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, −2, −3) and b = (−6, 0, k) are perpendicular.
a = (−1, −2, −3)b = (−6, 0, k)
k =
2.Vectors: dot & cross productHard
A force F = (30, −30, −50) N acts at point A(−2, 3, 2). Coordinates are in metres. Find the magnitude |M| of the moment of the force about the origin O(0, 0, 0).
F = (30, −30, −50) NA(−2, 3, 2) mO(0, 0, 0) m
|M| = N·m
3.Beam support reactionsMedium
The simply supported beam (pin at A, roller at B) carries a linearly varying load that falls from 6 kN/m at A to zero at B. A point load P also acts at 1 m from A. Find the vertical support reaction RB.
L = 4 mw0 = 6 kN/mP = 30 kN at 1 m
RB = kN
4.Beam support reactionsHard
The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 15 kN/m at the free end B, plus a point load P at 2.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 4 mw0 = 15 kN/mP = 70 kN at 2.5 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.
F: 35 kN ↓G: 10 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 20° 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 = 20°
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 25 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 = 25 mm
F = N
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 6 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 8 mw = 4 kN/m from 0 to 8 mx = 6 m
Mx = kN·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.
30 kN @ 0 m35 kN @ 6 m15 kN/m, 1–5 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)
11.Centroid & second moment of areaMedium
A 100 mm × 240 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 84 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 240 mmd = 30 mmhole centre: y = 84 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 150 mm × 300 mm plate has a rectangular cut-out 40 mm wide and 30 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 300 mmcut-out: 40 × 30 mmhole centre: y = 120 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: k = 2
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)·(−6) + (−2)·0 + (−3)·k = 0
6 + (−3)·k = 0
Solve for k
k = −6/(−3) = 2
Answer
k = 2
2.Vectors: dot & cross product
Answer: |M| = 103 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.
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 = ½·6·4 = 12 kN
x̄ = L/3 = 1.333 m
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·4 = 12·1.333 + 30·1
RB = 11.5 kN
RA = 42 − 11.5 = 30.5 kN
Answer
RB = 11.5 kN
4.Beam support reactions
Answer: MA = 255 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 = ½·15·4 = 30 kN
x̄ = 2L/3 = 2.667 m
Moment about the wall
MA = W·x̄ + P·a = 30·2.667 + 70·2.5 = 255 kN·m
Answer
MA = 255 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 D
CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.
ΣF⊥ = 0 → FDH = 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 20° = 150
Flink = 219.3 N
Clamping force
Q = Flink·cos 20° = 150/(2·tan 20°) = 206.1 N
Answer
Q = 206.1 N
The force ratio Q/F = 1.374 grows without limit as the links straighten.
8.Frames & machines
Answer: F = 1033 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 = 25/100 = 0.25, θ = 14.48°
Link force
Flink = Q/cos θ = 2066 N
Force at B
F = 2·Flink·sin θ = 2·2000·tan θ = 1033 N
Answer
F = 1033 N
9.Shear force & bending moment
Answer: Mx = 24 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 = 4·8 = 32 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 4·8·4
RB = 128/8 = 16 kN
ΣFy = 0: RA = 32 − 16 = 16 kN
Cut at x = 6 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 = 16·6 − 4·6·3 = 24 kN·m
Answer
Mx = 24 kN·m
10.Shear force & bending moment
Answer: F (x = 5 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 = 58.75 kN, R2 = 66.25 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 = −30, MC = −8.75, MD = −2.5, ME = −11.25, MF = −35 kN·m, MG = 0
Answer
Mark F: MF = −35 kN·m
11.Centroid & second moment of area
Answer: Ix = 114.2 × 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 = (24000·120 − 706.9·84)/(24000 − 706.9) = 121.1 mm
Full plate about ȳ
100·240³/12 + 24000·−1.092² = 115.2 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·−37.09² = 1.012 × 10⁶ mm⁴
Subtract
Ix = 114.2 × 10⁶ mm⁴
Answer
Ix = 114.2 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Ix = 336.3 × 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 = (45000·150 − 1200·120)/(45000 − 1200) = 150.8 mm