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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, 2, 0) and b = (−1, k, −5) are perpendicular.
a = (1, 2, 0)b = (−1, k, −5)
k =
2.Vectors: dot & cross productHard
A force F = (−60, 50, 30) N acts at point A(−4, −1, −2). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).
F = (−60, 50, 30) NA(−4, −1, −2) mO(0, 0, 0) m
My = N·m
3.Beam support reactionsMedium
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
L = 6 mP1 = 60 kN at 2 mP2 = 45 kN at 4 m
RA = 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 = 8 mP = 50 kN at 4 mw = 5 kN/m from 3 to 7 m
RB = kN
5.Truss: method of jointsMedium
Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.
span = 4 mheight = 3 mB is 3 m from Avertical load = 60 kN
FAB = kN
6.Truss: method of jointsHard
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 = 60 kNhorizontal load = 30 kN ←
FBC = kN
7.Frames & machinesMedium
The bell crank shown turns about the pin O. A vertical force of 100 N at A (240 mm from O) is balanced by a horizontal force at B (80 mm above O). Find the magnitude of the pin reaction FO.
F = 100 NOA = 240 mmOB = 80 mm
FO = N
8.Frames & machinesHard
The frame shown has pins at A, B and C; A and C are fixed to the ground 6 m apart, and B is 2 m high and 2.7 m to the right of A. Member AB carries a vertical load of 8 kN at K, 1.08 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 = 8 kNQ = 3 kNL = 6 mH = 2 mhorizontal distance A to B = 2.7 mhorizontal distance A to K = 1.08 m
Ax = kN
9.Shear force & bending momentMedium
Find the shear force Vx at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 5 mw = 10 kN/m from 0 to 5 mx = 4.5 m
Vx = kN
10.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.
L = 5 mw = 12 kN/m from 1 to 5 m
Mmax = kN·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 = 100 mmtf = 30 mmtw = 8 mmhw = 140 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 110 mm wide and 14 mm thick, an overall depth of 340 mm and a web 6 mm thick. It carries a bending moment of 40 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
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·(−1) + 2·k + 0·(−5) = 0
−1 + 2·k = 0
Solve for k
k = −(−1)/2 = 0.5
Answer
k = 0.5
2.Vectors: dot & cross product
Answer: My = 240 N·m
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.
Position vector from O to A
r = A − O = (−4, −1, −2) − (0, 0, 0) = (−4, −1, −2) m
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·6 = 60·2 + 45·4
RB = 300/6 = 50 kN
ΣFy = 0: RA = 105 − 50 = 55 kN
Answer
RA = 55 kN
4.Beam support reactions
Answer: RB = 37.5 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 = 5·4 = 20 kN, acting at x̄ = 5 m
ΣMA = 0: RB·8 = 50·4 + 5·4·5
RB = 300/8 = 37.5 kN
ΣFy = 0: RA = 70 − 37.5 = 32.5 kN
Answer
RB = 37.5 kN
5.Truss: method of joints
Answer: FAB = −21.21 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·4 = 60·3 → Cy = 45 kN
ΣFy = 0: Ay = 60 − 45 = 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 = 3/4.243 = 0.7071
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −15/0.7071 = −21.21 kN
Answer
FAB = −21.21 kN (compression)
6.Truss: method of joints
Answer: FBC = −8.333 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 = 60·2 − 30·3 → Cy = 5 kN
ΣFy = 0: Ay = 60 − 5 = 55 kN
ΣFx = 0: Ax = 30 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
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 = 100·240/80 = 300 N
Pin reaction
Ox = 300 N, Oy = 100 N
FO = √(300² + 100²) = 316.2 N
Answer
FO = 316.2 N
8.Frames & machines
Answer: Ax = 1.026 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 = 3.859 m, cos = 0.8552, sin = 0.5183
Whole frame: moments about A
FBC·0.5183·6 = 8·1.08 + 3·2
FBC = 4.708 kN
Horizontal balance
Ax + 3 − 4.708·0.8552 = 0
Ax = 1.026 kN → |Ax| = 1.026 kN
Answer
Ax = 1.026 kN
Ax points to the right.
9.Shear force & bending moment
Answer: Vx = −20 kN
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 = 10·5 = 50 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 10·5·2.5
RB = 125/5 = 25 kN
ΣFy = 0: RA = 50 − 25 = 25 kN
Cut at x = 4.5 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.
V = RA − (all downward loads left of the cut)
Vx = 25 − 10·4.5 = −20 kN
Answer
Vx = −20 kN
10.Shear force & bending moment
Answer: Mmax = 34.56 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 = 12·4 = 48 kN, acting at x̄ = 3 m
ΣMA = 0: RB·5 = 12·4·3
RB = 144/5 = 28.8 kN
ΣFy = 0: RA = 48 − 28.8 = 19.2 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
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 = 100·30 = 3000 mm², yf = 140 + 30/2 = 155 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.
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.