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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 value of k for which the vectors a = (1, −5, 6) and b = (−3, k, 3) are perpendicular.
a = (1, −5, 6)b = (−3, k, 3)
k =
2.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (2, 2, 1), b = (−2, 4, 0) and c = (2, 3, 4).
a = (2, 2, 1)b = (−2, 4, 0)c = (2, 3, 4)
V =
3.Beam support reactionsMedium
The simply supported beam has a pin at A and a roller at B. It carries a point load P at 2 m from A and a counter-clockwise couple M at 1.5 m from A. Find the vertical reaction RA, taking upward as positive.
L = 5 mP = 50 kN at 2 mM = 15 kN·m (counter-clockwise) at 1.5 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 50° to the horizontal, and a vertical load Q acts at 4 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.
L = 8 mP = 65 kN at 7 m, θ = 50°Q = 70 kN at 4 m
α = °
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 = 8 mheight = 3 mB is 3 m from Avertical load = 55 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 = 2 mB is 1 m from Avertical load = 30 kNhorizontal load = 10 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 500 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 500 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 0.5 m below A). The beam carries a uniform load of 2 kN/m over its full length and a point load of 5 kN at its free end B. Find the force FCD in the strut.
w = 2 kN/mP = 5 kNAD = 1.2 mAB = 4 mAC = 0.5 m
FCD = kN
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 6 mw = 10 kN/m from 0 to 6 mx = 2.5 m
Mx = kN·m
10.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.
L = 8 mP = 65 kN at 7 mw = 8 kN/m from 0 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 = 260 mmtf = 10 mmtw = 10 mmhw = 280 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 150 mm × 240 mm plate has a rectangular cut-out 40 mm wide and 30 mm high, centred on the vertical centre line with its centre 72 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 240 mmcut-out: 40 × 30 mmhole centre: y = 72 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: k = 3
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·(−3) + (−5)·k + 6·3 = 0
15 + (−5)·k = 0
Solve for k
k = −15/(−5) = 3
Answer
k = 3
2.Vectors: dot & cross product
Answer: V = 34
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 b × c
(b × c)x = bycz − bzcy = 4·4 − 0·3 = 16
(b × c)y = bzcx − bxcz = 0·2 − (−2)·4 = 8
(b × c)z = bxcy − bycx = (−2)·3 − 4·2 = −14
Dot product with a
a·(b × c) = 2·16 + 2·8 + 1·(−14) = 34
V = |34| = 34
Answer
V = 34
3.Beam support reactions
Answer: RA = 33 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.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·5 − 50·2 + 15 = 0
RB = 85/5 = 17 kN
Vertical equilibrium
RA = 50 − 17 = 33 kN
Answer
RA = 33 kN
A negative reaction acts downward.
4.Beam support reactions
Answer: α = 44.62 °
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 = 65·sin 50° = 49.79 kN
Ph = 65·cos 50° = 41.78 kN
Reactions
ΣMA = 0: RB·8 = 49.79·7 + 70·4 → RB = 78.57 kN
Ay = 49.79 + 70 − 78.57 = 41.22 kN
Ax = Ph = 41.78 kN
Direction at A
α = atan(Ay/Ax) = atan(41.22/41.78) = 44.62°
Answer
α = 44.62°
5.Truss: method of joints
Answer: FAB = −48.61 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 = 55·3 → Cy = 20.63 kN
ΣFy = 0: Ay = 55 − 20.63 = 34.38 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 3/4.243 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −34.38/0.7071 = −48.61 kN
Answer
FAB = −48.61 kN (compression)
6.Truss: method of joints
Answer: FAB = −24.22 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 = 30·1 + 10·2 → Cy = 8.333 kN
ΣFy = 0: Ay = 30 − 8.333 = 21.67 kN
ΣFx = 0: Ax = −10 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 2/2.236 = 0.8944
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −21.67/0.8944 = −24.22 kN
Answer
FAB = −24.22 kN (compression)
7.Frames & machines
Answer: Q = 1418 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° = 500
Flink = 1440 N
Clamping force
Q = Flink·cos 10° = 500/(2·tan 10°) = 1418 N
Answer
Q = 1418 N
The force ratio Q/F = 2.836 grows without limit as the links straighten.
8.Frames & machines
Answer: FCD = 78 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 = 2·4 = 8 kN at 2 m
Moments about A for the beam
FCD·0.3846·1.2 = 8·2 + 5·4 = 36 kN·m
FCD = 78 kN
Answer
FCD = 78 kN
9.Shear force & bending moment
Answer: Mx = 43.75 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 = 10·6 = 60 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 10·6·3
RB = 180/6 = 30 kN
ΣFy = 0: RA = 60 − 30 = 30 kN
Cut at x = 2.5 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 = 30·2.5 − 10·2.5·1.25 = 43.75 kN·m
Answer
Mx = 43.75 kN·m
10.Shear force & bending moment
Answer: Mmax = 100.6 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 = 8·8 = 64 kN, acting at x̄ = 4 m
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 = 260·10 = 2600 mm², yf = 280 + 10/2 = 285 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 = (36000·120 − 1200·72)/(36000 − 1200) = 121.7 mm