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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 scalar projection compb a of a = (6, 2, 0) onto b = (6, −4, −1), that is, the component of a in the direction of b.
a = (6, 2, 0)b = (6, −4, −1)
compb a =
2.Vectors: dot & cross productHard
A force F = (20, 10, −40) N acts at point A(−1, 1, −4). Coordinates are in metres. Find the y-component My of the moment of the force about point O(−2, 0, −3).
F = (20, 10, −40) NA(−1, 1, −4) mO(−2, 0, −3) m
My = N·m
3.Beam support reactionsMedium
The simply supported beam (pin at A, roller at B) carries a linearly varying load that rises from zero at A to 6 kN/m at B. A point load P also acts at 3 m from A. Find the vertical support reaction RB.
L = 10 mw0 = 6 kN/mP = 60 kN at 3 m
RB = kN
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
L = 6 mP = 55 kN at 1.5 mw = 3 kN/m from 3 to 5 m
RA = kN
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 = 6 mheight = 2 mB is 4 m from Avertical load = 35 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 = 6 mheight = 3 mB is 4 m from Avertical load = 60 kNhorizontal load = 30 kN ←
FAB = kN
7.Frames & machinesMedium
In the toggle clamp shown, two equal links AB and BC are pinned together at B, and each makes 10° with the line AC. A downward force of 400 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 400 Nθ = 10°
Q = N
8.Frames & machinesHard
The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 0.75 m below A). The beam carries a uniform load of 5 kN/m over its full length and a point load of 10 kN at its free end B. Find the force FCD in the strut.
w = 5 kN/mP = 10 kNAD = 0.8 mAB = 1.5 mAC = 0.75 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.
10 kN @ 3 m6 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 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 @ 12 m25 kN @ 2 m4 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
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 140 mmtf = 15 mmtw = 8 mmhw = 220 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 120 mm × 150 mm plate has a rectangular cut-out 40 mm wide and 20 mm high, centred on the 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 = 120 mmh = 150 mmcut-out: 40 × 20 mmhole centre: y = 60 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: compb a = 3.846
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 = 6·6 + 2·(−4) + 0·(−1) = 28
Length of b
|b| = √(6² + (−4)² + (−1)²) = 7.28
Scalar projection
compb a = a·b/|b| = 28/7.28 = 3.846
Answer
compb a = 3.846
2.Vectors: dot & cross product
Answer: My = 20 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.
Position vector from O to A
r = A − O = (−1, 1, −4) − (−2, 0, −3) = (1, 1, −1) 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.
Resultant of the triangular load
W = ½·6·10 = 30 kN
x̄ = 2L/3 = 6.667 m
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·10 = 30·6.667 + 60·3
RB = 38 kN
RA = 90 − 38 = 52 kN
Answer
RB = 38 kN
4.Beam support reactions
Answer: RA = 43.25 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 = 3·2 = 6 kN, acting at x̄ = 4 m
ΣMA = 0: RB·6 = 55·1.5 + 3·2·4
RB = 106.5/6 = 17.75 kN
ΣFy = 0: RA = 61 − 17.75 = 43.25 kN
Answer
RA = 43.25 kN
5.Truss: method of joints
Answer: FAB = −26.09 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 = 35·4 → Cy = 23.33 kN
ΣFy = 0: Ay = 35 − 23.33 = 11.67 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 2/4.472 = 0.4472
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −11.67/0.4472 = −26.09 kN
Answer
FAB = −26.09 kN (compression)
6.Truss: method of joints
Answer: FAB = −58.33 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 = 60·4 − 30·3 → Cy = 25 kN
ΣFy = 0: Ay = 60 − 25 = 35 kN
ΣFx = 0: Ax = 30 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 3/5 = 0.6
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −35/0.6 = −58.33 kN
Answer
FAB = −58.33 kN (compression)
7.Frames & machines
Answer: Q = 1134 N
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).
Equilibrium of pin B
2·Flink·sin 10° = 400
Flink = 1152 N
Clamping force
Q = Flink·cos 10° = 400/(2·tan 10°) = 1134 N
Answer
Q = 1134 N
The force ratio Q/F = 2.836 grows without limit as the links straighten.
8.Frames & machines
Answer: FCD = 37.7 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 = 5·1.5 = 7.5 kN at 0.75 m
Moments about A for the beam
FCD·0.6839·0.8 = 7.5·0.75 + 10·1.5 = 20.63 kN·m
FCD = 37.7 kN
Answer
FCD = 37.7 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 = 23 kN, R2 = 23 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 = 20, MC = 34, MD = 42 kN·m, ME = 34, MF = 20, MG = 0
Answer
Mark D: MD = 42 kN·m
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 = 32 kN, R2 = 73 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 = 56, MC = 46, MD = 20, ME = −22, MF = −80 kN·m, MG = 0
Answer
Mark F: MF = −80 kN·m
11.Centroid & second moment of area
Answer: Ix = 20.36 × 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 = 140·15 = 2100 mm², yf = 220 + 15/2 = 227.5 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.