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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
Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
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
Draw a free-body diagram and replace each support by its reactions.
Replace every distributed load by a single force at its centroid.
Take moments about one support to find the reaction at the other.
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
Find the support reactions for the whole truss.
Method of joints: start at a joint with at most two unknown member forces and use ΣFx = 0 and ΣFy = 0.
Method of sections: cut through the member you need and take moments about a point where the other unknown members meet.
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
Draw the whole structure and find the support reactions if there are no more than three unknowns.
Spot the two-force members: loaded only at two pins, their force acts along the line between the pins.
Take the frame apart and draw every member on its own, with equal and opposite pin forces where members meet.
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
Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
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
Find the angle θ between the vectors a = (2, 3, −5) and b = (−5, 6, 6), in degrees.
a = (2, 3, −5)b = (−5, 6, 6)
θ = °
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−4, −3, 4) and b = (0, 4, 3) and has a positive z-component. Find its y-component n̂y.
a = (−4, −3, 4)b = (0, 4, 3)n̂z > 0
n̂y =
3.Beam support reactionsMedium
The beam has a pin at A and a roller at B, 2.5 m from A, and overhangs B to the free end C. Find the vertical reaction RB, taking upward as positive.
AB = 2.5 mAC = 4 mP1 = 10 kN at 0.5 mP2 = 10 kN at 4 m
RB = kN
4.Beam support reactionsHard
A compound beam is fixed to the wall at A, has an internal hinge at C (3 m from A) and rests on a roller at D (6 m from A). It carries a point load P at 5.5 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 1 m from A. Find the magnitude of the reaction moment MA at the wall.
AC = 3 mAD = 6 mP = 15 kN at 5.5 mw = 6 kN/m from 3 to 6 mQ = 50 kN at 1 m
MA = kN·m
5.Truss: method of jointsMedium
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
span = 8 mheight = 2 mB is 7 m from Avertical load = 60 kN
FAC = kN
6.Truss: method of jointsHard
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
The bell crank shown turns about the pin O. A vertical force of 100 N at A (100 mm from O) is balanced by a horizontal force at B (100 mm above O). Find the magnitude of the pin reaction FO.
F = 100 NOA = 100 mmOB = 100 mm
FO = N
8.Frames & machinesHard
The A-frame shown stands on a pin at A and a roller at B, 6 m apart. The legs meet at the pin C, 2.4 m high, and are tied by the horizontal bar DE at 1.2 m. A horizontal force of 12 kN to the right acts at C. Find the tension T in the tie.
P = 12 kNAB = 6 mheight of C = 2.4 mheight of DE = 1.2 m
T = kN
9.Shear force & bending momentMedium
The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 1 m from A. Find the bending moment MA at the fixed support A. Use sagging moments positive and upward shear on the left face positive.
L = 2 mP = 30 kN at 1 mw = 4 kN/m from 0 to 2 m
MA = kN·m
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 4 mP = 45 kN at 1.5 mw = 10 kN/m from 0 to 4 mx = 1.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A circular tube has outside diameter D = 250 mm and inside diameter d = 234 mm. Find its radius of gyration r about the horizontal centroidal axis.
D = 250 mmd = 234 mm
r = mm
12.Centroid & second moment of areaHard
A 200 mm × 150 mm plate has a rectangular cut-out 50 mm wide and 20 mm high, centred on the vertical centre line with its centre 52.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 200 mmh = 150 mmcut-out: 50 × 20 mmhole centre: y = 52.5 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: θ = 111.2 °
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·(−5) + 3·6 + (−5)·6 = −22
Magnitudes
|a| = √(2² + 3² + (−5)²) = 6.164
|b| = √((−5)² + 6² + 6²) = 9.849
Angle
cos θ = −22/(6.164·9.849) = −0.3624
θ = arccos(−0.3624) = 111.2°
Answer
θ = 111.2°
2.Vectors: dot & cross product
Answer: n̂y = −0.3748
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 = (−3)·3 − 4·4 = −25
(a × b)y = azbx − axbz = 4·0 − (−4)·3 = 12
(a × b)z = axby − aybx = (−4)·4 − (−3)·0 = −16
Normalise
The z-component of a × b is negative, so use −(a × b).
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.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·2.5 = 10·0.5 + 10·4
RB = 45/2.5 = 18 kN
Vertical equilibrium
RA = 10 + 10 − 18 = 2 kN
Answer
RB = 18 kN
4.Beam support reactions
Answer: MA = 84.5 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.
W = 6·3 = 18 kN at 1.5 m from C
ΣMC = 0: RD·3 = 15·2.5 + 18·1.5
RD = 21.5 kN
VC = 33 − 21.5 = 11.5 kN
Part AC
The hinge force VC acts downward on AC (equal and opposite to its action on CD).
RA = 50 + 11.5 = 61.5 kN
MA = 50·1 + 11.5·3 = 84.5 kN·m
Answer
MA = 84.5 kN·m
5.Truss: method of joints
Answer: FAC = 26.25 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.
Support reactions
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
ΣMA = 0: Cy·8 = 60·7 → Cy = 52.5 kN
ΣFy = 0: Ay = 60 − 52.5 = 7.5 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
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).
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 = 100·100/100 = 100 N
Pin reaction
Ox = 100 N, Oy = 100 N
FO = √(100² + 100²) = 141.4 N
Answer
FO = 141.4 N
8.Frames & machines
Answer: T = 12 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·6 = 12·2.4
By = 4.8 kN
Leg BC: moments about C
4.8·3 = T·1.2
T = 12 kN
Answer
T = 12 kN
A positive T means the tie is in tension, as assumed.
9.Shear force & bending moment
Answer: MA = −38 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.
Keep the free-end part
P: −30·1 = −30 kN·m
w: −4·2²/2 = −8 kN·m
MA = −30 − 8 = −38 kN·m
Answer
MA = −38 kN·m
Negative means hogging: tension on the top face.
10.Shear force & bending moment
Answer: Mx = 60.94 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 = 10·4 = 40 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 45·1.5 + 10·4·2
RB = 147.5/4 = 36.88 kN
ΣFy = 0: RA = 85 − 36.88 = 48.13 kN
Cut at x = 1.5 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 = 48.13·1.5 − 10·1.5·0.75 = 60.94 kN·m
Answer
Mx = 60.94 kN·m
11.Centroid & second moment of area
Answer: r = 85.61 mm
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.
Net area and second moment of area
A = π(D² − d²)/4 = 6082 mm²
I = π(D⁴ − d⁴)/64 = 4.457 × 10⁷ mm⁴
Radius of gyration
r = √(I/A) = √(4.457 × 10⁷/6082) = 85.61 mm
Answer
r = 85.61 mm
12.Centroid & second moment of area
Answer: Ix = 55.69 × 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 = (30000·75 − 1000·52.5)/(30000 − 1000) = 75.78 mm