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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 angle θ between the vectors a = (1, −6, 3) and b = (0, −1, 3), in degrees.
a = (1, −6, 3)b = (0, −1, 3)
θ = °
2.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−2, 0, −5), b = (−2, 5, 2) and c = (−3, 0, 4).
a = (−2, 0, −5)b = (−2, 5, 2)c = (−3, 0, 4)
V =
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 = 10 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 RB.
L = 4 mP = 60 kN at 2 mw = 5 kN/m from 1 to 3 m
RB = kN
5.Truss: method of jointsMedium
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
F: 30 kN ↓
1member AB
2member BC
3member CD
4member DE
5member FG
6member GH
7member AF
8member HE
9member BF
10member CG
11member DH
12member BG
13member DG
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.
D: 15 kN ↓F: 35 kN →H: 30 kN ↓
1member AB
2member BC
3member CD
4member DE
5member FG
6member GH
7member AF
8member HE
9member BF
10member CG
11member DH
12member BG
13member DG
7.Frames & machinesMedium
A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 0.75 m below A. A load of 12 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.
P = 12 kNAD = 1.6 mAB = 2.4 mAC = 0.75 m
A = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 3 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 2.5 m above A. A load of 15 kN hangs at B. Find the tension T in the cable.
W = 15 kNWb = 3 kNAB = 2 mAC = 2.5 m
T = kN
9.Shear force & bending momentMedium
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 = 10 mw = 5 kN/m from 0 to 10 mx = 0.5 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 = 4 mP = 40 kN at 2.5 mw = 4 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 18 mm thick, an overall depth of 500 mm and a web 12 mm thick. Find its elastic section modulus Sx for bending about the horizontal centroidal axis x.
B = 140 mmH = 500 mmtf = 18 mmtw = 12 mm
Sx = × 10³ mm³
12.Centroid & second moment of areaHard
A welded I-section has a top flange 190 × 18 mm, a web 190 mm high and 8 mm thick, and a bottom flange 260 × 18 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.
top flange: 190 × 18 mmweb: 8 × 190 mmbottom flange: 260 × 18 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: θ = 45.62 °
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 = 1·0 + (−6)·(−1) + 3·3 = 15
Magnitudes
|a| = √(1² + (−6)² + 3²) = 6.782
|b| = √(0² + (−1)² + 3²) = 3.162
Angle
cos θ = 15/(6.782·3.162) = 0.6994
θ = arccos(0.6994) = 45.62°
Answer
θ = 45.62°
2.Vectors: dot & cross product
Answer: V = 115
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 = 5·4 − 2·0 = 20
(b × c)y = bzcx − bxcz = 2·(−3) − (−2)·4 = 2
(b × c)z = bxcy − bycx = (−2)·0 − 5·(−3) = 15
Dot product with a
a·(b × c) = (−2)·20 + 0·2 + (−5)·15 = −115
V = |−115| = 115
Answer
V = 115
3.Beam support reactions
Answer: RB = 36.25 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 = 10·2 + 50·2.5
RB = 145/4 = 36.25 kN
ΣFy = 0: RA = 60 − 36.25 = 23.75 kN
Answer
RB = 36.25 kN
4.Beam support reactions
Answer: RB = 35 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 = 5·2 = 10 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 60·2 + 5·2·2
RB = 140/4 = 35 kN
ΣFy = 0: RA = 70 − 35 = 35 kN
Answer
RB = 35 kN
5.Truss: method of joints
Answer: member CG
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).
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.
Strut force
FCD = 12·2.4/(0.4244·1.6) = 42.41 kN
Components at A
Ax = FCD·cos α = 38.4 kN (towards the wall)
Ay = 12 − FCD·sin α = −6 kN
Magnitude
A = √(38.4² + −6²) = 38.87 kN
Answer
A = 38.87 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: T = 21.13 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.
Cable angle
BC = √(2² + 2.5²) = 3.202 m, sin β = 0.7809
Moments about A
T·0.7809·2 = 15·2 + 3·1
T = 21.13 kN
Answer
T = 21.13 kN
9.Shear force & bending moment
Answer: Mx = 11.88 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 = 5·10 = 50 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 5·10·5
RB = 250/10 = 25 kN
ΣFy = 0: RA = 50 − 25 = 25 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 = 25·0.5 − 5·0.5·0.25 = 11.88 kN·m
Answer
Mx = 11.88 kN·m
10.Shear force & bending moment
Answer: Mmax = 45 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 = 4·4 = 16 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 40·2.5 + 4·4·2
RB = 132/4 = 33 kN
ΣFy = 0: RA = 56 − 33 = 23 kN
Where is V = 0?
V(2.5⁻) = RA − w·2.5 = 13 kN
V(2.5⁺) = 13 − 40 = −27 kN
The shear changes sign at the load, so the maximum is under the load.
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 three rectangles
Measure every y from the bottom edge.
top flange: A = 3420 mm², y = 217 mm
web: A = 1520 mm², y = 113 mm
bottom flange: A = 4680 mm², y = 9 mm
Centroid
ȳ = (3420·217 + 1520·113 + 4680·9)/9620 = 99.38 mm
Parallel-axis theorem
top flange: 190·18³/12 + 3420·117.6² = 4.741 × 10⁷ mm⁴