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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
A constant force F = (5, 10, 5) N acts on an object while it moves in a straight line from point A(3, 2, 0) to point B(2, 3, −2). Coordinates are in metres. Find the work W done by the force.
F = (5, 10, 5) NA(3, 2, 0) mB(2, 3, −2) m
W = J
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (2, 1, −4) and b = (−4, 4, −1) and has a positive z-component. Find its x-component n̂x.
a = (2, 1, −4)b = (−4, 4, −1)n̂z > 0
n̂x =
3.Beam support reactionsMedium
The beam has a pin at A and a roller at B, 8 m from A, and overhangs B to the free end C. Find the vertical reaction RB, taking upward as positive.
AB = 8 mAC = 10 mP1 = 55 kN at 3 mP2 = 15 kN at 10 m
RB = kN
4.Beam support reactionsHard
The beam has a pin at A and a roller at B, 4 m from A, and overhangs B to the free end C. It carries a point load P at 2.5 m from A and a uniformly distributed load w from 3 m to the free end at 6 m. Find the vertical reaction RA, taking upward as positive.
AB = 4 mAC = 6 mP = 20 kN at 2.5 mw = 4 kN/m from 3 to 6 m
RA = 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.
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 8 mheight = 4 mB is 5 m from Avertical load = 60 kNhorizontal load = 10 kN ←
FBC = kN
7.Frames & machinesMedium
A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 0.8 m below A. A load of 3 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.
P = 3 kNAD = 1 mAB = 3 mAC = 0.8 m
A = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 0.5 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 2 m above A. A load of 5 kN hangs at B. Find the tension T in the cable.
W = 5 kNWb = 0.5 kNAB = 6 mAC = 2 m
T = kN
9.Shear force & bending momentMedium
The simply supported beam carries a clockwise couple M0 = 20 kN·m at x = 1 m from A, and no other load. Find the bending moment Mx just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.
L = 6 mM0 = 20 kN·m clockwise, at 1 m
Mx = kN·m
10.Shear force & bending momentHard
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 = 8 mP = 40 kN at 5 mw = 3 kN/m from 0 to 8 mx = 2.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A channel section lies with its open side up: a base plate 120 mm wide with two upright legs, 140 mm high overall. Base and legs are all 12 mm thick. Find the height ȳ of the centroid above the bottom edge.
b = 120 mmh = 140 mmt = 12 mm
ȳ = mm
12.Centroid & second moment of areaHard
A 120 mm × 150 mm plate has a rectangular cut-out 30 mm wide and 40 mm high, centred on the vertical centre line with its centre 52.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 150 mmcut-out: 30 × 40 mmhole centre: y = 52.5 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: W = −5 J
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.
Displacement
d = B − A = (2, 3, −2) − (3, 2, 0) = (−1, 1, −2) m
Work
W = F·d = 5·(−1) + 10·1 + 5·(−2) = −5 J
Answer
W = −5 J
Negative work: the force acts mostly against the motion.
2.Vectors: dot & cross product
Answer: n̂x = 0.5698
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.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·8 = 55·3 + 15·10
RB = 315/8 = 39.38 kN
Vertical equilibrium
RA = 55 + 15 − 39.38 = 30.63 kN
Answer
RB = 39.38 kN
4.Beam support reactions
Answer: RA = 6 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 = 4·3 = 12 kN, x̄ = (3 + 6)/2 = 4.5 m
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·4 = 20·2.5 + 12·4.5
RB = 104/4 = 26 kN
RA = 32 − 26 = 6 kN
Answer
RA = 6 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 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).
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 = 60·5 − 10·4 → Cy = 32.5 kN
ΣFy = 0: Ay = 60 − 32.5 = 27.5 kN
ΣFx = 0: Ax = 10 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
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.
Strut force
FCD = 3·3/(0.6247·1) = 14.41 kN
Components at A
Ax = FCD·cos α = 11.25 kN (towards the wall)
Ay = 3 − FCD·sin α = −6 kN
Magnitude
A = √(11.25² + −6²) = 12.75 kN
Answer
A = 12.75 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: T = 16.6 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.
Cable angle
BC = √(6² + 2²) = 6.325 m, sin β = 0.3162
Moments about A
T·0.3162·6 = 5·6 + 0.5·3
T = 16.6 kN
Answer
T = 16.6 kN
9.Shear force & bending moment
Answer: Mx = 16.67 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. An applied couple makes the bending moment diagram jump by the size of the couple.
Support reactions
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
RB = M0/L = 20/6 = 3.333 kN (upward)
RA = −RB = −3.333 kN (downward)
Cut just right of the couple
The left part now contains RA and the couple.
Mx = RA·a + M0 = −3.333·1 + 20 = 16.67 kN·m
Answer
Mx = 16.67 kN·m
10.Shear force & bending moment
Answer: Mx = 58.13 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 = 3·8 = 24 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 40·5 + 3·8·4
RB = 296/8 = 37 kN
ΣFy = 0: RA = 64 − 37 = 27 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 = 27·2.5 − 3·2.5·1.25 = 58.13 kN·m
Answer
Mx = 58.13 kN·m
11.Centroid & second moment of area
Answer: ȳ = 53.66 mm
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 rectangles
Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.
base: A1 = 120·12 = 1440 mm², y1 = 6 mm
two legs: A2 = 2·12·128 = 3072 mm², y2 = 12 + 128/2 = 76 mm
Centroid
ȳ = (1440·6 + 3072·76)/4512 = 53.66 mm
Answer
ȳ = 53.66 mm
12.Centroid & second moment of area
Answer: Ix = 32.94 × 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 = (18000·75 − 1200·52.5)/(18000 − 1200) = 76.61 mm