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 scalar projection compb a of a = (3, −7, −2) onto b = (0, 1, 3), that is, the component of a in the direction of b.
a = (3, −7, −2)b = (0, 1, 3)
compb a =
2.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, 2, −2) and b = (1, −3, 1) and has a positive z-component. Find its y-component n̂y.
a = (−1, 2, −2)b = (1, −3, 1)n̂z > 0
n̂y =
3.Beam support reactionsMedium
The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w from the wall to 4.5 m and a point load P at 5.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
L = 6 mP = 40 kN at 5.5 mw = 8 kN/m from 0 to 4.5 m
MA = kN·m
4.Beam support reactionsHard
The simply supported beam has a pin at A and a roller at B. A load P acts at 1.5 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 4 m from A. Find the magnitude of the total reaction force RA at the pin A.
L = 5 mP = 65 kN at 1.5 m, θ = 40°Q = 55 kN at 4 m
RA = 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 = 8 mheight = 2 mB is 1 m from Avertical load = 15 kN
FAB = 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.
D: 15 kN ↓H: 15 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
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 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 = 5°
Q = N
8.Frames & machinesHard
The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 1 m below A). The beam carries a uniform load of 1.5 kN/m over its full length and a point load of 3 kN at its free end B. Find the force FCD in the strut.
w = 1.5 kN/mP = 3 kNAD = 1 mAB = 2 mAC = 1 m
FCD = kN
9.Shear force & bending momentMedium
The simply supported beam carries a clockwise couple M0 = 55 kN·m at x = 1.5 m from A, and no other load. Find the bending moment Mx just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.
L = 4 mM0 = 55 kN·m clockwise, at 1.5 m
Mx = kN·m
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 10 mP = 30 kN at 3 mw = 6 kN/m from 0 to 10 mx = 3 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A circular tube has outside diameter D = 230 mm and inside diameter d = 206 mm. Find its radius of gyration r about the horizontal centroidal axis.
D = 230 mmd = 206 mm
r = mm
12.Centroid & second moment of areaHard
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 15 kN·m. Find the size of the bending stress σtop at the top edge.
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 = 3·0 + (−7)·1 + (−2)·3 = −13
Length of b
|b| = √(0² + 1² + 3²) = 3.162
Scalar projection
compb a = a·b/|b| = −13/3.162 = −4.111
Answer
compb a = −4.111
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: n̂y = −0.2357
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.
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 = 8·4.5 = 36 kN, x̄ = 2.25 m
Moment about the wall
MA = P·a + W·x̄ = 40·5.5 + 36·2.25 = 301 kN·m
Answer
MA = 301 kN·m
4.Beam support reactions
Answer: RA = 64.02 kN
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.
Components of P
Pv = 65·sin 40° = 41.78 kN
Ph = 65·cos 40° = 49.79 kN
Reactions
ΣMA = 0: RB·5 = 41.78·1.5 + 55·4 → RB = 56.53 kN
Ay = 41.78 + 55 − 56.53 = 40.25 kN
Ax = Ph = 49.79 kN
Resultant at A
RA = √(49.79² + 40.25²) = 64.02 kN
Answer
RA = 64.02 kN
5.Truss: method of joints
Answer: FAB = −14.67 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·8 = 15·1 → Cy = 1.875 kN
ΣFy = 0: Ay = 15 − 1.875 = 13.13 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 2/2.236 = 0.8944
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −13.13/0.8944 = −14.67 kN
Answer
FAB = −14.67 kN (compression)
6.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).
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° = 100
Flink = 573.7 N
Clamping force
Q = Flink·cos 5° = 100/(2·tan 5°) = 571.5 N
Answer
Q = 571.5 N
The force ratio Q/F = 5.715 grows without limit as the links straighten.
8.Frames & machines
Answer: FCD = 12.73 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.
Resultant of the uniform load
w·L = 1.5·2 = 3 kN at 1 m
Moments about A for the beam
FCD·0.7071·1 = 3·1 + 3·2 = 9 kN·m
FCD = 12.73 kN
Answer
FCD = 12.73 kN
9.Shear force & bending moment
Answer: Mx = −20.63 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. An applied couple makes the bending moment diagram jump by the size of the couple.
Support reactions
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
RB = M0/L = 55/4 = 13.75 kN (upward)
RA = −RB = −13.75 kN (downward)
Cut just left of the couple
The left part contains only RA.
Mx = RA·a = −13.75·1.5 = −20.63 kN·m
Answer
Mx = −20.63 kN·m
10.Shear force & bending moment
Answer: Mx = 126 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 = 6·10 = 60 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 30·3 + 6·10·5
RB = 390/10 = 39 kN
ΣFy = 0: RA = 90 − 39 = 51 kN
Cut at x = 3 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 = 51·3 − 6·3·1.5 = 126 kN·m
Answer
Mx = 126 kN·m
11.Centroid & second moment of area
Answer: r = 77.19 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
A = π(D² − d²)/4 = 8218 mm²
I = π(D⁴ − d⁴)/64 = 4.897 × 10⁷ mm⁴
Radius of gyration
r = √(I/A) = √(4.897 × 10⁷/8218) = 77.19 mm
Answer
r = 77.19 mm
12.Centroid & second moment of area
Answer: σtop = 48.54 MPa
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 = 160 + 30/2 = 175 mm