = W·x̄ + P·a = 36·2 + 35·5 = 247 kN·m

Answer

MA = 247 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: 27.19 kN (T), BC: 35.63 kN (T), CD: 35.63 kN (T), DE: 32.81 kN (T), FG: 27.19 kN (C), GH: 32.81 kN (C), AF: 45.31 kN (C), HE: 54.69 kN (C), BF: 36.25 kN (T), CG: 0, DH: 43.75 kN (T), BG: 14.06 kN (C), DG: 4.688 kN (C)

Answer

Zero-force member: CG
6.Truss: method of joints

Answer: FAC = 18.28 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.

ABBCACACB15 kN30 kNAy​Cy​Ax​

Support reactions

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

ΣMA = 0: Cy·8 = 15·5 + 30·4 → Cy = 24.38 kN
ΣFy = 0: Ay = 15 − 24.38 = −9.375 kN
ΣFx = 0: Ax = −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 = 4/5 = 0.8, cos γ = 3/5 = 0.6
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −24.38/0.8 = −30.47 kN
ΣFx = 0: −FAC − FBC·cos γ = 0 → FAC = −(−30.47)·0.6 = 18.28 kN

Answer

FAC = 18.28 kN (tension)
7.Frames & machines

Answer: T = 0.2 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 = 4·0.25/2 = 0.5 kN

Leg BC: moments about C

0.5·1 = T·2.5
T = 0.2 kN

Answer

T = 0.2 kN
8.Frames & machines

Answer: A = 41.25 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.

Cable tension

T = (8 + 0.5/2)/0.1961 = 42.07 kN

Components at A

Ax = T·cos β = 41.25 kN
Ay = 8 + 0.5 − T·sin β = 0.25 kN

Magnitude

A = √(41.25² + 0.25²) = 41.25 kN

Answer

A = 41.25 kN
9.Shear force & bending moment

Answer: Vx = 8 kN

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 = 4·8 = 32 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 4·8·4
RB = 128/8 = 16 kN
ΣFy = 0: RA = 32 − 16 = 16 kN
w = 4 kN/mABRA​ = 16RB​ = 16section2 m6 m

Cut at x = 2 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.

V = RA − (all downward loads left of the cut)
Vx = 16 − 4·2 = 8 kN

Answer

Vx = 8 kN
10.Shear force & bending moment

Answer: Mx = 34.5 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·4 = 24 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 45·1 + 6·4·2
RB = 93/4 = 23.25 kN
ΣFy = 0: RA = 69 − 23.25 = 45.75 kN
w = 6 kN/mP = 45 kNABRA​ = 45.75RB​ = 23.25section1 m1 m2 m

Cut at x = 2 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 = 45.75·2 − 45·(2 − 1) − 6·2·1 = 34.5 kN·m

Answer

Mx = 34.5 kN·m
11.Centroid & second moment of area

Answer: Ix = 15.87 × 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 = 140·15 = 2100 mm², yf = 200 + 15/2 = 207.5 mm
web: Aw = 8·200 = 1600 mm², yw = 200/2 = 100 mm

Centroid

ȳ = (Afyf + Awyw)/(Af + Aw) = (2100·207.5 + 1600·100)/3700 = 161 mm

Parallel-axis theorem

flange: 140·15³/12 + 2100·(207.5 − 161)² = 4.577 × 10⁶ mm⁴
web: 8·200³/12 + 1600·(100 − 161)² = 1.129 × 10⁷ mm⁴
Ix = 4.577 × 10⁶ + 1.129 × 10⁷ = 1.587 × 10⁷ mm⁴ = 15.87 × 10⁶ mm⁴

Answer

Ix = 15.87 × 10⁶ mm⁴
12.Centroid & second moment of area

Answer: Ix = 60.12 × 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 = 3420 mm², y = 211 mm
web: A = 2280 mm², y = 107 mm
bottom flange: A = 1920 mm², y = 6 mm

Centroid

ȳ = (3420·211 + 2280·107 + 1920·6)/7620 = 128.2 mm

Parallel-axis theorem

top flange: 190·18³/12 + 3420·82.77² = 2.352 × 10⁷ mm⁴
web: 12·190³/12 + 2280·21.23² = 7.886 × 10⁶ mm⁴
bottom flange: 160·12³/12 + 1920·122.2² = 2.871 × 10⁷ mm⁴
Ix = 2.352 × 10⁷ + 7.886 × 10⁶ + 2.871 × 10⁷
Ix = 6.012 × 10⁷ mm⁴ = 60.12 × 10⁶ mm⁴

Answer

Ix = 60.12 × 10⁶ mm⁴

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