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 angle θ between the vectors a = (0, 6, −2) and b = (−3, −2, −1), in degrees.

a = (0, 6, −2)b = (−3, −2, −1)

θ = °

2.Vectors: dot & cross productHard

Find the area A of the triangle with corners A(2, −2, 3), B(−3, 1, 3) and C(2, 1, 1). Coordinates are in metres.

A(2, −2, 3)B(−3, 1, 3)C(2, 1, 1)

A = m²

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 3 m from the wall. Find the magnitude of the reaction moment M_A at the wall.

L = 10 mP = 20 kN at 3 mw = 10 kN/m from 0 to 10 m

M_A = 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 R_B.

L = 6 mP = 40 kN at 3 mw = 6 kN/m from 4 to 5 m

R_B = kN

5.Truss: method of jointsMedium

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

span = 8 mheight = 4 mB is 3 m from Avertical load = 35 kN

F_BC = kN

6.Truss: method of jointsHard

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

span = 4 mheight = 4 mB is 1 m from Avertical load = 60 kNhorizontal load = 15 kN →

F_AC = kN

7.Frames & machinesMedium

The frame shown has pins at A, B and C; A and C are fixed to the ground 6 m apart, and B is 4 m above the ground and 2.7 m to the right of A. A vertical load of 8 kN acts on member AB at K, 2.025 m (horizontally) from A. Find the force F_BC in member BC.

P = 8 kNL = 6 mH = 4 mhorizontal distance A to B = 2.7 mhorizontal distance A to K = 2.025 m

F_BC = kN

8.Frames & machinesHard

A uniform horizontal boom AB of weight 0.5 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 4 m above A. A load of 8 kN hangs at B. Find the tension T in the cable.

W = 8 kNW_b = 0.5 kNAB = 6 mAC = 4 m

T = kN

9.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. It carries a point load P₁ between the supports and a point load P₂ at C. Find the bending moment M_x under the load P₁. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP₁ = 70 kN at 3 mP₂ = 35 kN at 10 msupports at x = 0 and x = 8 m

M_x = kN·m

10.Shear force & bending momentHard

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.

60 kN @ 12 m45 kN @ 6 m5 kN/m, 0–10 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)

11.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 240 mm wide with two upright legs, 140 mm high overall. Base and legs are all 8 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 240 mmh = 140 mmt = 8 mm

ȳ = mm

12.Centroid & second moment of areaHard

A 100 mm × 240 mm plate has a rectangular cut-out 50 mm wide and 20 mm high, centred on the vertical centre line with its centre 84 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 240 mmcut-out: 50 × 20 mmhole centre: y = 84 mm

I_x = × 10⁶ mm⁴

3Solutions

1.Vectors: dot & cross product

Answer: θ = 115 °

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 = 0·(−3) + 6·(−2) + (−2)·(−1) = −10

Magnitudes

|a| = √(0² + 6² + (−2)²) = 6.325

|b| = √((−3)² + (−2)² + (−1)²) = 3.742

Angle

cos θ = −10/(6.325·3.742) = −0.4226

θ = arccos(−0.4226) = 115°

Answer

θ = 115°

2.Vectors: dot & cross product

Answer: A = 9.5 m²

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.

Edge vectors

AB = B − A = (−5, 3, 0)

AC = C − A = (0, 3, −2)

Cross product

(AB × AC)_x = AB_yAC_z − AB_zAC_y = 3·(−2) − 0·3 = −6

(AB × AC)_y = AB_zAC_x − AB_xAC_z = 0·0 − (−5)·(−2) = −10

(AB × AC)_z = AB_xAC_y − AB_yAC_x = (−5)·3 − 3·0 = −15

|AB × AC| = √((−6)² + (−10)² + (−15)²) = 19

Area of the triangle

A = ½·19 = 9.5 m²

Answer

A = 9.5 m²

3.Beam support reactions

Answer: M_A = 560 kN·m

Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction R_A and a reaction moment M_A. ΣF_y = 0 gives R_A, and moments about the wall give M_A = 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 = 10·10 = 100 kN, x̄ = 5 m

Moment about the wall

M_A = P·a + W·x̄ = 20·3 + 100·5 = 560 kN·m

Answer

M_A = 560 kN·m

4.Beam support reactions

Answer: R_B = 24.5 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 = 6·1 = 6 kN, acting at x̄ = 4.5 m

