Statics: cram pack

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

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 dot product a·b of a = (−4, 7, −3) and b = (0, −6, 4).

a = (−4, 7, −3)b = (0, −6, 4)

a·b =

2.Vectors: dot & cross productHard

The unit vector n̂ is perpendicular to both a = (1, 0, −2) and b = (−4, 4, 5) and has a positive z-component. Find its y-component n̂_y.

a = (1, 0, −2)b = (−4, 4, 5)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 clockwise couple M at 1.5 m from A. Find the vertical reaction R_A, taking upward as positive.

L = 4 mP = 80 kN at 2 mM = 25 kN·m (clockwise) at 1.5 m

R_A = kN

4.Beam support reactionsHard

The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction R_A.

L = 10 mP = 30 kN at 9 mw = 10 kN/m from 3 to 8 m

R_A = kN

5.Truss: method of jointsMedium

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

span = 4 mheight = 2 mB is 3 m from Avertical load = 45 kN

F_AB = 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: 35 kN ↓F: 10 kN →G: 35 kN ↓H: 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

7.Frames & machinesMedium

The A-frame shown stands on a pin at A and a roller at B, 3.2 m apart. The legs meet at the pin C, 2 m high, and the horizontal tie DE is 0.8 m above the ground. A load of 8 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.

P = 8 kNAB = 3.2 mheight of C = 2 mheight of DE = 0.8 m

C = kN

8.Frames & machinesHard

A uniform horizontal boom AB of weight 2 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 1.5 m above A. A load of 4 kN hangs at B. Find the tension T in the cable.

W = 4 kNW_b = 2 kNAB = 2 mAC = 1.5 m

T = 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.

40 kN @ 4 m6 kN/m, 0–12 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

10.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance x_c from A to the point of contraflexure between the supports (where the bending moment changes sign).

L = 6 mw = 5 kN/m from 0 to 6 msupports at x = 0 and x = 5 m

x_c = 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 = 240 mmt_f = 20 mmt_w = 18 mmh_w = 140 mm

I_x = × 10⁶ mm⁴

12.Centroid & second moment of areaHard

A welded I-section has a top flange 170 × 12 mm, a web 200 mm high and 8 mm thick, and a bottom flange 250 × 14 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 170 × 12 mmweb: 8 × 200 mmbottom flange: 250 × 14 mm

I_x = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: a·b = −54

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.

Multiply matching components and add

a·b = (−4)·0 + 7·(−6) + (−3)·4 = 0 + (−42) + (−12) = −54

Answer

a·b = −54

Negative: the angle between the vectors is more than 90°.

2.Vectors: dot & cross product

Answer: n̂_y = 0.318

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 = 0·5 − (−2)·4 = 8

(a × b)_y = a_zb_x − a_xb_z = (−2)·(−4) − 1·5 = 3

(a × b)_z = a_xb_y − a_yb_x = 1·4 − 0·(−4) = 4

Normalise

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

|a × b| = √(8² + 3² + 4²) = 9.434

n̂ = (8, 3, 4)/9.434 = (0.848, 0.318, 0.424)

n̂_y = 0.318

Answer

n̂_y = 0.318

3.Beam support reactions

Answer: R_A = 33.75 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.

Moments about A

Counter-clockwise positive.

ΣM_A = 0: R_B·4 − 80·2 − 25 = 0

R_B = 185/4 = 46.25 kN

Vertical equilibrium

R_A = 80 − 46.25 = 33.75 kN

Answer

R_A = 33.75 kN

A negative reaction acts downward.

