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 = (0, 6, 6) onto b = (3, −5, −6), that is, the component of a in the direction of b.

a = (0, 6, 6)b = (3, −5, −6)

comp_b a =

2.Vectors: dot & cross productHard

A force F = (60, 60, 20) N acts at point A(3, 0, 2). Coordinates are in metres. Find the y-component M_y of the moment of the force about point O(0, 2, −1).

F = (60, 60, 20) NA(3, 0, 2) mO(0, 2, −1) m

M_y = N·m

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 5 m and a point load P at 9 m from the wall. Find the vertical reaction R_A at the wall.

L = 10 mP = 80 kN at 9 mw = 6 kN/m from 0 to 5 m

R_A = kN

4.Beam support reactionsHard

The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 4 kN/m at the free end B, plus a point load P at 6 m from the wall. Find the magnitude of the reaction moment M_A at the wall.

L = 10 mw₀ = 4 kN/mP = 40 kN at 6 m

M_A = kN·m

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 = 6 mheight = 3 mB is 2 m from Avertical load = 30 kN

F_AC = 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 = 6 mheight = 4 mB is 2 m from Avertical load = 35 kNhorizontal load = 10 kN ←

F_AC = kN

7.Frames & machinesMedium

The A-frame shown stands on a pin at A and a roller at B, 3.2 m apart, with its apex pin C 2.4 m high and the horizontal tie DE 0.6 m above the ground. A vertical load of 12 kN acts on leg AC at K, 0.8 m (horizontally) from A. Find the tension T in the tie.

P = 12 kNAB = 3.2 mheight of C = 2.4 mheight of DE = 0.6 mhorizontal distance A to K = 0.8 m

T = kN

8.Frames & machinesHard

The A-frame shown stands on a pin at A and a roller at B, 4.8 m apart, with its apex pin C 4 m high and the horizontal tie DE 0.8 m above the ground. A vertical load of 10 kN acts on leg AC at K, 1.8 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.

P = 10 kNAB = 4.8 mheight of C = 4 mheight of DE = 0.8 mhorizontal distance A to K = 1.8 m

C = kN

9.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 4 kN/m from 0 to 4 mx = 3 m

M_x = kN·m

10.Shear force & bending momentHard

Find the bending moment M_x at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 50 kN at 2 mw = 5 kN/m from 0 to 10 mx = 2 m

M_x = kN·m

11.Centroid & second moment of areaMedium

A 200 mm × 240 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 144 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 240 mmd = 40 mmhole centre: y = 144 mm

I_x = × 10⁶ mm⁴

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 bending stress σ_bottom at the bottom edge.

b_f = 160 mmt_f = 15 mmt_w = 8 mmh_w = 120 mmM = 15 kN·m

σ_bottom = MPa

3Solutions

1.Vectors: dot & cross product

Answer: comp_b a = −7.889

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·(−5) + 6·(−6) = −66

Length of b

|b| = √(3² + (−5)² + (−6)²) = 8.367

Scalar projection

comp_b a = a·b/|b| = −66/8.367 = −7.889

Answer

comp_b a = −7.889

Negative: a has a component pointing against b.

2.Vectors: dot & cross product

Answer: M_y = 120 N·m

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.

Position vector from O to A

r = A − O = (3, 0, 2) − (0, 2, −1) = (3, −2, 3) m

Moment M = r × F

(r × F)_x = r_yF_z − r_zF_y = (−2)·20 − 3·60 = −220 N·m

(r × F)_y = r_zF_x − r_xF_z = 3·60 − 3·20 = 120 N·m

(r × F)_z = r_xF_y − r_yF_x = 3·60 − (−2)·60 = 300 N·m

Requested component

M_y = 120 N·m

Answer

M_y = 120 N·m

3.Beam support reactions

Answer: R_A = 110 kN

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 = 6·5 = 30 kN, x̄ = 2.5 m

Vertical equilibrium

R_A = P + W = 80 + 30 = 110 kN

Answer

R_A = 110 kN

4.Beam support reactions

Answer: M_A = 373.3 kN·m

Resultant of the triangular load

W = ½·4·10 = 20 kN

x̄ = 2L/3 = 6.667 m

Moment about the wall

M_A = W·x̄ + P·a = 20·6.667 + 40·6 = 373.3 kN·m

Answer

M_A = 373.3 kN·m

5.Truss: method of joints

Answer: F_AC = 13.33 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·6 = 30·2 → C_y = 10 kN

