In the print window, choose “Save as PDF” to keep a copy on your device.
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 value of k for which the vectors a = (1, −6, 6) and b = (k, 0, 3) are perpendicular.
a = (1, −6, 6)b = (k, 0, 3)
k =
2.Vectors: dot & cross productHard
A force F = (0, −10, −60) N acts at point A(1, 4, 1). Coordinates are in metres. Find the x-component Mx of the moment of the force about the origin O(0, 0, 0).
F = (0, −10, −60) NA(1, 4, 1) mO(0, 0, 0) m
Mx = N·m
3.Beam support reactionsMedium
The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 2 m and a point load P at 1.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 5 mP = 50 kN at 1.5 mw = 3 kN/m from 0 to 2 m
MA = kN·m
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
L = 10 mP = 55 kN at 5 mw = 6 kN/m from 8 to 10 m
RB = 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 = 6 mheight = 3 mB is 2 m from Avertical load = 25 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 = 2 mB is 3 m from Avertical load = 10 kNhorizontal load = 25 kN ←
FAB = kN
7.Frames & machinesMedium
The A-frame shown stands on a pin at A and a roller at B, 2.4 m apart, with its apex pin C 2 m high and the horizontal tie DE 1 m above the ground. A vertical load of 20 kN acts on leg AC at K, 0.9 m (horizontally) from A. Find the tension T in the tie.
P = 20 kNAB = 2.4 mheight of C = 2 mheight of DE = 1 mhorizontal distance A to K = 0.9 m
T = kN
8.Frames & machinesHard
The A-frame shown stands on a pin at A and a roller at B, 5 m apart, with its apex pin C 3 m high and the horizontal tie DE 1 m above the ground. A vertical load of 5 kN acts on leg AC at K, 1.25 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 5 kNAB = 5 mheight of C = 3 mheight of DE = 1 mhorizontal distance A to K = 1.25 m
C = 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.
40 kN @ 10 m35 kN @ 4 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.
25 kN @ 12 m10 kN @ 4 m5 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)
11.Centroid & second moment of areaMedium
A 150 mm × 150 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 150 mmd = 60 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 200 mm wide and 14 mm thick, an overall depth of 280 mm and a web 14 mm thick. It carries a bending moment of 80 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
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 equals zero
a·b = 1·k + (−6)·0 + 6·3 = 0
18 + 1·k = 0
Solve for k
k = −18/1 = −18
Answer
k = −18
2.Vectors: dot & cross product
Answer: Mx = −230 N·m
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.
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 = 3·2 = 6 kN, x̄ = 1 m
Moment about the wall
MA = P·a + W·x̄ = 50·1.5 + 6·1 = 81 kN·m
Answer
MA = 81 kN·m
4.Beam support reactions
Answer: RB = 38.3 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.
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.
UDL resultant: W = w·ℓ = 6·2 = 12 kN, acting at x̄ = 9 m
ΣMA = 0: RB·10 = 55·5 + 6·2·9
RB = 383/10 = 38.3 kN
ΣFy = 0: RA = 67 − 38.3 = 28.7 kN
Answer
RB = 38.3 kN
5.Truss: method of joints
Answer: FBC = −13.89 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·6 = 25·2 → Cy = 8.333 kN
ΣFy = 0: Ay = 25 − 8.333 = 16.67 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
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 = 10·3 − 25·2 → Cy = −4 kN
ΣFy = 0: Ay = 10 − −4 = 14 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 = 2/3.606 = 0.5547
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −14/0.5547 = −25.24 kN
Answer
FAB = −25.24 kN (compression)
7.Frames & machines
Answer: T = 9 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 = 20·0.9/2.4 = 7.5 kN
Leg BC: moments about C
7.5·1.2 = T·1
T = 9 kN
Answer
T = 9 kN
8.Frames & machines
Answer: C = 2.001 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.
Support reaction at B
By = 5·1.25/5 = 1.25 kN
Tie force
T = 1.25·2.5/2 = 1.563 kN
Pin force on BC
Cx = 1.563 kN, Cy = −1.25 kN
C = √(1.563² + 1.25²) = 2.001 kN
Answer
C = 2.001 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 = 30 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 = 60, MC = 120 kN·m, MD = 110, ME = 100, MF = 90, MG = 0
Answer
Mark C: MC = 120 kN·m
10.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 = 26 kN, R2 = 59 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 = 42, MC = 64 kN·m, MD = 46, ME = 8, MF = −50, MG = 0
Answer
Mark C: MC = 64 kN·m
11.Centroid & second moment of area
Answer: Ix = 40.82 × 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 = (22500·75 − 2827·60)/(22500 − 2827) = 77.16 mm
Full plate about ȳ
150·150³/12 + 22500·−2.156² = 42.29 × 10⁶ mm⁴
Hole about ȳ
π·60⁴/64 + 2827·−17.16² = 1.468 × 10⁶ mm⁴
Subtract
Ix = 40.82 × 10⁶ mm⁴
Answer
Ix = 40.82 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: σj = 85.55 MPa
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.
Outer rectangle minus the two side cut-outs
Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.