Statics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

Vectors: dot & cross product

Magnitude: |a| = √(a_x² + a_y² + a_z²)
Unit vector: û = a/|a|
Components from an angle: F_x = F·cos θ, F_y = F·sin θ
Dot product: a·b = a_xb_x + a_yb_y + a_zb_z = |a||b|cos θ
Scalar projection of a onto b: a·b/|b|
Work: W = F·d
Cross product: a × b = (a_yb_z − a_zb_y, a_zb_x − a_xb_z, a_xb_y − a_yb_x)
Areas: parallelogram |a × b|, triangle ½|a × b|
Moment about O: M_O = 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: ΣF_x = 0, ΣF_y = 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 ΣF_y = 0 to find the remaining reaction, and check with a second moment equation.

Truss: method of joints

Method of joints: ΣF_x = 0, ΣF_y = 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 ΣF_x = 0 and ΣF_y = 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: ΣF_x = 0, ΣF_y = 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 = √(F_x² + F_y²)
Lever, pliers, bell crank: F_in·a = F_out·b
Toggle: F_link = 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 ΣF_x = 0 and ΣF_y = 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/y_max
Simply supported, uniform load: M_max = 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 = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

2Practice problems

1.Vectors: dot & cross productMedium

Find the scalar projection comp_b a of a = (4, −2, −6) onto b = (1, 3, −5), that is, the component of a in the direction of b.

a = (4, −2, −6)b = (1, 3, −5)

comp_b a =

2.Vectors: dot & cross productHard

The unit vector n̂ is perpendicular to both a = (2, 0, −3) and b = (2, 1, 0) and has a positive z-component. Find its y-component n̂_y.

a = (2, 0, −3)b = (2, 1, 0)n̂_z > 0

n̂_y =

3.Beam support reactionsMedium

The simply supported beam has a pin at A and a roller at B. It carries a point load P at 2 m from A and a counter-clockwise couple M at 3 m from A. Find the vertical reaction R_B, taking upward as positive.

L = 10 mP = 70 kN at 2 mM = 20 kN·m (counter-clockwise) at 3 m

R_B = kN

4.Beam support reactionsHard

A compound beam is fixed to the wall at A, has an internal hinge at C (2.5 m from A) and rests on a roller at D (6 m from A). It carries a point load P at 5.5 m from A. Find the reaction R_D at the roller.

AC = 2.5 mAD = 6 mP = 55 kN at 5.5 m

R_D = kN

5.Truss: method of jointsMedium

Use the method of joints to find the force F_AC in member AC. Give tension as positive and compression as negative.

span = 5 mheight = 4 mB is 3 m from Avertical load = 25 kN

F_AC = 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.

B: 25 kN ↓C: 30 kN ↓F: 40 kN →G: 20 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 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 5° with the line AC. A downward force of 150 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.

F = 150 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 0.6 m below A). The beam carries a uniform load of 5 kN/m over its full length and a point load of 6 kN at its free end B. Find the force F_CD in the strut.

w = 5 kN/mP = 6 kNAD = 0.8 mAB = 2.5 mAC = 0.6 m

F_CD = kN

9.Shear force & bending momentMedium

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

50 kN @ 4 m15 kN/m, 0–6 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

10.Shear force & bending momentHard

Find the bending moment M_x at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 20 kN at 1.5 mw = 4 kN/m from 0 to 4 mx = 1.5 m

M_x = kN·m

11.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 100 mmt_f = 15 mmt_w = 18 mmh_w = 240 mm

I_x = × 10⁶ mm⁴

12.Centroid & second moment of areaHard

A welded I-section has a top flange 180 × 20 mm, a web 380 mm high and 12 mm thick, and a bottom flange 150 × 16 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 180 × 20 mmweb: 12 × 380 mmbottom flange: 150 × 16 mm

I_x = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: comp_b a = 4.733

Dot product: a·b = a_xb_x + a_yb_y + a_zb_z = |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 = 4·1 + (−2)·3 + (−6)·(−5) = 28

Length of b

|b| = √(1² + 3² + (−5)²) = 5.916

Scalar projection

comp_b a = a·b/|b| = 28/5.916 = 4.733

Answer

comp_b a = 4.733

2.Vectors: dot & cross product

Answer: n̂_y = −0.8571

Cross product: a × b = (a_yb_z − a_zb_y, a_zb_x − a_xb_z, a_xb_y − a_yb_x) 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 M_O = r × F, with r = A − O.

Cross product

(a × b)_x = a_yb_z − a_zb_y = 0·0 − (−3)·1 = 3

(a × b)_y = a_zb_x − a_xb_z = (−3)·2 − 2·0 = −6

(a × b)_z = a_xb_y − a_yb_x = 2·1 − 0·2 = 2

Normalise

The z-component of a × b is already positive.

|a × b| = √(3² + (−6)² + 2²) = 7

n̂ = (3, −6, 2)/7 = (0.4286, −0.8571, 0.2857)

n̂_y = −0.8571

Answer

n̂_y = −0.8571

3.Beam support reactions

Answer: R_B = 12 kN

Moments about A

Counter-clockwise positive.