ΣF_y = 0: A_y = 30 − 10 = 20 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 = 3/5 = 0.6, cos γ = 4/5 = 0.8

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −10/0.6 = −16.67 kN

ΣF_x = 0: −F_AC − F_BC·cos γ = 0 → F_AC = −(−16.67)·0.8 = 13.33 kN

Answer

F_AC = 13.33 kN (tension)

6.Truss: method of joints

Answer: F_AC = 5 kN

Support reactions

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

ΣM_A = 0: C_y·6 = 35·2 − 10·4 → C_y = 5 kN

ΣF_y = 0: A_y = 35 − 5 = 30 kN

ΣF_x = 0: A_x = 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 = 4/5.657 = 0.7071, cos γ = 4/5.657 = 0.7071

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −5/0.7071 = −7.071 kN

ΣF_x = 0: −F_AC − F_BC·cos γ = 0 → F_AC = −(−7.071)·0.7071 = 5 kN

Answer

F_AC = 5 kN (tension)

7.Frames & machines

Answer: T = 2.667 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.

Whole frame: moments about A

B_y = 12·0.8/3.2 = 3 kN

Leg BC: moments about C

3·1.6 = T·1.8

T = 2.667 kN

Answer

T = 2.667 kN

8.Frames & machines

Answer: C = 4.688 kN

Support reaction at B

B_y = 10·1.8/4.8 = 3.75 kN

Tie force

T = 3.75·2.4/3.2 = 2.813 kN

Pin force on BC

C_x = 2.813 kN, C_y = −3.75 kN

C = √(2.813² + 3.75²) = 4.688 kN

Answer

C = 4.688 kN

9.Shear force & bending moment

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

ΣM_A = 0: R_B·4 = 4·4·2

R_B = 32/4 = 8 kN

ΣF_y = 0: R_A = 16 − 8 = 8 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 = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 8·3 − 4·3·1.5 = 6 kN·m

Answer

M_x = 6 kN·m

10.Shear force & bending moment

Answer: M_x = 120 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 50·2 + 5·10·5

R_B = 350/10 = 35 kN

ΣF_y = 0: R_A = 100 − 35 = 65 kN

Cut at x = 2 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 = 65·2 − 5·2·1 = 120 kN·m

Answer

M_x = 120 kN·m

11.Centroid & second moment of area

Answer: I_x = 229.5 × 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 = (48000·120 − 1257·144)/(48000 − 1257) = 119.4 mm

Full plate about ȳ

200·240³/12 + 48000·0.6452² = 230.4 × 10⁶ mm⁴

Hole about ȳ

π·40⁴/64 + 1257·24.65² = 0.8889 × 10⁶ mm⁴

Subtract

I_x = 229.5 × 10⁶ mm⁴

Answer

I_x = 229.5 × 10⁶ mm⁴

12.Centroid & second moment of area

Answer: σ_bottom = 375.6 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: A_f = 160·15 = 2400 mm², y_f = 120 + 15/2 = 127.5 mm

web: A_w = 8·120 = 960 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2400·127.5 + 960·60)/3360 = 108.2 mm

Parallel-axis theorem

flange: 160·15³/12 + 2400·(127.5 − 108.2)² = 937653 mm⁴

web: 8·120³/12 + 960·(60 − 108.2)² = 3.384 × 10⁶ mm⁴

I_x = 937653 + 3.384 × 10⁶ = 4.321 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 108.2 mm

σ = M·y/I = 1.5 × 10⁷·108.2/4.321 × 10⁶ = 375.6 MPa

Answer

σ_bottom = 375.6 MPa

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