ΣM_A = 0: R_B·6 = 40·3 + 6·1·4.5

R_B = 147/6 = 24.5 kN

ΣF_y = 0: R_A = 46 − 24.5 = 21.5 kN

Answer

R_B = 24.5 kN

5.Truss: method of joints

Answer: F_BC = −21.01 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·8 = 35·3 → C_y = 13.13 kN

ΣF_y = 0: A_y = 35 − 13.13 = 21.88 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 = 4/6.403 = 0.6247, cos γ = 5/6.403 = 0.7809

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −13.13/0.6247 = −21.01 kN

Answer

F_BC = −21.01 kN (compression)

6.Truss: method of joints

Answer: F_AC = 22.5 kN

Support reactions

Equilibrium of the whole truss. Moments about A, counter-clockwise positive.

ΣM_A = 0: C_y·4 = 60·1 + 15·4 → C_y = 30 kN

ΣF_y = 0: A_y = 60 − 30 = 30 kN

ΣF_x = 0: A_x = −15 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 = 4/5 = 0.8, cos γ = 3/5 = 0.6

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −30/0.8 = −37.5 kN

ΣF_x = 0: −F_AC − F_BC·cos γ = 0 → F_AC = −(−37.5)·0.6 = 22.5 kN

Answer

F_AC = 22.5 kN (tension)

7.Frames & machines

Answer: F_BC = 3.5 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.

Two-force member BC

BC = √(3.3² + 4²) = 5.186 m

Whole frame: moments about A

The ground pushes on BC at C along CB; its horizontal component passes through A, only the vertical one has a moment.

F_BC·(4/5.186)·6 = 8·2.025

F_BC = 3.5 kN (compression)

Answer

F_BC = 3.5 kN

8.Frames & machines

Answer: T = 14.87 kN

Cable angle

BC = √(6² + 4²) = 7.211 m, sin β = 0.5547

Moments about A

T·0.5547·6 = 8·6 + 0.5·3

T = 14.87 kN

Answer

T = 14.87 kN

9.Shear force & bending moment

Answer: M_x = 105 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. The overhang load also acts about A.

ΣM_A = 0: R_B·8 = 70·3 + 35·10

R_B = 560/8 = 70 kN

ΣF_y = 0: R_A = 105 − 70 = 35 kN

Cut under P₁ and keep the left part

M_x = R_A·3 = 35·3 = 105 kN·m

Answer

M_x = 105 kN·m

10.Shear force & bending moment

Answer: F (x = 10 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 = 31 kN, R₂ = 124 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 = 52, M_C = 84, M_D = 96, M_E = −2, M_F = −120 kN·m, M_G = 0

Answer

Mark F: M_F = −120 kN·m

11.Centroid & second moment of area

Answer: ȳ = 40.67 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.

Split into rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 240·8 = 1920 mm², y₁ = 4 mm

two legs: A₂ = 2·8·132 = 2112 mm², y₂ = 8 + 132/2 = 74 mm

Centroid

ȳ = (1920·4 + 2112·74)/4032 = 40.67 mm

Answer

ȳ = 40.67 mm

12.Centroid & second moment of area

Answer: I_x = 113.8 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (24000·120 − 1000·84)/(24000 − 1000) = 121.6 mm

Full plate about ȳ

100·240³/12 + 24000·−1.565² = 115.3 × 10⁶ mm⁴

Hole about ȳ

50·20³/12 + 1000·−37.57² = 1.444 × 10⁶ mm⁴

Subtract

I_x = 113.8 × 10⁶ mm⁴

Answer

I_x = 113.8 × 10⁶ mm⁴

ExamProblems · examproblems.com/courses/statics

Statics: cram pack

1Formulas and methods

Vectors: dot & cross product

Method

Beam support reactions

Method

Truss: method of joints

Method

Frames & machines

Method

Shear force & bending moment

Method

Centroid & second moment of area

2Practice problems

3Solutions

Dot product

Magnitudes

Angle

Answer

Edge vectors

Cross product

Area of the triangle

Answer

Resultant of the distributed load

Moment about the wall

Answer

Free-body diagram

Equilibrium

Answer

Support reactions

Joint C

Answer

Support reactions

Joint C

Answer

Two-force member BC

Whole frame: moments about A

Answer

Cable angle

Moments about A

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Bending moment at every mark

Answer

Split into rectangles

Centroid

Answer

Centroid

Full plate about ȳ

Hole about ȳ

Subtract

Answer

Cut under P₁ and keep the left part