4.Beam support reactions

Answer: R_A = 25.5 kN

Static equilibrium: a body at rest has ΣF_x = 0, ΣF_y = 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 = 10·5 = 50 kN, acting at x̄ = 5.5 m

ΣM_A = 0: R_B·10 = 30·9 + 10·5·5.5

R_B = 545/10 = 54.5 kN

ΣF_y = 0: R_A = 80 − 54.5 = 25.5 kN

Answer

R_A = 25.5 kN

5.Truss: method of joints

Answer: F_AB = −20.28 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·4 = 45·3 → C_y = 33.75 kN

ΣF_y = 0: A_y = 45 − 33.75 = 11.25 kN

Joint A

Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.

sin α = 2/3.606 = 0.5547

ΣF_y = 0: A_y + F_AB·sin α = 0

F_AB = −11.25/0.5547 = −20.28 kN

Answer

F_AB = −20.28 kN (compression)

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: 46.87 kN (T), BC: 46.87 kN (T), CD: 41.87 kN (T), DE: 41.87 kN (T), FG: 57.5 kN (C), GH: 57.5 kN (C), AF: 61.46 kN (C), HE: 69.79 kN (C), BF: 35 kN (T), CG: 35 kN (C), DH: 0, FC: 17.71 kN (T), HC: 26.04 kN (T)

Answer

Zero-force member: DH

7.Frames & machines

Answer: C = 6.667 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.

Tie force from leg BC

B_y = 4 kN

T = 4·1.6/1.2 = 5.333 kN

Components of the pin force on BC

C_x = T = 5.333 kN, C_y = −B_y = −4 kN

Magnitude

C = √(5.333² + 4²) = 6.667 kN

Answer

C = 6.667 kN

8.Frames & machines

Answer: T = 8.333 kN

Cable angle

BC = √(2² + 1.5²) = 2.5 m, sin β = 0.6

Moments about A

T·0.6·2 = 4·2 + 2·1

T = 8.333 kN

Answer

T = 8.333 kN

9.Shear force & bending moment

Answer: C (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 = 62.67 kN, R₂ = 49.33 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 = 113.3, M_C = 202.7 kN·m, M_D = 188, M_E = 149.3, M_F = 86.67, M_G = 0

Answer

Mark C: M_C = 202.7 kN·m

10.Shear force & bending moment

Answer: x_c = 4.8 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 = 5·6 = 30 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·5 = 5·6·3

R_B = 90/5 = 18 kN

ΣF_y = 0: R_A = 30 − 18 = 12 kN

Zero moment

Between the supports M(x) = R_A·x − w·x²/2. Apart from x = 0 it is zero at x = 2R_A/w.

R_A·x_c − w·x_c²/2 = 0

x_c = 2·12/5 = 4.8 m

Answer

x_c = 4.8 m

11.Centroid & second moment of area

Answer: I_x = 14.85 × 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 = 240·20 = 4800 mm², y_f = 140 + 20/2 = 150 mm

web: A_w = 18·140 = 2520 mm², y_w = 140/2 = 70 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4800·150 + 2520·70)/7320 = 122.5 mm

Parallel-axis theorem

flange: 240·20³/12 + 4800·(150 − 122.5)² = 3.801 × 10⁶ mm⁴

web: 18·140³/12 + 2520·(70 − 122.5)² = 1.105 × 10⁷ mm⁴

I_x = 3.801 × 10⁶ + 1.105 × 10⁷ = 1.485 × 10⁷ mm⁴ = 14.85 × 10⁶ mm⁴

Answer

I_x = 14.85 × 10⁶ mm⁴

12.Centroid & second moment of area

Answer: I_x = 64.9 × 10⁶ mm⁴

Split into three rectangles

Measure every y from the bottom edge.

top flange: A = 2040 mm², y = 220 mm

web: A = 1600 mm², y = 114 mm

bottom flange: A = 3500 mm², y = 7 mm

Centroid

ȳ = (2040·220 + 1600·114 + 3500·7)/7140 = 91.83 mm

Parallel-axis theorem

top flange: 170·12³/12 + 2040·128.2² = 3.353 × 10⁷ mm⁴

web: 8·200³/12 + 1600·22.17² = 6.119 × 10⁶ mm⁴

bottom flange: 250·14³/12 + 3500·84.83² = 2.525 × 10⁷ mm⁴

I_x = 3.353 × 10⁷ + 6.119 × 10⁶ + 2.525 × 10⁷

I_x = 6.49 × 10⁷ mm⁴ = 64.9 × 10⁶ mm⁴

Answer

I_x = 64.9 × 10⁶ mm⁴

ExamProblems · examproblems.com/courses/statics