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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 = (−4, −5, −2) and b = (k, 6, 0) are perpendicular.
a = (−4, −5, −2)b = (k, 6, 0)
k =
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−4, 4, 4) and b = (1, −4, 5).
a = (−4, 4, 4)b = (1, −4, 5)
A =
3.Beam support reactionsMedium
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
L = 5 mP1 = 15 kN at 1.5 mP2 = 55 kN at 2.5 m
RB = kN
4.Beam support reactionsHard
The beam has a pin at A and a roller at B, 3 m from A, and overhangs B to the free end C. It carries a point load P at 1.5 m from A and a uniformly distributed load w from 2.5 m to the free end at 5 m. Find the vertical reaction RB, taking upward as positive.
AB = 3 mAC = 5 mP = 70 kN at 1.5 mw = 6 kN/m from 2.5 to 5 m
RB = kN
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.
D: 10 kN ↓F: 15 kN ↓G: 35 kN ↓H: 35 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
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 6 mheight = 4 mB is 1 m from Avertical load = 10 kNhorizontal load = 15 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 25° with the line AC. A downward force of 300 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 300 Nθ = 25°
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 0.8 m below A). The beam carries a uniform load of 5 kN/m over its full length and a point load of 6 kN at its free end B. Find the force FCD in the strut.
w = 5 kN/mP = 6 kNAD = 2 mAB = 4 mAC = 0.8 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.
55 kN @ 4 m50 kN @ 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)
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.
25 kN @ 0 m40 kN @ 6 m30 kN @ 2 m4 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 150 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 168 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 240 mmd = 50 mmhole centre: y = 168 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 equals zero
a·b = (−4)·k + (−5)·6 + (−2)·0 = 0
−30 + (−4)·k = 0
Solve for k
k = −(−30)/(−4) = −7.5
Answer
k = −7.5
2.Vectors: dot & cross product
Answer: A = 44.9
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 = 4·5 − 4·(−4) = 36
(a × b)y = azbx − axbz = 4·1 − (−4)·5 = 24
(a × b)z = axby − aybx = (−4)·(−4) − 4·1 = 12
Area
A = |a × b| = √(36² + 24² + 12²) = 44.9
Answer
A = 44.9
3.Beam support reactions
Answer: RB = 32 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·5 = 15·1.5 + 55·2.5
RB = 160/5 = 32 kN
ΣFy = 0: RA = 70 − 32 = 38 kN
Answer
RB = 32 kN
4.Beam support reactions
Answer: RB = 53.75 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.
Resultant of the distributed load
W = 6·2.5 = 15 kN, x̄ = (2.5 + 5)/2 = 3.75 m
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·3 = 70·1.5 + 15·3.75
RB = 161.3/3 = 53.75 kN
RA = 85 − 53.75 = 31.25 kN
Answer
RB = 53.75 kN
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).
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 = 10·1 − 15·4 → Cy = −8.333 kN
ΣFy = 0: Ay = 10 − −8.333 = 18.33 kN
ΣFx = 0: Ax = 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 = 4/6.403 = 0.6247, cos γ = 5/6.403 = 0.7809
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 25° = 300
Flink = 354.9 N
Clamping force
Q = Flink·cos 25° = 300/(2·tan 25°) = 321.7 N
Answer
Q = 321.7 N
The force ratio Q/F = 1.072 grows without limit as the links straighten.
8.Frames & machines
Answer: FCD = 86.16 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 = 5·4 = 20 kN at 2 m
Moments about A for the beam
FCD·0.3714·2 = 20·2 + 6·4 = 64 kN·m
FCD = 86.16 kN
Answer
FCD = 86.16 kN
9.Shear force & bending moment
Answer: C (x = 4 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 = 45 kN, R2 = 60 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 = 90, MC = 180 kN·m, MD = 160, ME = 140, MF = 120, MG = 0
Answer
Mark C: MC = 180 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 = 51.75 kN, R2 = 59.25 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 = −25, MC = −0.25, MD = −9.5, ME = −22.75, MF = −40 kN·m, MG = 0
Answer
Mark F: MF = −40 kN·m
11.Centroid & second moment of area
Answer: Ix = 167.7 × 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 = (36000·120 − 1963·168)/(36000 − 1963) = 117.2 mm
Full plate about ȳ
150·240³/12 + 36000·2.769² = 173.1 × 10⁶ mm⁴
Hole about ȳ
π·50⁴/64 + 1963·50.77² = 5.368 × 10⁶ mm⁴
Subtract
Ix = 167.7 × 10⁶ mm⁴
Answer
Ix = 167.7 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Mallow = 6.335 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.