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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 = (4, 6, −7) onto b = (−4, 5, 4), that is, the component of a in the direction of b.
a = (4, 6, −7)b = (−4, 5, 4)
compb a =
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−1, 0, 1) and b = (1, 6, 0).
a = (−1, 0, 1)b = (1, 6, 0)
A =
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 = 4 mP1 = 30 kN at 2 mP2 = 50 kN at 2.5 m
RA = kN
4.Beam support reactionsHard
The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is 6 kN/m at the wall and falls to zero at the free end B, plus a point load P at 9 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 10 mw0 = 6 kN/mP = 30 kN at 9 m
MA = kN·m
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.
B: 35 kN ↓D: 20 kN ↓
1member AB
2member BC
3member CD
4member DE
5member FG
6member GH
7member AF
8member HE
9member BF
10member CG
11member DH
12member BG
13member DG
6.Truss: method of jointsHard
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
In the toggle clamp shown, two equal links AB and BC are pinned together at B, and each makes 15° with the line AC. A downward force of 150 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 150 Nθ = 15°
Q = N
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 100 mm long between pins. The joint B sits 12 mm above the line AC. Which downward force F at B produces a horizontal pressing force of 5000 N at the slider C? Neglect friction and weights.
Q = 5000 NAB = BC = 100 mmh = 12 mm
F = N
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 8 mw = 6 kN/m from 0 to 8 mx = 3 m
Mx = kN·m
10.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.
L = 8 mw = 6 kN/m from 1 to 8 m
Mmax = kN·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 = 10 mmtw = 18 mmhw = 120 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 150 mm × 150 mm plate has a rectangular cut-out 80 mm wide and 40 mm high, centred on the vertical centre line with its centre 97.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 150 mmcut-out: 80 × 40 mmhole centre: y = 97.5 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: compb a = −1.854
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 = 4·(−4) + 6·5 + (−7)·4 = −14
Length of b
|b| = √((−4)² + 5² + 4²) = 7.55
Scalar projection
compb a = a·b/|b| = −14/7.55 = −1.854
Answer
compb a = −1.854
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: A = 8.544
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 = 0·0 − 1·6 = −6
(a × b)y = azbx − axbz = 1·1 − (−1)·0 = 1
(a × b)z = axby − aybx = (−1)·6 − 0·1 = −6
Area
A = |a × b| = √((−6)² + 1² + (−6)²) = 8.544
Answer
A = 8.544
3.Beam support reactions
Answer: RA = 33.75 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.
ΣMA = 0: RB·4 = 30·2 + 50·2.5
RB = 185/4 = 46.25 kN
ΣFy = 0: RA = 80 − 46.25 = 33.75 kN
Answer
RA = 33.75 kN
4.Beam support reactions
Answer: MA = 370 kN·m
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 triangular load
W = ½·6·10 = 30 kN
x̄ = L/3 = 3.333 m
Moment about the wall
MA = W·x̄ + P·a = 30·3.333 + 30·9 = 370 kN·m
Answer
MA = 370 kN·m
5.Truss: method of joints
Answer: member CG
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 C
BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
ΣF⊥ = 0 → FCG = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
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 G
FG and GH lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
ΣF⊥ = 0 → FCG = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
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 15° = 150
Flink = 289.8 N
Clamping force
Q = Flink·cos 15° = 150/(2·tan 15°) = 279.9 N
Answer
Q = 279.9 N
The force ratio Q/F = 1.866 grows without limit as the links straighten.
8.Frames & machines
Answer: F = 1209 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).
Link angle
sin θ = 12/100 = 0.12, θ = 6.892°
Link force
Flink = Q/cos θ = 5036 N
Force at B
F = 2·Flink·sin θ = 2·5000·tan θ = 1209 N
Answer
F = 1209 N
9.Shear force & bending moment
Answer: Mx = 45 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.
Support reactions
Equilibrium of the whole beam.
UDL resultant: W = w·ℓ = 6·8 = 48 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 6·8·4
RB = 192/8 = 24 kN
ΣFy = 0: RA = 48 − 24 = 24 kN
Cut at x = 3 m and keep the left part
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 24·3 − 6·3·1.5 = 45 kN·m
Answer
Mx = 45 kN·m
10.Shear force & bending moment
Answer: Mmax = 46.51 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. 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.
UDL resultant: W = w·ℓ = 6·7 = 42 kN, acting at x̄ = 4.5 m
ΣMA = 0: RB·8 = 6·7·4.5
RB = 189/8 = 23.63 kN
ΣFy = 0: RA = 42 − 23.63 = 18.38 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
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 = 140·10 = 1400 mm², yf = 120 + 10/2 = 125 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 = (22500·75 − 3200·97.5)/(22500 − 3200) = 71.27 mm