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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, −4, 2) onto b = (4, 3, −5), that is, the component of a in the direction of b.
a = (−4, −4, 2)b = (4, 3, −5)
compb a =
2.Vectors: dot & cross productHard
A force F = (60, −10, 50) N acts at point A(2, 4, 3). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(0, 1, −3).
F = (60, −10, 50) NA(2, 4, 3) mO(0, 1, −3) m
Mz = 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 4.5 m and a point load P at 5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 6 mP = 55 kN at 5 mw = 8 kN/m from 0 to 4.5 m
MA = kN·m
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. A load P acts at 2.5 m from A, inclined at 45° to the horizontal, and a vertical load Q acts at 0.5 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.
L = 4 mP = 70 kN at 2.5 m, θ = 45°Q = 80 kN at 0.5 m
α = °
5.Truss: method of jointsMedium
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
span = 4 mheight = 3 mB is 1 m from Avertical load = 20 kN
FAC = kN
6.Truss: method of jointsHard
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 6 mheight = 2 mB is 5 m from Avertical load = 15 kNhorizontal load = 20 kN →
FBC = kN
7.Frames & machinesMedium
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 8 mm above the line AC. Which downward force F at B produces a horizontal pressing force of 2000 N at the slider C? Neglect friction and weights.
Q = 2000 NAB = BC = 100 mmh = 8 mm
F = N
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 5 mw = 12 kN/m from 0 to 5 mx = 3.5 m
Mx = kN·m
10.Shear force & bending momentHard
The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment Mmax between the supports.
L = 6 mw = 2 kN/m from 0 to 6 msupports at x = 0 and x = 5 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A 100 mm × 240 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 144 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 240 mmd = 40 mmhole centre: y = 144 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 30 MPa in tension and 120 MPa in compression. Find the largest sagging bending moment Mallow the beam can carry.
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 + (−4)·3 + 2·(−5) = −38
Length of b
|b| = √(4² + 3² + (−5)²) = 7.071
Scalar projection
compb a = a·b/|b| = −38/7.071 = −5.374
Answer
compb a = −5.374
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: Mz = −200 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 = 8·4.5 = 36 kN, x̄ = 2.25 m
Moment about the wall
MA = P·a + W·x̄ = 55·5 + 36·2.25 = 356 kN·m
Answer
MA = 356 kN·m
4.Beam support reactions
Answer: α = 60.8 °
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 = 70·sin 45° = 49.5 kN
Ph = 70·cos 45° = 49.5 kN
Reactions
ΣMA = 0: RB·4 = 49.5·2.5 + 80·0.5 → RB = 40.94 kN
Ay = 49.5 + 80 − 40.94 = 88.56 kN
Ax = Ph = 49.5 kN
Direction at A
α = atan(Ay/Ax) = atan(88.56/49.5) = 60.8°
Answer
α = 60.8°
5.Truss: method of joints
Answer: FAC = 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·4 = 20·1 → Cy = 5 kN
ΣFy = 0: Ay = 20 − 5 = 15 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 = 3/4.243 = 0.7071, cos γ = 3/4.243 = 0.7071
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 = 15·5 + 20·2 → Cy = 19.17 kN
ΣFy = 0: Ay = 15 − 19.17 = −4.167 kN
ΣFx = 0: Ax = −20 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
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 = 321 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 = 8/100 = 0.08, θ = 4.589°
Link force
Flink = Q/cos θ = 2006 N
Force at B
F = 2·Flink·sin θ = 2·2000·tan θ = 321 N
Answer
F = 321 N
9.Shear force & bending moment
Answer: Mx = 31.5 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.
Support reactions
Equilibrium of the whole beam.
UDL resultant: W = w·ℓ = 12·5 = 60 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 12·5·2.5
RB = 150/5 = 30 kN
ΣFy = 0: RA = 60 − 30 = 30 kN
Cut at x = 3.5 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 = 30·3.5 − 12·3.5·1.75 = 31.5 kN·m
Answer
Mx = 31.5 kN·m
10.Shear force & bending moment
Answer: Mmax = 5.76 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, including the load on the overhang.
UDL resultant: W = w·ℓ = 2·6 = 12 kN, acting at x̄ = 3 m
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 = (24000·120 − 1257·144)/(24000 − 1257) = 118.7 mm
Full plate about ȳ
100·240³/12 + 24000·1.326² = 115.2 × 10⁶ mm⁴
Hole about ȳ
π·40⁴/64 + 1257·25.33² = 0.9317 × 10⁶ mm⁴
Subtract
Ix = 114.3 × 10⁶ mm⁴
Answer
Ix = 114.3 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Mallow = 3.381 kN·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.