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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 dot product a·b of a = (−4, −5, −4) and b = (−4, 2, −7).
a = (−4, −5, −4)b = (−4, 2, −7)
a·b =
2.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (5, 5, −2), b = (−3, −1, 3) and c = (−4, −1, −5).
a = (5, 5, −2)b = (−3, −1, 3)c = (−4, −1, −5)
V =
3.Beam support reactionsMedium
The beam has a pin at A and a roller at B, 6 m from A, and overhangs B to the free end C. Find the vertical reaction RA, taking upward as positive.
AB = 6 mAC = 8 mP1 = 15 kN at 5 mP2 = 40 kN at 8 m
RA = 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 4.5 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 1.5 m from A. Find the magnitude of the reaction moment MA at the wall.
AC = 3 mAD = 6 mP = 30 kN at 4.5 mw = 8 kN/m from 3 to 6 mQ = 50 kN at 1.5 m
MA = kN·m
5.Truss: method of jointsMedium
Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.
span = 8 mheight = 3 mB is 1 m from Avertical load = 25 kN
FAB = kN
6.Truss: method of jointsHard
Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.
span = 5 mheight = 4 mB is 3 m from Avertical load = 55 kNhorizontal load = 25 kN ←
FAB = kN
7.Frames & machinesMedium
The A-frame shown stands on a pin at A and a roller at B, 2 m apart. The legs meet at the pin C, 3 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 = 2 mheight of C = 3 mheight of DE = 0.8 m
C = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 1 kN is pinned to a wall at A and held by a cable from B to the point C on the wall, 2 m above A. A load of 2 kN hangs at B. Find the magnitude of the pin reaction A.
W = 2 kNWb = 1 kNAB = 2 mAC = 2 m
A = kN
9.Shear force & bending momentMedium
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.
55 kN @ 3 m8 kN/m, 0–6 m
1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)
10.Shear force & bending momentHard
The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance xc from A to the point of contraflexure between the supports (where the bending moment changes sign).
L = 10 mw = 4 kN/m from 0 to 10 msupports at x = 0 and x = 8 m
xc = m
11.Centroid & second moment of areaMedium
A 200 mm × 150 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 105 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 200 mmh = 150 mmd = 40 mmhole centre: y = 105 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 120 mm × 300 mm plate has a rectangular cut-out 40 mm wide and 30 mm high, centred on the vertical centre line with its centre 105 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 300 mmcut-out: 40 × 30 mmhole centre: y = 105 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: a·b = 34
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.
Positive: the angle between the vectors is less than 90°.
2.Vectors: dot & cross product
Answer: V = 93
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.
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·6 = 15·5 + 40·8
RB = 395/6 = 65.83 kN
Vertical equilibrium
RA = 15 + 40 − 65.83 = −10.83 kN
Answer
RA = −10.83 kN
RA is negative: the pin at A has to hold the beam down.
4.Beam support reactions
Answer: MA = 156 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 = 8·3 = 24 kN at 1.5 m from C
ΣMC = 0: RD·3 = 30·1.5 + 24·1.5
RD = 27 kN
VC = 54 − 27 = 27 kN
Part AC
The hinge force VC acts downward on AC (equal and opposite to its action on CD).
RA = 50 + 27 = 77 kN
MA = 50·1.5 + 27·3 = 156 kN·m
Answer
MA = 156 kN·m
5.Truss: method of joints
Answer: FAB = −23.06 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 = 25·1 → Cy = 3.125 kN
ΣFy = 0: Ay = 25 − 3.125 = 21.88 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 3/3.162 = 0.9487
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −21.88/0.9487 = −23.06 kN
Answer
FAB = −23.06 kN (compression)
6.Truss: method of joints
Answer: FAB = −52.5 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·5 = 55·3 − 25·4 → Cy = 13 kN
ΣFy = 0: Ay = 55 − 13 = 42 kN
ΣFx = 0: Ax = 25 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 4/5 = 0.8
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −42/0.8 = −52.5 kN
Answer
FAB = −52.5 kN (compression)
7.Frames & machines
Answer: C = 2.197 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 = 2·1/2.2 = 0.9091 kN
Components of the pin force on BC
Cx = T = 0.9091 kN, Cy = −By = −2 kN
Magnitude
C = √(0.9091² + 2²) = 2.197 kN
Answer
C = 2.197 kN
8.Frames & machines
Answer: A = 2.55 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 = (2 + 1/2)/0.7071 = 3.536 kN
Components at A
Ax = T·cos β = 2.5 kN
Ay = 2 + 1 − T·sin β = 0.5 kN
Magnitude
A = √(2.5² + 0.5²) = 2.55 kN
Answer
A = 2.55 kN
9.Shear force & bending moment
Answer: D (x = 3 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 = 51.5 kN, R2 = 51.5 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 = 47.5, MC = 87, MD = 118.5 kN·m, ME = 87, MF = 47.5, MG = 0
Answer
Mark D: MD = 118.5 kN·m
10.Shear force & bending moment
Answer: xc = 7.5 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, including the load on the overhang.
UDL resultant: W = w·ℓ = 4·10 = 40 kN, acting at x̄ = 5 m
ΣMA = 0: RB·8 = 4·10·5
RB = 200/8 = 25 kN
ΣFy = 0: RA = 40 − 25 = 15 kN
Zero moment
Between the supports M(x) = RA·x − w·x²/2. Apart from x = 0 it is zero at x = 2RA/w.
RA·xc − w·xc²/2 = 0
xc = 2·15/4 = 7.5 m
Answer
xc = 7.5 m
11.Centroid & second moment of area
Answer: Ix = 54.94 × 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 − 1257·105)/(30000 − 1257) = 73.69 mm
Full plate about ȳ
200·150³/12 + 30000·1.312² = 56.3 × 10⁶ mm⁴
Hole about ȳ
π·40⁴/64 + 1257·31.31² = 1.358 × 10⁶ mm⁴
Subtract
Ix = 54.94 × 10⁶ mm⁴
Answer
Ix = 54.94 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Ix = 267.4 × 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 = (36000·150 − 1200·105)/(36000 − 1200) = 151.6 mm