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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 = (5, 6, 3) and b = (3, 3, −1), in degrees.
a = (5, 6, 3)b = (3, 3, −1)
θ = °
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (4, −3, 5) and b = (0, 4, 2) and has a positive z-component. Find its y-component n̂y.
a = (4, −3, 5)b = (0, 4, 2)n̂z > 0
n̂y =
3.Beam support reactionsMedium
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
L = 10 mP1 = 10 kN at 1 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. Find the reaction RD at the roller.
AC = 3 mAD = 6 mP = 65 kN at 4.5 mw = 10 kN/m from 3 to 6 m
RD = kN
5.Truss: method of jointsMedium
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 8 mheight = 3 mB is 7 m from Avertical load = 45 kN
FBC = 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 = 3 mB is 2 m from Avertical load = 40 kNhorizontal load = 5 kN ←
FAB = kN
7.Frames & machinesMedium
The frame shown has pins at A, B and C; A and C are fixed to the ground 3 m apart, and B is 3 m above the ground and 1.35 m to the right of A. A vertical load of 15 kN acts on member AB at K, 0.675 m (horizontally) from A. Find the force FBC in member BC.
P = 15 kNL = 3 mH = 3 mhorizontal distance A to B = 1.35 mhorizontal distance A to K = 0.675 m
FBC = kN
8.Frames & machinesHard
The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 0.6 m below A). The beam carries a uniform load of 1.5 kN/m over its full length and a point load of 15 kN at its free end B. Find the force FCD in the strut.
w = 1.5 kN/mP = 15 kNAD = 0.5 mAB = 1.5 mAC = 0.6 m
FCD = 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 1.5 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 = 2.5 mP = 75 kN at 1.5 mw = 8 kN/m from 0 to 2.5 m
MA = kN·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.
40 kN @ 0 m45 kN @ 12 m4 kN/m, 2–10 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)
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 = 280 mmtf = 25 mmtw = 14 mmhw = 240 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 100 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 90 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 150 mmcut-out: 50 × 20 mmhole centre: y = 90 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: θ = 34.65 °
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 = 5·3 + 6·3 + 3·(−1) = 30
Magnitudes
|a| = √(5² + 6² + 3²) = 8.367
|b| = √(3² + 3² + (−1)²) = 4.359
Angle
cos θ = 30/(8.367·4.359) = 0.8226
θ = arccos(0.8226) = 34.65°
Answer
θ = 34.65°
2.Vectors: dot & cross product
Answer: n̂y = −0.2535
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. Taking moments about one support removes that support’s reaction from the equation.
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 = 10·1 + 40·8
RB = 330/10 = 33 kN
ΣFy = 0: RA = 50 − 33 = 17 kN
Answer
RA = 17 kN
4.Beam support reactions
Answer: RD = 47.5 kN
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 = 10·3 = 30 kN at 1.5 m from C
ΣMC = 0: RD·3 = 65·1.5 + 30·1.5
RD = 47.5 kN
VC = 95 − 47.5 = 47.5 kN
Answer
RD = 47.5 kN
5.Truss: method of joints
Answer: FBC = −41.5 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·8 = 45·7 → Cy = 39.38 kN
ΣFy = 0: Ay = 45 − 39.38 = 5.625 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 = 3/3.162 = 0.9487, cos γ = 1/3.162 = 0.3162
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 = 40·2 − 5·3 → Cy = 13 kN
ΣFy = 0: Ay = 40 − 13 = 27 kN
ΣFx = 0: Ax = 5 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 3/3.606 = 0.8321
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −27/0.8321 = −32.45 kN
Answer
FAB = −32.45 kN (compression)
7.Frames & machines
Answer: FBC = 3.852 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.
Two-force member BC
BC = √(1.65² + 3²) = 3.424 m
Whole frame: moments about A
The ground pushes on BC at C along CB; its horizontal component passes through A, only the vertical one has a moment.
FBC·(3/3.424)·3 = 15·0.675
FBC = 3.852 kN (compression)
Answer
FBC = 3.852 kN
8.Frames & machines
Answer: FCD = 62.97 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 = 1.5·1.5 = 2.25 kN at 0.75 m
Moments about A for the beam
FCD·0.7682·0.5 = 2.25·0.75 + 15·1.5 = 24.19 kN·m
FCD = 62.97 kN
Answer
FCD = 62.97 kN
9.Shear force & bending moment
Answer: MA = −137.5 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.
Keep the free-end part
P: −75·1.5 = −112.5 kN·m
w: −8·2.5²/2 = −25 kN·m
MA = −112.5 − 25 = −137.5 kN·m
Answer
MA = −137.5 kN·m
Negative means hogging: tension on the top face.
10.Shear force & bending moment
Answer: F (x = 10 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 = 54.75 kN, R2 = 62.25 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 = −80, MC = −58.5, MD = −53, ME = −63.5, MF = −90 kN·m, MG = 0
Answer
Mark F: MF = −90 kN·m
11.Centroid & second moment of area
Answer: Ix = 56.35 × 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 = 280·25 = 7000 mm², yf = 240 + 25/2 = 252.5 mm
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 = (15000·75 − 1000·90)/(15000 − 1000) = 73.93 mm