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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 value of k for which the vectors a = (2, −2, 5) and b = (−5, k, −5) are perpendicular.
a = (2, −2, 5)b = (−5, k, −5)
k =
2.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (0, 4, −1), b = (−3, −3, 5) and c = (4, 0, 5).
a = (0, 4, −1)b = (−3, −3, 5)c = (4, 0, 5)
V =
3.Beam support reactionsMedium
The beam has a pin at A and a roller at B, 5 m from A, and overhangs B to the free end C. Find the vertical reaction RA, taking upward as positive.
AB = 5 mAC = 8 mP1 = 20 kN at 3 mP2 = 65 kN at 8 m
RA = kN
4.Beam support reactionsHard
A compound beam is fixed to the wall at A, has an internal hinge at C (4 m from A) and rests on a roller at D (8 m from A). It carries a point load P at 6 m from A and a point load Q at 3 m from A. Find the magnitude of the reaction moment MA at the wall.
AC = 4 mAD = 8 mP = 75 kN at 6 mQ = 65 kN at 3 m
MA = kN·m
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 = 8 mheight = 4 mB is 7 m from Avertical load = 35 kN
FAB = 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.
The bell crank shown turns about the pin O. A vertical force of 120 N at A (120 mm from O) is balanced by a horizontal force at B (150 mm above O). Find the magnitude of the pin reaction FO.
F = 120 NOA = 120 mmOB = 150 mm
FO = N
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 200 mm long between pins. The joint B sits 12 mm above the line AC. Which downward force F at B produces a horizontal pressing force of 3000 N at the slider C? Neglect friction and weights.
Q = 3000 NAB = BC = 200 mmh = 12 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.
40 kN @ 4 m12 kN/m, 0–6 m
1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 4 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 6 mP = 10 kN at 4 mw = 4 kN/m from 0 to 6 mx = 4 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A symmetric I-section has flanges 250 mm wide and 22 mm thick, an overall depth of 380 mm and a web 14 mm thick. Find its elastic section modulus Sx for bending about the horizontal centroidal axis x.
B = 250 mmH = 380 mmtf = 22 mmtw = 14 mm
Sx = × 10³ mm³
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 150 mm wide and 20 mm thick, an overall depth of 360 mm and a web 12 mm thick. It carries a bending moment of 80 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
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.
Dot product equals zero
a·b = 2·(−5) + (−2)·k + 5·(−5) = 0
−35 + (−2)·k = 0
Solve for k
k = −(−35)/(−2) = −17.5
Answer
k = −17.5
2.Vectors: dot & cross product
Answer: V = 128
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 b × c
(b × c)x = bycz − bzcy = (−3)·5 − 5·0 = −15
(b × c)y = bzcx − bxcz = 5·4 − (−3)·5 = 35
(b × c)z = bxcy − bycx = (−3)·0 − (−3)·4 = 12
Dot product with a
a·(b × c) = 0·(−15) + 4·35 + (−1)·12 = 128
V = |128| = 128
Answer
V = 128
3.Beam support reactions
Answer: RA = −31 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.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·5 = 20·3 + 65·8
RB = 580/5 = 116 kN
Vertical equilibrium
RA = 20 + 65 − 116 = −31 kN
Answer
RA = −31 kN
RA is negative: the pin at A has to hold the beam down.
4.Beam support reactions
Answer: MA = 345 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·4 = 75·2
RD = 37.5 kN
VC = 75 − 37.5 = 37.5 kN
Part AC
The hinge force VC acts downward on AC (equal and opposite to its action on CD).
RA = 65 + 37.5 = 102.5 kN
MA = 65·3 + 37.5·4 = 345 kN·m
Answer
MA = 345 kN·m
5.Truss: method of joints
Answer: FAB = −8.818 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 = 35·7 → Cy = 30.63 kN
ΣFy = 0: Ay = 35 − 30.63 = 4.375 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 4/8.062 = 0.4961
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −4.375/0.4961 = −8.818 kN
Answer
FAB = −8.818 kN (compression)
6.Truss: method of joints
Answer: member BF
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).
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).
Force at B
Q = 120·120/150 = 96 N
Pin reaction
Ox = 96 N, Oy = 120 N
FO = √(96² + 120²) = 153.7 N
Answer
FO = 153.7 N
8.Frames & machines
Answer: F = 360.6 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 = 12/200 = 0.06, θ = 3.44°
Link force
Flink = Q/cos θ = 3005 N
Force at B
F = 2·Flink·sin θ = 2·3000·tan θ = 360.6 N
Answer
F = 360.6 N
9.Shear force & bending moment
Answer: E (x = 4 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 = 49.33 kN, R2 = 62.67 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 = 43.33, MC = 74.67, MD = 94, ME = 101.3 kN·m, MF = 56.67, MG = 0
Answer
Mark E: ME = 101.3 kN·m
10.Shear force & bending moment
Answer: Mx = 29.33 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 = 4·6 = 24 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 10·4 + 4·6·3
RB = 112/6 = 18.67 kN
ΣFy = 0: RA = 34 − 18.67 = 15.33 kN
Cut at x = 4 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 = 15.33·4 − 4·4·2 = 29.33 kN·m
Answer
Mx = 29.33 kN·m
11.Centroid & second moment of area
Answer: Sx = 2090 × 10³ mm³
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.
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.