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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 scalar projection compb a of a = (−7, −4, −1) onto b = (−2, −5, −1), that is, the component of a in the direction of b.
a = (−7, −4, −1)b = (−2, −5, −1)
compb a =
2.Vectors: dot & cross productHard
For a = (−1, −5, 2) and b = (−2, 2, 6), find the y-component (a × b)y of the cross product a × b.
a = (−1, −5, 2)b = (−2, 2, 6)
(a × b)y =
3.Beam support reactionsMedium
The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w along its whole length and a point load P at 1.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 5 mP = 75 kN at 1.5 mw = 8 kN/m from 0 to 5 m
MA = kN·m
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 = 5 kN/m from 0 to 5 m
RA = kN
5.Truss: method of jointsMedium
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
span = 5 mheight = 3 mB is 1 m from Avertical load = 55 kN
FAC = 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 = 4 mheight = 4 mB is 1 m from Avertical load = 35 kNhorizontal load = 15 kN →
FAB = kN
7.Frames & machinesMedium
The bell crank shown turns about the pin O. A vertical force of 250 N at A (150 mm from O) is balanced by a horizontal force at B (100 mm above O). Find the magnitude of the pin reaction FO.
F = 250 NOA = 150 mmOB = 100 mm
FO = N
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 150 mm long between pins. The joint B sits 5 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 = 150 mmh = 5 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.
55 kN @ 2 m40 kN @ 4 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 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.
55 kN @ 0 m50 kN @ 6 m8 kN/m, 1–4 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)
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 = 25 mmtw = 14 mmhw = 300 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 35 MPa in tension and 100 MPa in compression. Find the largest sagging bending moment Mallow the beam can carry.
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
a·b = (−7)·(−2) + (−4)·(−5) + (−1)·(−1) = 35
Length of b
|b| = √((−2)² + (−5)² + (−1)²) = 5.477
Scalar projection
compb a = a·b/|b| = 35/5.477 = 6.39
Answer
compb a = 6.39
2.Vectors: dot & cross product
Answer: (a × b)y = 2
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, component by component
(a × b)x = aybz − azby = (−5)·6 − 2·2 = −34
(a × b)y = azbx − axbz = 2·(−2) − (−1)·6 = 2
(a × b)z = axby − aybx = (−1)·2 − (−5)·(−2) = −12
Requested component
(a × b)y = 2
Answer
(a × b)y = 2
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
3.Beam support reactions
Answer: MA = 212.5 kN·m
Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction RA and a reaction moment MA. ΣFy = 0 gives RA, and moments about the wall give MA = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.
Resultant of the distributed load
W = 8·5 = 40 kN, x̄ = 2.5 m
Moment about the wall
MA = P·a + W·x̄ = 75·1.5 + 40·2.5 = 212.5 kN·m
Answer
MA = 212.5 kN·m
4.Beam support reactions
Answer: RA = 22.08 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 = 5·5 = 25 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·6 = 45·5 + 5·5·2.5
RB = 287.5/6 = 47.92 kN
ΣFy = 0: RA = 70 − 47.92 = 22.08 kN
Answer
RA = 22.08 kN
5.Truss: method of joints
Answer: FAC = 14.67 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 = 55·1 → Cy = 11 kN
ΣFy = 0: Ay = 55 − 11 = 44 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
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 = 35·1 + 15·4 → Cy = 23.75 kN
ΣFy = 0: Ay = 35 − 23.75 = 11.25 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 = 4/4.123 = 0.9701
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −11.25/0.9701 = −11.6 kN
Answer
FAB = −11.6 kN (compression)
7.Frames & machines
Answer: FO = 450.7 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).
Force at B
Q = 250·150/100 = 375 N
Pin reaction
Ox = 375 N, Oy = 250 N
FO = √(375² + 250²) = 450.7 N
Answer
FO = 450.7 N
8.Frames & machines
Answer: F = 133.4 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 = 5/150 = 0.03333, θ = 1.91°
Link force
Flink = Q/cos θ = 2001 N
Force at B
F = 2·Flink·sin θ = 2·2000·tan θ = 133.4 N
Answer
F = 133.4 N
9.Shear force & bending moment
Answer: C (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 = 72.5 kN, R2 = 22.5 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 = 145, MC = 180 kN·m, MD = 135, ME = 90, MF = 45, MG = 0
Answer
Mark C: MC = 180 kN·m
10.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 = 52 kN, R2 = 77 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 = −55, MC = −62, MD = −77, ME = −100 kN·m, MF = −50, MG = 0
Answer
Mark E: ME = −100 kN·m
11.Centroid & second moment of area
Answer: Ix = 97.05 × 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 = 240·25 = 6000 mm², yf = 300 + 25/2 = 312.5 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.