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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 = (25, 5, −15) N acts on an object while it moves in a straight line from point A(1, −3, 2) to point B(2, 1, 0). Coordinates are in metres. Find the work W done by the force.
F = (25, 5, −15) NA(1, −3, 2) mB(2, 1, 0) m
W = J
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (3, 1, −2) and b = (3, −3, −4) and has a positive z-component. Find its x-component n̂x.
a = (3, 1, −2)b = (3, −3, −4)n̂z > 0
n̂x =
3.Beam support reactionsMedium
The simply supported beam (pin at A, roller at B) carries a linearly varying load that rises from zero at A to 8 kN/m at B. A point load P also acts at 2 m from A. Find the vertical support reaction RB.
L = 6 mw0 = 8 kN/mP = 55 kN at 2 m
RB = kN
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
L = 10 mP = 50 kN at 5 mw = 3 kN/m from 7 to 10 m
RB = kN
5.Truss: method of jointsMedium
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 5 mheight = 3 mB is 2 m from Avertical load = 45 kN
FBC = kN
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.
D: 30 kN ↓F: 20 kN →G: 10 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
7.Frames & machinesMedium
The A-frame shown stands on a pin at A and a roller at B, 3.2 m apart. The legs meet at the pin C, 4.5 m high, and the horizontal tie DE is 1 m above the ground. A load of 6 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 6 kNAB = 3.2 mheight of C = 4.5 mheight of DE = 1 m
C = kN
8.Frames & machinesHard
The frame shown has pins at A, B and C; A and C are fixed to the ground 3 m apart, and B is 4 m high and 1.8 m to the right of A. Member AB carries a vertical load of 6 kN at K, 1.08 m (horizontally) from A, and a horizontal load of 4 kN to the right acts at B on member AB. Find the size of the horizontal reaction Ax at A.
P = 6 kNQ = 4 kNL = 3 mH = 4 mhorizontal distance A to B = 1.8 mhorizontal distance A to K = 1.08 m
Ax = 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.
50 kN @ 6 m6 kN/m, 4–8 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 simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.
L = 6 mw = 16 kN/m from 1 to 4 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A symmetric I-section has flanges 120 mm wide and 16 mm thick, an overall depth of 460 mm and a web 14 mm thick. Find its second moment of area Ix about the horizontal centroidal axis x.
B = 120 mmH = 460 mmtf = 16 mmtw = 14 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 5 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 = (2, 1, 0) − (1, −3, 2) = (1, 4, −2) m
Work
W = F·d = 25·1 + 5·4 + (−15)·(−2) = 75 J
Answer
W = 75 J
2.Vectors: dot & cross product
Answer: n̂x = 0.5976
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 = 1·(−4) − (−2)·(−3) = −10
(a × b)y = azbx − axbz = (−2)·3 − 3·(−4) = 6
(a × b)z = axby − aybx = 3·(−3) − 1·3 = −12
Normalise
The z-component of a × b is negative, so use −(a × b).
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 triangular load
W = ½·8·6 = 24 kN
x̄ = 2L/3 = 4 m
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·6 = 24·4 + 55·2
RB = 34.33 kN
RA = 79 − 34.33 = 44.67 kN
Answer
RB = 34.33 kN
4.Beam support reactions
Answer: RB = 32.65 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.
UDL resultant: W = w·ℓ = 3·3 = 9 kN, acting at x̄ = 8.5 m
ΣMA = 0: RB·10 = 50·5 + 3·3·8.5
RB = 326.5/10 = 32.65 kN
ΣFy = 0: RA = 59 − 32.65 = 26.35 kN
Answer
RB = 32.65 kN
5.Truss: method of joints
Answer: FBC = −25.46 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·5 = 45·2 → Cy = 18 kN
ΣFy = 0: Ay = 45 − 18 = 27 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 = 3/4.243 = 0.7071, cos γ = 3/4.243 = 0.7071
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 B
AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.
ΣF⊥ = 0 → FBF = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
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 = 3 kN
T = 3·1.6/3.5 = 1.371 kN
Components of the pin force on BC
Cx = T = 1.371 kN, Cy = −By = −3 kN
Magnitude
C = √(1.371² + 3²) = 3.299 kN
Answer
C = 3.299 kN
8.Frames & machines
Answer: Ax = 1.752 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.
Direction of BC
BC = 4.176 m, cos = 0.2873, sin = 0.9578
Whole frame: moments about A
FBC·0.9578·3 = 6·1.08 + 4·4
FBC = 7.823 kN
Horizontal balance
Ax + 4 − 7.823·0.2873 = 0
Ax = −1.752 kN → |Ax| = 1.752 kN
Answer
Ax = 1.752 kN
The minus sign means Ax points to the left.
9.Shear force & bending moment
Answer: D (x = 6 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 = 37 kN, R2 = 37 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 = 74, MC = 148, MD = 210 kN·m, ME = 148, MF = 74, MG = 0
Answer
Mark D: MD = 210 kN·m
10.Shear force & bending moment
Answer: Mmax = 52.5 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. 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.
UDL resultant: W = w·ℓ = 16·3 = 48 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·6 = 16·3·2.5
RB = 120/6 = 20 kN
ΣFy = 0: RA = 48 − 20 = 28 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.
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.
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 = 220·30 = 6600 mm², yf = 260 + 30/2 = 275 mm