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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 = (−1, 1, −1) and b = (4, 3, −7).
a = (−1, 1, −1)b = (4, 3, −7)
a·b =
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−5, −5, 4) and b = (4, 5, −5) and has a positive z-component. Find its y-component n̂y.
a = (−5, −5, 4)b = (4, 5, −5)n̂z > 0
n̂y =
3.Beam support reactionsMedium
The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w along its whole length and a point load P at 2.5 m from the wall. Find the vertical reaction RA at the wall.
L = 4 mP = 45 kN at 2.5 mw = 8 kN/m from 0 to 4 m
RA = kN
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. A load P acts at 7 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 3 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.
L = 8 mP = 10 kN at 7 m, θ = 40°Q = 30 kN at 3 m
α = °
5.Truss: method of jointsMedium
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
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 6 m from Avertical load = 40 kNhorizontal load = 15 kN →
FAB = kN
7.Frames & machinesMedium
The A-frame shown stands on a pin at A and a roller at B, 2.4 m apart, with its apex pin C 2.4 m high and the horizontal tie DE 0.6 m above the ground. A vertical load of 4 kN acts on leg AC at K, 0.48 m (horizontally) from A. Find the tension T in the tie.
P = 4 kNAB = 2.4 mheight of C = 2.4 mheight of DE = 0.6 mhorizontal distance A to K = 0.48 m
T = kN
8.Frames & machinesHard
The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 1 m below A). The beam carries a uniform load of 5 kN/m over its full length and a point load of 12 kN at its free end B. Find the force FCD in the strut.
w = 5 kN/mP = 12 kNAD = 1.5 mAB = 2.4 mAC = 1 m
FCD = 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.
60 kN @ 4 m6 kN/m, 0–10 m
1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)
10.Shear force & bending momentHard
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 @ 6 m25 kN @ 1 m15 kN/m, 0–4 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)
11.Centroid & second moment of areaMedium
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 120 mmtf = 30 mmtw = 16 mmhw = 160 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 25 kN·m. Find the size of the bending stress σtop at the top edge.
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.
Multiply matching components and add
a·b = (−1)·4 + 1·3 + (−1)·(−7) = (−4) + 3 + 7 = 6
Answer
a·b = 6
Positive: the angle between the vectors is less than 90°.
2.Vectors: dot & cross product
Answer: n̂y = 0.7863
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 = (−5)·(−5) − 4·5 = 5
(a × b)y = azbx − axbz = 4·4 − (−5)·(−5) = −9
(a × b)z = axby − aybx = (−5)·5 − (−5)·4 = −5
Normalise
The z-component of a × b is negative, so use −(a × b).
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 distributed load
W = 8·4 = 32 kN, x̄ = 2 m
Vertical equilibrium
RA = P + W = 45 + 32 = 77 kN
Answer
RA = 77 kN
4.Beam support reactions
Answer: α = 68.61 °
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.
Components of P
Pv = 10·sin 40° = 6.428 kN
Ph = 10·cos 40° = 7.66 kN
Reactions
ΣMA = 0: RB·8 = 6.428·7 + 30·3 → RB = 16.87 kN
Ay = 6.428 + 30 − 16.87 = 19.55 kN
Ax = Ph = 7.66 kN
Direction at A
α = atan(Ay/Ax) = atan(19.55/7.66) = 68.61°
Answer
α = 68.61°
5.Truss: method of joints
Answer: member DH
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 D
CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.
ΣF⊥ = 0 → FDH = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
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 = 40·6 + 15·3 → Cy = 35.63 kN
ΣFy = 0: Ay = 40 − 35.63 = 4.375 kN
ΣFx = 0: Ax = −15 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 3/6.708 = 0.4472
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −4.375/0.4472 = −9.783 kN
Answer
FAB = −9.783 kN (compression)
7.Frames & machines
Answer: T = 0.5333 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 = 4·0.48/2.4 = 0.8 kN
Leg BC: moments about C
0.8·1.2 = T·1.8
T = 0.5333 kN
Answer
T = 0.5333 kN
8.Frames & machines
Answer: FCD = 51.92 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 = 5·2.4 = 12 kN at 1.2 m
Moments about A for the beam
FCD·0.5547·1.5 = 12·1.2 + 12·2.4 = 43.2 kN·m
FCD = 51.92 kN
Answer
FCD = 51.92 kN
9.Shear force & bending moment
Answer: C (x = 4 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 = 75 kN, R2 = 45 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 = 138, MC = 252 kN·m, MD = 222, ME = 168, MF = 90, MG = 0
Answer
Mark C: MC = 252 kN·m
10.Shear force & bending moment
Answer: E (x = 4 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 = 23.75 kN, R2 = 111.3 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 = 16.25, MC = −7.5, MD = −46.25, ME = −100 kN·m, MF = −50, MG = 0
Answer
Mark E: ME = −100 kN·m
11.Centroid & second moment of area
Answer: Ix = 19.23 × 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 = 120·30 = 3600 mm², yf = 160 + 30/2 = 175 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 two rectangles
Measure every y from the bottom edge.
flange: Af = 300·15 = 4500 mm², yf = 140 + 15/2 = 147.5 mm