In the print window, choose “Save as PDF” to keep a copy on your device.
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 dot product a·b of a = (6, −6, −4) and b = (−6, 2, 2).
a = (6, −6, −4)b = (−6, 2, 2)
a·b =
2.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (0, 4, −5) and b = (1, 4, −2).
a = (0, 4, −5)b = (1, 4, −2)
A =
3.Beam support reactionsMedium
The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w along its whole length and a point load P at 4.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 6 mP = 65 kN at 4.5 mw = 4 kN/m from 0 to 6 m
MA = kN·m
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
L = 5 mP = 55 kN at 2.5 mw = 5 kN/m from 1 to 4 m
RA = 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.
B: 35 kN ↓D: 30 kN ↓F: 35 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
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 4 mheight = 3 mB is 3 m from Avertical load = 35 kNhorizontal load = 25 kN ←
FBC = kN
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, 1 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 = 4 mAC = 1 m
A = kN
8.Frames & machinesHard
The A-frame shown stands on a pin at A and a roller at B, 3 m apart, with its apex pin C 2.5 m high and the horizontal tie DE 0.6 m above the ground. A vertical load of 2 kN acts on leg AC at K, 0.6 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 2 kNAB = 3 mheight of C = 2.5 mheight of DE = 0.6 mhorizontal distance A to K = 0.6 m
C = kN
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 5 mw = 3 kN/m from 0 to 5 mx = 1 m
Mx = kN·m
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 10 mP = 40 kN at 1 mw = 4 kN/m from 0 to 10 mx = 1 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A hollow rectangular (box) section has outer width B = 220 mm, outer height H = 225 mm and a uniform wall thickness t = 18 mm. Find its radius of gyration r about the horizontal centroidal axis.
B = 220 mmH = 225 mmt = 18 mm
r = 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 35 MPa in tension and 90 MPa in compression. Find the largest hogging bending moment Mallow the beam can carry (give its size).
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.
Negative: the angle between the vectors is more than 90°.
2.Vectors: dot & cross product
Answer: A = 13.6
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 = 4·(−2) − (−5)·4 = 12
(a × b)y = azbx − axbz = (−5)·1 − 0·(−2) = −5
(a × b)z = axby − aybx = 0·4 − 4·1 = −4
Area
A = |a × b| = √(12² + (−5)² + (−4)²) = 13.6
Answer
A = 13.6
3.Beam support reactions
Answer: MA = 364.5 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 distributed load
W = 4·6 = 24 kN, x̄ = 3 m
Moment about the wall
MA = P·a + W·x̄ = 65·4.5 + 24·3 = 364.5 kN·m
Answer
MA = 364.5 kN·m
4.Beam support reactions
Answer: RA = 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 = 5·3 = 15 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 55·2.5 + 5·3·2.5
RB = 175/5 = 35 kN
ΣFy = 0: RA = 70 − 35 = 35 kN
Answer
RA = 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 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).
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 = 35·3 − 25·3 → Cy = 7.5 kN
ΣFy = 0: Ay = 35 − 7.5 = 27.5 kN
ΣFx = 0: Ax = 25 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 = 3/3.162 = 0.9487, cos γ = 1/3.162 = 0.3162
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 = 12·4/(0.53·1.6) = 56.6 kN
Components at A
Ax = FCD·cos α = 48 kN (towards the wall)
Ay = 12 − FCD·sin α = −18 kN
Magnitude
A = √(48² + −18²) = 51.26 kN
Answer
A = 51.26 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: C = 0.5096 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.
Support reaction at B
By = 2·0.6/3 = 0.4 kN
Tie force
T = 0.4·1.5/1.9 = 0.3158 kN
Pin force on BC
Cx = 0.3158 kN, Cy = −0.4 kN
C = √(0.3158² + 0.4²) = 0.5096 kN
Answer
C = 0.5096 kN
9.Shear force & bending moment
Answer: Mx = 6 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 = 3·5 = 15 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 3·5·2.5
RB = 37.5/5 = 7.5 kN
ΣFy = 0: RA = 15 − 7.5 = 7.5 kN
Cut at x = 1 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 = 7.5·1 − 3·1·0.5 = 6 kN·m
Answer
Mx = 6 kN·m
10.Shear force & bending moment
Answer: Mx = 54 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 = 4·10 = 40 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 40·1 + 4·10·5
RB = 240/10 = 24 kN
ΣFy = 0: RA = 80 − 24 = 56 kN
Cut at x = 1 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 = 56·1 − 4·1·0.5 = 54 kN·m
Answer
Mx = 54 kN·m
11.Centroid & second moment of area
Answer: r = 84.57 mm
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. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.
Net area and second moment of area
b = 184 mm, h = 189 mm
A = BH − bh = 220·225 − 184·189 = 14724 mm²
I = (BH³ − bh³)/12 = 1.053 × 10⁸ mm⁴
Radius of gyration
r = √(I/A) = √(1.053 × 10⁸/14724) = 84.57 mm
Answer
r = 84.57 mm
12.Centroid & second moment of area
Answer: Mallow = 31.74 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.