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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, −1, 4) and b = (k, −5, 0) are perpendicular.
a = (2, −1, 4)b = (k, −5, 0)
k =
2.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (1, −1, −1), b = (3, −5, −3) and c = (1, −5, −3).
a = (1, −1, −1)b = (3, −5, −3)c = (1, −5, −3)
V =
3.Beam support reactionsMedium
The simply supported beam (pin at A, roller at B) carries a linearly varying load that falls from 12 kN/m at A to zero at B. Find the vertical support reaction RA.
L = 10 mw0 = 12 kN/m
RA = kN
4.Beam support reactionsHard
The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 20 kN/m at the free end B, plus a point load P at 3.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 5 mw0 = 20 kN/mP = 60 kN at 3.5 m
MA = kN·m
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 = 3 mB is 2 m from Avertical load = 35 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.
B: 25 kN ↓C: 10 kN ↓G: 35 kN ↓H: 40 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
In the toggle clamp shown, two equal links AB and BC are pinned together at B, and each makes 5° with the line AC. A downward force of 250 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 250 Nθ = 5°
Q = N
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 3 m high and 1.8 m to the right of A. Member AB carries a vertical load of 6 kN at K, 0.45 m (horizontally) from A, and a horizontal load of 10 kN to the right acts at B on member AB. Find the size of the horizontal reaction Ax at A.
P = 6 kNQ = 10 kNL = 3 mH = 3 mhorizontal distance A to B = 1.8 mhorizontal distance A to K = 0.45 m
Ax = kN
9.Shear force & bending momentMedium
The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.
L = 6 mP1 = 25 kN at 3 mP2 = 75 kN at 4 m
Mmax = kN·m
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 4 mP = 10 kN at 3 mw = 3 kN/m from 0 to 4 mx = 0.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A 100 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 168 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 100 mmh = 240 mmd = 50 mmhole centre: y = 168 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 40 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 = 2·k + (−1)·(−5) + 4·0 = 0
5 + 2·k = 0
Solve for k
k = −5/2 = −2.5
Answer
k = −2.5
2.Vectors: dot & cross product
Answer: V = 4
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.
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 = ½·12·10 = 60 kN
x̄ = L/3 = 3.333 m
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·10 = 60·3.333
RB = 20 kN
RA = 60 − 20 = 40 kN
Answer
RA = 40 kN
4.Beam support reactions
Answer: MA = 376.7 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 triangular load
W = ½·20·5 = 50 kN
x̄ = 2L/3 = 3.333 m
Moment about the wall
MA = W·x̄ + P·a = 50·3.333 + 60·3.5 = 376.7 kN·m
Answer
MA = 376.7 kN·m
5.Truss: method of joints
Answer: FBC = −19.44 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 = 35·2 → Cy = 11.67 kN
ΣFy = 0: Ay = 35 − 11.67 = 23.33 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
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 D
CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.
ΣF⊥ = 0 → FDH = 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).
Equilibrium of pin B
2·Flink·sin 5° = 250
Flink = 1434 N
Clamping force
Q = Flink·cos 5° = 250/(2·tan 5°) = 1429 N
Answer
Q = 1429 N
The force ratio Q/F = 5.715 grows without limit as the links straighten.
8.Frames & machines
Answer: Ax = 5.64 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 = 3.231 m, cos = 0.3714, sin = 0.9285
Whole frame: moments about A
FBC·0.9285·3 = 6·0.45 + 10·3
FBC = 11.74 kN
Horizontal balance
Ax + 10 − 11.74·0.3714 = 0
Ax = −5.64 kN → |Ax| = 5.64 kN
Answer
Ax = 5.64 kN
The minus sign means Ax points to the left.
9.Shear force & bending moment
Answer: Mmax = 125 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.
ΣMA = 0: RB·6 = 25·3 + 75·4
RB = 375/6 = 62.5 kN
ΣFy = 0: RA = 100 − 62.5 = 37.5 kN
Moment under each load
Keep the part left of each load.
under P1: M = RA·3 = 37.5·3 = 112.5 kN·m
under P2: M = RB·2 = 62.5·2 = 125 kN·m
Largest value
Mmax = 125 kN·m (under P2)
Answer
Mmax = 125 kN·m
10.Shear force & bending moment
Answer: Mx = 3.875 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 = 3·4 = 12 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 10·3 + 3·4·2
RB = 54/4 = 13.5 kN
ΣFy = 0: RA = 22 − 13.5 = 8.5 kN
Cut at x = 0.5 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 = 8.5·0.5 − 3·0.5·0.25 = 3.875 kN·m
Answer
Mx = 3.875 kN·m
11.Centroid & second moment of area
Answer: Ix = 110 × 10⁶ mm⁴
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Centroid
ȳ = (24000·120 − 1963·168)/(24000 − 1963) = 115.7 mm
Full plate about ȳ
100·240³/12 + 24000·4.277² = 115.6 × 10⁶ mm⁴
Hole about ȳ
π·50⁴/64 + 1963·52.28² = 5.673 × 10⁶ mm⁴
Subtract
Ix = 110 × 10⁶ mm⁴
Answer
Ix = 110 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Mallow = 5.801 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.