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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 = (1, 7, 6) onto b = (−1, 2, −4), that is, the component of a in the direction of b.
a = (1, 7, 6)b = (−1, 2, −4)
compb a =
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, −4, −3) and b = (2, −1, 2) and has a positive z-component. Find its y-component n̂y.
a = (−1, −4, −3)b = (2, −1, 2)n̂z > 0
n̂y =
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 = 8 mP1 = 35 kN at 1 mP2 = 30 kN at 3 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 = 20 kN at 2 mw = 2 kN/m from 2 to 4 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.
D: 30 kN ↓G: 30 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
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 ↓D: 10 kN ↓F: 20 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 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 10° with the line AC. A downward force of 100 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 100 Nθ = 10°
Q = N
8.Frames & machinesHard
The frame shown has pins at A, B and C; A and C are fixed to the ground 4 m apart, and B is 2 m high and 0.6 m to the right of A. Member AB carries a vertical load of 12 kN at K, 0.3 m (horizontally) from A, and a horizontal load of 3 kN to the right acts at B on member AB. Find the size of the horizontal reaction Ax at A.
P = 12 kNQ = 3 kNL = 4 mH = 2 mhorizontal distance A to B = 0.6 mhorizontal distance A to K = 0.3 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 = 4 mP1 = 80 kN at 1.5 mP2 = 70 kN at 3 m
Mmax = kN·m
10.Shear force & bending momentHard
The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance xc from A to the point of contraflexure between the supports (where the bending moment changes sign).
L = 6 mw = 3 kN/m from 0 to 6 msupports at x = 0 and x = 4 m
xc = 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 = 260 mmtf = 30 mmtw = 12 mmhw = 200 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A welded I-section has a top flange 200 × 10 mm, a web 220 mm high and 16 mm thick, and a bottom flange 160 × 18 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.
top flange: 200 × 10 mmweb: 16 × 220 mmbottom flange: 160 × 18 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: compb a = −2.4
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·(−1) + 7·2 + 6·(−4) = −11
Length of b
|b| = √((−1)² + 2² + (−4)²) = 4.583
Scalar projection
compb a = a·b/|b| = −11/4.583 = −2.4
Answer
compb a = −2.4
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: n̂y = −0.2709
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. 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·8 = 35·1 + 30·3
RB = 125/8 = 15.63 kN
ΣFy = 0: RA = 65 − 15.63 = 49.38 kN
Answer
RB = 15.63 kN
4.Beam support reactions
Answer: RB = 13 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 = 2·2 = 4 kN, acting at x̄ = 3 m
ΣMA = 0: RB·4 = 20·2 + 2·2·3
RB = 52/4 = 13 kN
ΣFy = 0: RA = 24 − 13 = 11 kN
Answer
RB = 13 kN
5.Truss: method of joints
Answer: member BF
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 B
AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.
ΣF⊥ = 0 → FBF = 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 G
FG and GH 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).
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 10° = 100
Flink = 287.9 N
Clamping force
Q = Flink·cos 10° = 100/(2·tan 10°) = 283.6 N
Answer
Q = 283.6 N
The force ratio Q/F = 2.836 grows without limit as the links straighten.
8.Frames & machines
Answer: Ax = 1.08 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 = 3.945 m, cos = 0.8619, sin = 0.507
Whole frame: moments about A
FBC·0.507·4 = 12·0.3 + 3·2
FBC = 4.734 kN
Horizontal balance
Ax + 3 − 4.734·0.8619 = 0
Ax = 1.08 kN → |Ax| = 1.08 kN
Answer
Ax = 1.08 kN
Ax points to the right.
9.Shear force & bending moment
Answer: Mmax = 101.3 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·4 = 80·1.5 + 70·3
RB = 330/4 = 82.5 kN
ΣFy = 0: RA = 150 − 82.5 = 67.5 kN
Moment under each load
Keep the part left of each load.
under P1: M = RA·1.5 = 67.5·1.5 = 101.3 kN·m
under P2: M = RB·1 = 82.5·1 = 82.5 kN·m
Largest value
Mmax = 101.3 kN·m (under P1)
Answer
Mmax = 101.3 kN·m
10.Shear force & bending moment
Answer: xc = 3 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, including the load on the overhang.
UDL resultant: W = w·ℓ = 3·6 = 18 kN, acting at x̄ = 3 m
ΣMA = 0: RB·4 = 3·6·3
RB = 54/4 = 13.5 kN
ΣFy = 0: RA = 18 − 13.5 = 4.5 kN
Zero moment
Between the supports M(x) = RA·x − w·x²/2. Apart from x = 0 it is zero at x = 2RA/w.
RA·xc − w·xc²/2 = 0
xc = 2·4.5/3 = 3 m
Answer
xc = 3 m
11.Centroid & second moment of area
Answer: Ix = 32.86 × 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 = 260·30 = 7800 mm², yf = 200 + 30/2 = 215 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 three rectangles
Measure every y from the bottom edge.
top flange: A = 2000 mm², y = 243 mm
web: A = 3520 mm², y = 128 mm
bottom flange: A = 2880 mm², y = 9 mm
Centroid
ȳ = (2000·243 + 3520·128 + 2880·9)/8400 = 114.6 mm
Parallel-axis theorem
top flange: 200·10³/12 + 2000·128.4² = 3.3 × 10⁷ mm⁴