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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 dot product a·b of a = (0, 2, 6) and b = (−1, −4, −3).
a = (0, 2, 6)b = (−1, −4, −3)
a·b =
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−4, 2, −3) and b = (2, −2, −3).
a = (−4, 2, −3)b = (2, −2, −3)
A =
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 5 m and a point load P at 6 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 8 mP = 60 kN at 6 mw = 10 kN/m from 0 to 5 m
MA = kN·m
4.Beam support reactionsHard
A compound beam is fixed to the wall at A, has an internal hinge at C (3.5 m from A) and rests on a roller at D (6 m from A). It carries a point load P at 4 m from A and a point load Q at 1.5 m from A. Find the magnitude of the reaction moment MA at the wall.
AC = 3.5 mAD = 6 mP = 70 kN at 4 mQ = 70 kN at 1.5 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.
B: 40 kN ↓F: 20 kN ↓H: 40 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.
B: 15 kN ↓F: 30 kN ↓H: 35 kN ↓H: 30 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
7.Frames & machinesMedium
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 400 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 400 Nθ = 20°
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 2 kN/m over its full length and a point load of 2 kN at its free end B. Find the force FCD in the strut.
w = 2 kN/mP = 2 kNAD = 1.6 mAB = 2.5 mAC = 0.8 m
FCD = kN
9.Shear force & bending momentMedium
The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. It carries a point load P1 between the supports and a point load P2 at C. Find the bending moment MB at support B. Use sagging moments positive and upward shear on the left face positive.
L = 7.5 mP1 = 80 kN at 4 mP2 = 60 kN at 7.5 msupports at x = 0 and x = 6 m
MB = 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 = 5 mP = 35 kN at 1 mw = 5 kN/m from 0 to 5 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 180 mm wide with two upright legs, 100 mm high overall. Base and legs are all 12 mm thick. Find its second moment of area Ix about the horizontal axis x through its centroid.
b = 180 mmh = 100 mmt = 12 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 120 mm × 150 mm plate has a rectangular cut-out 40 mm wide and 40 mm high, centred on the vertical centre line with its centre 45 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 150 mmcut-out: 40 × 40 mmhole centre: y = 45 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: a·b = −26
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.
Negative: the angle between the vectors is more than 90°.
2.Vectors: dot & cross product
Answer: A = 22
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 = 2·(−3) − (−3)·(−2) = −12
(a × b)y = azbx − axbz = (−3)·2 − (−4)·(−3) = −18
(a × b)z = axby − aybx = (−4)·(−2) − 2·2 = 4
Area
A = |a × b| = √((−12)² + (−18)² + 4²) = 22
Answer
A = 22
3.Beam support reactions
Answer: MA = 485 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 distributed load
W = 10·5 = 50 kN, x̄ = 2.5 m
Moment about the wall
MA = P·a + W·x̄ = 60·6 + 50·2.5 = 485 kN·m
Answer
MA = 485 kN·m
4.Beam support reactions
Answer: MA = 301 kN·m
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. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.
Part CD
The hinge has no moment, so moments about C on CD alone must balance.
ΣMC = 0: RD·2.5 = 70·0.5
RD = 14 kN
VC = 70 − 14 = 56 kN
Part AC
The hinge force VC acts downward on AC (equal and opposite to its action on CD).
RA = 70 + 56 = 126 kN
MA = 70·1.5 + 56·3.5 = 301 kN·m
Answer
MA = 301 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 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).
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° = 400
Flink = 584.8 N
Clamping force
Q = Flink·cos 20° = 400/(2·tan 20°) = 549.5 N
Answer
Q = 549.5 N
The force ratio Q/F = 1.374 grows without limit as the links straighten.
8.Frames & machines
Answer: FCD = 15.72 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 = 2·2.5 = 5 kN at 1.25 m
Moments about A for the beam
FCD·0.4472·1.6 = 5·1.25 + 2·2.5 = 11.25 kN·m
FCD = 15.72 kN
Answer
FCD = 15.72 kN
9.Shear force & bending moment
Answer: MB = −90 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. The overhang load also acts about A.
ΣMA = 0: RB·6 = 80·4 + 60·7.5
RB = 770/6 = 128.3 kN
ΣFy = 0: RA = 140 − 128.3 = 11.67 kN
Cut at B and keep the overhang
Only P2 acts on the overhang, with lever arm a. It bends the beam upward over the support (hogging).
MB = −P2·a = −60·1.5 = −90 kN·m
Answer
MB = −90 kN·m
10.Shear force & bending moment
Answer: Mx = 33.13 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 = 5·5 = 25 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 35·1 + 5·5·2.5
RB = 97.5/5 = 19.5 kN
ΣFy = 0: RA = 60 − 19.5 = 40.5 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)
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 = 180·12 = 2160 mm², y1 = 6 mm
two legs: A2 = 2·12·88 = 2112 mm², y2 = 12 + 88/2 = 56 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 = (18000·75 − 1600·45)/(18000 − 1600) = 77.93 mm