ΣM_A = 0: R_B·10 − 70·2 + 20 = 0

R_B = 120/10 = 12 kN

Vertical equilibrium

R_A = 70 − 12 = 58 kN

Answer

R_B = 12 kN

A negative reaction acts downward.

4.Beam support reactions

Answer: R_D = 47.14 kN

Static equilibrium: a body at rest has ΣF_x = 0, ΣF_y = 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. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.

Part CD

The hinge has no moment, so moments about C on CD alone must balance.

ΣM_C = 0: R_D·3.5 = 55·3

R_D = 47.14 kN

V_C = 55 − 47.14 = 7.857 kN

Answer

R_D = 47.14 kN

5.Truss: method of joints

Answer: F_AC = 7.5 kN

Method of joints: every pin of a truss is in equilibrium, so ΣF_x = 0 and ΣF_y = 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.

ΣM_A = 0: C_y·5 = 25·3 → C_y = 15 kN

ΣF_y = 0: A_y = 25 − 15 = 10 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 = 4/4.472 = 0.8944, cos γ = 2/4.472 = 0.4472

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −15/0.8944 = −16.77 kN

ΣF_x = 0: −F_AC − F_BC·cos γ = 0 → F_AC = −(−16.77)·0.4472 = 7.5 kN

Answer

F_AC = 7.5 kN (tension)

6.Truss: method of joints

Answer: member DH

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 D

CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.

ΣF⊥ = 0 → F_DH = 0

Check with the method of joints

Solving all joints gives these member forces (T = tension, C = compression).

AB: 77.5 kN (T), BC: 77.5 kN (T), CD: 52.5 kN (T), DE: 52.5 kN (T), FG: 90 kN (C), GH: 90 kN (C), AF: 53.03 kN (C), HE: 74.25 kN (C), BF: 25 kN (T), CG: 20 kN (C), DH: 0, FC: 17.68 kN (T), HC: 53.03 kN (T)

Answer

Zero-force member: DH

7.Frames & machines

Answer: Q = 857.3 N

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 F_in·a = F_out·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·F_link·sin 5° = 150

F_link = 860.5 N

Clamping force

Q = F_link·cos 5° = 150/(2·tan 5°) = 857.3 N

Answer

Q = 857.3 N

The force ratio Q/F = 5.715 grows without limit as the links straighten.

8.Frames & machines

Answer: F_CD = 63.8 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 ΣF_x = 0, ΣF_y = 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 = 5·2.5 = 12.5 kN at 1.25 m

Moments about A for the beam

F_CD·0.6·0.8 = 12.5·1.25 + 6·2.5 = 30.63 kN·m

F_CD = 63.8 kN

Answer

F_CD = 63.8 kN

9.Shear force & bending moment

Answer: E (x = 4 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 61.67 kN, R₂ = 78.33 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 54.17, M_C = 93.33, M_D = 117.5, M_E = 126.7 kN·m, M_F = 70.83, M_G = 0

Answer

Mark E: M_E = 126.7 kN·m

10.Shear force & bending moment

Answer: M_x = 26.25 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 = 4·4 = 16 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 20·1.5 + 4·4·2

R_B = 62/4 = 15.5 kN

ΣF_y = 0: R_A = 36 − 15.5 = 20.5 kN

Cut at x = 1.5 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 = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 20.5·1.5 − 4·1.5·0.75 = 26.25 kN·m

Answer

M_x = 26.25 kN·m

11.Centroid & second moment of area

Answer: I_x = 38.86 × 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: A_f = 100·15 = 1500 mm², y_f = 240 + 15/2 = 247.5 mm

web: A_w = 18·240 = 4320 mm², y_w = 240/2 = 120 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1500·247.5 + 4320·120)/5820 = 152.9 mm

Parallel-axis theorem

flange: 100·15³/12 + 1500·(247.5 − 152.9)² = 1.346 × 10⁷ mm⁴

web: 18·240³/12 + 4320·(120 − 152.9)² = 2.54 × 10⁷ mm⁴

I_x = 1.346 × 10⁷ + 2.54 × 10⁷ = 3.886 × 10⁷ mm⁴ = 38.86 × 10⁶ mm⁴

Answer

I_x = 38.86 × 10⁶ mm⁴

12.Centroid & second moment of area

Answer: I_x = 287.5 × 10⁶ mm⁴

Split into three rectangles

Measure every y from the bottom edge.

top flange: A = 3600 mm², y = 406 mm

web: A = 4560 mm², y = 206 mm

bottom flange: A = 2400 mm², y = 8 mm

Centroid

ȳ = (3600·406 + 4560·206 + 2400·8)/10560 = 229.2 mm

Parallel-axis theorem

top flange: 180·20³/12 + 3600·176.8² = 1.127 × 10⁸ mm⁴

web: 12·380³/12 + 4560·23.18² = 5.732 × 10⁷ mm⁴

bottom flange: 150·16³/12 + 2400·221.2² = 1.175 × 10⁸ mm⁴

I_x = 1.127 × 10⁸ + 5.732 × 10⁷ + 1.175 × 10⁸

I_x = 2.875 × 10⁸ mm⁴ = 287.5 × 10⁶ mm⁴

Answer

I_x = 287.5 × 10⁶ mm⁴

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