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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 = (3, 6, 0) and b = (4, 1, −6).
a = (3, 6, 0)b = (4, 1, −6)
a·b =
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−4, 5, 4) and b = (1, 3, −4).
a = (−4, 5, 4)b = (1, 3, −4)
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 = 4 mP1 = 15 kN at 2 mP2 = 50 kN at 2.5 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 RA.
L = 6 mP = 45 kN at 5 mw = 10 kN/m from 3 to 4 m
RA = 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 = 6 mheight = 4 mB is 2 m from Avertical load = 40 kN
FBC = kN
6.Truss: method of jointsHard
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
span = 4 mheight = 4 mB is 1 m from Avertical load = 10 kNhorizontal load = 15 kN →
FAC = kN
7.Frames & machinesMedium
In the toggle clamp shown, two equal links AB and BC are pinned together at B, and each makes 8° 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 = 8°
Q = N
8.Frames & machinesHard
The A-frame shown stands on a pin at A and a roller at B, 3.2 m apart, with its apex pin C 3.6 m high and the horizontal tie DE 1 m above the ground. A vertical load of 12 kN acts on leg AC at K, 0.96 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 12 kNAB = 3.2 mheight of C = 3.6 mheight of DE = 1 mhorizontal distance A to K = 0.96 m
C = 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.
20 kN @ 4 m45 kN @ 2 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
The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.
L = 4 mP = 30 kN at 2 mw = 10 kN/m from 0 to 4 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A symmetric I-section has flanges 140 mm wide and 20 mm thick, an overall depth of 460 mm and a web 6 mm thick. Find its second moment of area Ix about the horizontal centroidal axis x.
B = 140 mmH = 460 mmtf = 20 mmtw = 6 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 40 kN·m. Find the bending stress σbottom at the bottom 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.
Multiply matching components and add
a·b = 3·4 + 6·1 + 0·(−6) = 12 + 6 + 0 = 18
Answer
a·b = 18
Positive: the angle between the vectors is less than 90°.
2.Vectors: dot & cross product
Answer: A = 38.17
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 = 5·(−4) − 4·3 = −32
(a × b)y = azbx − axbz = 4·1 − (−4)·(−4) = −12
(a × b)z = axby − aybx = (−4)·3 − 5·1 = −17
Area
A = |a × b| = √((−32)² + (−12)² + (−17)²) = 38.17
Answer
A = 38.17
3.Beam support reactions
Answer: RB = 38.75 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.
ΣMA = 0: RB·4 = 15·2 + 50·2.5
RB = 155/4 = 38.75 kN
ΣFy = 0: RA = 65 − 38.75 = 26.25 kN
Answer
RB = 38.75 kN
4.Beam support reactions
Answer: RA = 11.67 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 = 10·1 = 10 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·6 = 45·5 + 10·1·3.5
RB = 260/6 = 43.33 kN
ΣFy = 0: RA = 55 − 43.33 = 11.67 kN
Answer
RA = 11.67 kN
5.Truss: method of joints
Answer: FBC = −18.86 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 = 40·2 → Cy = 13.33 kN
ΣFy = 0: Ay = 40 − 13.33 = 26.67 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 = 4/5.657 = 0.7071, cos γ = 4/5.657 = 0.7071
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·4 = 10·1 + 15·4 → Cy = 17.5 kN
ΣFy = 0: Ay = 10 − 17.5 = −7.5 kN
ΣFx = 0: Ax = −15 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
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 8° = 400
Flink = 1437 N
Clamping force
Q = Flink·cos 8° = 400/(2·tan 8°) = 1423 N
Answer
Q = 1423 N
The force ratio Q/F = 3.558 grows without limit as the links straighten.
8.Frames & machines
Answer: C = 4.227 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.
Support reaction at B
By = 12·0.96/3.2 = 3.6 kN
Tie force
T = 3.6·1.6/2.6 = 2.215 kN
Pin force on BC
Cx = 2.215 kN, Cy = −3.6 kN
C = √(2.215² + 3.6²) = 4.227 kN
Answer
C = 4.227 kN
9.Shear force & bending moment
Answer: C (x = 2 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 = 36.67 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 = 36.67, MC = 73.33 kN·m, MD = 65, ME = 56.67, MF = 28.33, MG = 0
Answer
Mark C: MC = 73.33 kN·m
10.Shear force & bending moment
Answer: Mmax = 50 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 = 10·4 = 40 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 30·2 + 10·4·2
RB = 140/4 = 35 kN
ΣFy = 0: RA = 70 − 35 = 35 kN
Where is V = 0?
V(2⁻) = RA − w·2 = 15 kN
V(2⁺) = 15 − 30 = −15 kN
The shear changes sign at the load, so the maximum is under the load.
Maximum moment
Mmax = RA·2 − w·2²/2 = 35·2 − 10·2²/2 = 50 kN·m
Answer
Mmax = 50 kN·m
It occurs at x = 2 m from A.
11.Centroid & second moment of area
Answer: Ix = 308.3 × 10⁶ mm⁴
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 = 140·15 = 2100 mm², yf = 180 + 15/2 = 187.5 mm