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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 = (−1, −6, 4) and b = (5, k, 0) are perpendicular.
a = (−1, −6, 4)b = (5, k, 0)
k =
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (4, −1, 1) and b = (4, −6, −3).
a = (4, −1, 1)b = (4, −6, −3)
A =
3.Beam support reactionsMedium
The simply supported beam has a pin at A and a roller at B. It carries a point load P at 1.5 m from A and a counter-clockwise couple M at 4 m from A. Find the vertical reaction RB, taking upward as positive.
L = 5 mP = 55 kN at 1.5 mM = 10 kN·m (counter-clockwise) at 4 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 = 25 kN at 1 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 FAC in member AC. Give tension as positive and compression as negative.
span = 6 mheight = 4 mB is 1 m from Avertical load = 25 kN
FAC = 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 frame shown has pins at A, B and C; A and C are fixed to the ground 8 m apart, and B is 4 m above the ground and 1.2 m to the right of A. A vertical load of 4 kN acts on member AB at K, 0.9 m (horizontally) from A. Find the force FBC in member BC.
P = 4 kNL = 8 mH = 4 mhorizontal distance A to B = 1.2 mhorizontal distance A to K = 0.9 m
FBC = kN
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 150 mm long between pins. The joint B sits 8 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 = 8 mm
F = N
9.Shear force & bending momentMedium
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. It carries a point load P1 between the supports and a point load P2 at C. Find the bending moment Mx under the load P1. Use sagging moments positive and upward shear on the left face positive.
L = 5.5 mP1 = 40 kN at 1 mP2 = 60 kN at 5.5 msupports at x = 0 and x = 4 m
Mx = kN·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 = 5 mP = 80 kN at 4 mw = 6 kN/m from 0 to 5 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A channel section lies with its open side up: a base plate 160 mm wide with two upright legs, 90 mm high overall. Base and legs are all 8 mm thick. Find the height ȳ of the centroid above the bottom edge.
b = 160 mmh = 90 mmt = 8 mm
ȳ = mm
12.Centroid & second moment of areaHard
A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 50 MPa in tension and 90 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 equals zero
a·b = (−1)·5 + (−6)·k + 4·0 = 0
−5 + (−6)·k = 0
Solve for k
k = −(−5)/(−6) = −0.8333
Answer
k = −0.8333
2.Vectors: dot & cross product
Answer: A = 27.15
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)·(−3) − 1·(−6) = 9
(a × b)y = azbx − axbz = 1·4 − 4·(−3) = 16
(a × b)z = axby − aybx = 4·(−6) − (−1)·4 = −20
Area
A = |a × b| = √(9² + 16² + (−20)²) = 27.15
Answer
A = 27.15
3.Beam support reactions
Answer: RB = 14.5 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 − 55·1.5 + 10 = 0
RB = 72.5/5 = 14.5 kN
Vertical equilibrium
RA = 55 − 14.5 = 40.5 kN
Answer
RB = 14.5 kN
A negative reaction acts downward.
4.Beam support reactions
Answer: RB = 10.15 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 = 25·1 + 3·3·8.5
RB = 101.5/10 = 10.15 kN
ΣFy = 0: RA = 34 − 10.15 = 23.85 kN
Answer
RB = 10.15 kN
5.Truss: method of joints
Answer: FAC = 5.208 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 = 25·1 → Cy = 4.167 kN
ΣFy = 0: Ay = 25 − 4.167 = 20.83 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 = 4/6.403 = 0.6247, cos γ = 5/6.403 = 0.7809
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 C
BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
ΣF⊥ = 0 → FCG = 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.
Two-force member BC
BC = √(6.8² + 4²) = 7.889 m
Whole frame: moments about A
The ground pushes on BC at C along CB; its horizontal component passes through A, only the vertical one has a moment.
FBC·(4/7.889)·8 = 4·0.9
FBC = 0.8875 kN (compression)
Answer
FBC = 0.8875 kN
8.Frames & machines
Answer: F = 213.6 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 = 8/150 = 0.05333, θ = 3.057°
Link force
Flink = Q/cos θ = 2003 N
Force at B
F = 2·Flink·sin θ = 2·2000·tan θ = 213.6 N
Answer
F = 213.6 N
9.Shear force & bending moment
Answer: Mx = 7.5 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.
Support reactions
Equilibrium of the whole beam. The overhang load also acts about A.
ΣMA = 0: RB·4 = 40·1 + 60·5.5
RB = 370/4 = 92.5 kN
ΣFy = 0: RA = 100 − 92.5 = 7.5 kN
Cut under P1 and keep the left part
Mx = RA·1 = 7.5·1 = 7.5 kN·m
Answer
Mx = 7.5 kN·m
10.Shear force & bending moment
Answer: Mmax = 76 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 = 6·5 = 30 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 80·4 + 6·5·2.5
RB = 395/5 = 79 kN
ΣFy = 0: RA = 110 − 79 = 31 kN
Where is V = 0?
V(4⁻) = RA − w·4 = 7 kN
V(4⁺) = 7 − 80 = −73 kN
The shear changes sign at the load, so the maximum is under the load.
Maximum moment
Mmax = RA·4 − w·4²/2 = 31·4 − 6·4²/2 = 76 kN·m
Answer
Mmax = 76 kN·m
It occurs at x = 4 m from A.
11.Centroid & second moment of area
Answer: ȳ = 26.78 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.
Split into rectangles
Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.
base: A1 = 160·8 = 1280 mm², y1 = 4 mm
two legs: A2 = 2·8·82 = 1312 mm², y2 = 8 + 82/2 = 49 mm
Centroid
ȳ = (1280·4 + 1312·49)/2592 = 26.78 mm
Answer
ȳ = 26.78 mm
12.Centroid & second moment of area
Answer: Mallow = 9.867 kN·m
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.