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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, −15, −5) N acts on an object while it moves in a straight line from point A(−3, 2, 0) to point B(−1, −3, −2). Coordinates are in metres. Find the work W done by the force.
F = (5, −15, −5) NA(−3, 2, 0) mB(−1, −3, −2) m
W = J
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−6, −3, −6) and b = (−6, −1, 0).
a = (−6, −3, −6)b = (−6, −1, 0)
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 = 10 mP1 = 15 kN at 7 mP2 = 60 kN at 9 m
RB = kN
4.Beam support reactionsHard
A compound beam is fixed to the wall at A, has an internal hinge at C (2.5 m from A) and rests on a roller at D (6 m from A). It carries a point load P at 5.5 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 1 m from A. Find the reaction RD at the roller.
AC = 2.5 mAD = 6 mP = 60 kN at 5.5 mw = 3 kN/m from 2.5 to 6 mQ = 50 kN at 1 m
RD = kN
5.Truss: method of jointsMedium
Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.
span = 6 mheight = 2 mB is 5 m from Avertical load = 60 kN
FAB = kN
6.Truss: method of jointsHard
Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.
span = 8 mheight = 3 mB is 3 m from Avertical load = 30 kNhorizontal load = 15 kN ←
FAB = kN
7.Frames & machinesMedium
The A-frame shown stands on a pin at A and a roller at B, 5 m apart. The legs meet at the pin C, 4 m high, and the horizontal tie DE is 1.2 m above the ground. A load of 4 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 4 kNAB = 5 mheight of C = 4 mheight of DE = 1.2 m
C = kN
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 120 mm long between pins. The joint B sits 30 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 = 120 mmh = 30 mm
F = N
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.
30 kN @ 8 m10 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 beam rests on a pin at A and a roller at B and overhangs B by 1.5 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 = 5.5 mw = 2 kN/m from 0 to 5.5 msupports at x = 0 and x = 4 m
xc = m
11.Centroid & second moment of areaMedium
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 240 mmtf = 20 mmtw = 18 mmhw = 220 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 160 mm wide and 24 mm thick, an overall depth of 320 mm and a web 10 mm thick. It carries a bending moment of 120 kN·m about the axis x. Find the size of the maximum bending stress σmax in the section.
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 = (−1, −3, −2) − (−3, 2, 0) = (2, −5, −2) m
Work
W = F·d = 5·2 + (−15)·(−5) + (−5)·(−2) = 95 J
Answer
W = 95 J
2.Vectors: dot & cross product
Answer: A = 38.42
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. 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·10 = 15·7 + 60·9
RB = 645/10 = 64.5 kN
ΣFy = 0: RA = 75 − 64.5 = 10.5 kN
Answer
RB = 64.5 kN
4.Beam support reactions
Answer: RD = 56.68 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. 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.
W = 3·3.5 = 10.5 kN at 1.75 m from C
ΣMC = 0: RD·3.5 = 60·3 + 10.5·1.75
RD = 56.68 kN
VC = 70.5 − 56.68 = 13.82 kN
Answer
RD = 56.68 kN
5.Truss: method of joints
Answer: FAB = −26.93 kN
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 = 60·5 → Cy = 50 kN
ΣFy = 0: Ay = 60 − 50 = 10 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 2/5.385 = 0.3714
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −10/0.3714 = −26.93 kN
Answer
FAB = −26.93 kN (compression)
6.Truss: method of joints
Answer: FAB = −34.47 kN
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 = 30·3 − 15·3 → Cy = 5.625 kN
ΣFy = 0: Ay = 30 − 5.625 = 24.38 kN
ΣFx = 0: Ax = 15 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 3/4.243 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −24.38/0.7071 = −34.47 kN
Answer
FAB = −34.47 kN (compression)
7.Frames & machines
Answer: C = 2.681 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.
Tie force from leg BC
By = 2 kN
T = 2·2.5/2.8 = 1.786 kN
Components of the pin force on BC
Cx = T = 1.786 kN, Cy = −By = −2 kN
Magnitude
C = √(1.786² + 2²) = 2.681 kN
Answer
C = 2.681 kN
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 = 30/120 = 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: E (x = 8 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 = 11.67 kN, R2 = 28.33 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 = 23.33, MC = 46.67, MD = 70, ME = 93.33 kN·m, MF = 56.67, MG = 0
Answer
Mark E: ME = 93.33 kN·m
10.Shear force & bending moment
Answer: xc = 3.438 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 = 2·5.5 = 11 kN, acting at x̄ = 2.75 m
ΣMA = 0: RB·4 = 2·5.5·2.75
RB = 30.25/4 = 7.563 kN
ΣFy = 0: RA = 11 − 7.563 = 3.438 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 = 2·3.438/2 = 3.438 m
Answer
xc = 3.438 m
11.Centroid & second moment of area
Answer: Ix = 47.38 × 10⁶ mm⁴
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 = 240·20 = 4800 mm², yf = 220 + 20/2 = 230 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.
Outer rectangle minus the two side cut-outs
Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.