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
A constant force F = (20, −30, −5) N acts on an object while it moves in a straight line from point A(−3, −5, 5) to point B(−5, −3, −2). Coordinates are in metres. Find the work W done by the force.
F = (20, −30, −5) NA(−3, −5, 5) mB(−5, −3, −2) m
W = J
2.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(4, −3, 2), B(4, −4, −4) and C(−4, 1, −1). Coordinates are in metres.
A(4, −3, 2)B(4, −4, −4)C(−4, 1, −1)
A = m²
3.Beam support reactionsMedium
The simply supported beam has a pin at A and a roller at B. It carries a point load P at 2.5 m from A and a clockwise couple M at 4.5 m from A. Find the vertical reaction RA, taking upward as positive.
L = 5 mP = 50 kN at 2.5 mM = 25 kN·m (clockwise) at 4.5 m
RA = kN
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. A load P acts at 7 m from A, inclined at 30° to the horizontal, and a vertical load Q acts at 2 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.
L = 8 mP = 70 kN at 7 m, θ = 30°Q = 50 kN at 2 m
α = °
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: 30 kN ↓C: 20 kN ↓D: 25 kN ↓F: 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 FC
13member HC
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 12° with the line AC. A downward force of 200 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 200 Nθ = 12°
Q = N
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 150 mm long between pins. The joint B sits 10 mm above the line AC. Which downward force F at B produces a horizontal pressing force of 4000 N at the slider C? Neglect friction and weights.
Q = 4000 NAB = BC = 150 mmh = 10 mm
F = N
9.Shear force & bending momentMedium
Find the shear force Vx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 10 mw = 3 kN/m from 0 to 10 mx = 1.5 m
Vx = kN
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 distance xc from A to the point of contraflexure between the supports (where the bending moment changes sign).
L = 5 mw = 10 kN/m from 0 to 5 msupports at x = 0 and x = 4 m
xc = m
11.Centroid & second moment of areaMedium
A 120 mm × 150 mm plate has a circular hole of diameter 30 mm on its 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 mmd = 30 mmhole centre: y = 45 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 30 kN·m. Find the size of the bending stress σtop at the top edge.
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 = (−5, −3, −2) − (−3, −5, 5) = (−2, 2, −7) m
Work
W = F·d = 20·(−2) + (−30)·2 + (−5)·(−7) = −65 J
Answer
W = −65 J
Negative work: the force acts mostly against the motion.
2.Vectors: dot & cross product
Answer: A = 27.83 m²
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·5 − 50·2.5 − 25 = 0
RB = 150/5 = 30 kN
Vertical equilibrium
RA = 50 − 30 = 20 kN
Answer
RA = 20 kN
A negative reaction acts downward.
4.Beam support reactions
Answer: α = 34.64 °
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 30° = 35 kN
Ph = 70·cos 30° = 60.62 kN
Reactions
ΣMA = 0: RB·8 = 35·7 + 50·2 → RB = 43.12 kN
Ay = 35 + 50 − 43.12 = 41.88 kN
Ax = Ph = 60.62 kN
Direction at A
α = atan(Ay/Ax) = atan(41.88/60.62) = 34.64°
Answer
α = 34.64°
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).
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 12° = 200
Flink = 481 N
Clamping force
Q = Flink·cos 12° = 200/(2·tan 12°) = 470.5 N
Answer
Q = 470.5 N
The force ratio Q/F = 2.352 grows without limit as the links straighten.
8.Frames & machines
Answer: F = 534.5 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 = 10/150 = 0.06667, θ = 3.823°
Link force
Flink = Q/cos θ = 4009 N
Force at B
F = 2·Flink·sin θ = 2·4000·tan θ = 534.5 N
Answer
F = 534.5 N
9.Shear force & bending moment
Answer: Vx = 10.5 kN
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 = 3·10 = 30 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 3·10·5
RB = 150/10 = 15 kN
ΣFy = 0: RA = 30 − 15 = 15 kN
Cut at x = 1.5 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.
V = RA − (all downward loads left of the cut)
Vx = 15 − 3·1.5 = 10.5 kN
Answer
Vx = 10.5 kN
10.Shear force & bending moment
Answer: xc = 3.75 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 = 10·5 = 50 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·4 = 10·5·2.5
RB = 125/4 = 31.25 kN
ΣFy = 0: RA = 50 − 31.25 = 18.75 kN
Zero moment
Between the supports M(x) = RA·x − w·x²/2. Apart from x = 0 it is zero at x = 2RA/w.
RA·xc − w·xc²/2 = 0
xc = 2·18.75/10 = 3.75 m
Answer
xc = 3.75 m
11.Centroid & second moment of area
Answer: Ix = 33.05 × 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 − 706.9·45)/(18000 − 706.9) = 76.23 mm
Full plate about ȳ
120·150³/12 + 18000·−1.226² = 33.78 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·−31.23² = 0.729 × 10⁶ mm⁴
Subtract
Ix = 33.05 × 10⁶ mm⁴
Answer
Ix = 33.05 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: σtop = 28.41 MPa
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 two rectangles
Measure every y from the bottom edge.
flange: Af = 200·25 = 5000 mm², yf = 280 + 25/2 = 292.5 mm