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 value of k for which the vectors a = (6, −6, 5) and b = (−3, k, −5) are perpendicular.

a = (6, −6, 5)b = (−3, k, −5)

k =

2.Vectors: dot & cross productHard

For a = (−4, 1, 3) and b = (3, −3, −4), find the x-component (a × b)_x of the cross product a × b.

a = (−4, 1, 3)b = (3, −3, −4)

(a × b)_x =

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 4 m from A and a counter-clockwise couple M at 2.5 m from A. Find the vertical reaction R_B, taking upward as positive.

L = 6 mP = 35 kN at 4 mM = 40 kN·m (counter-clockwise) at 2.5 m

R_B = kN

4.Beam support reactionsHard

The beam has a pin at A and a roller at B, 8 m from A, and overhangs B to the free end C. It carries a point load P at 3 m from A and a uniformly distributed load w from 5 m to the free end at 10 m. Find the vertical reaction R_A, taking upward as positive.

AB = 8 mAC = 10 mP = 20 kN at 3 mw = 5 kN/m from 5 to 10 m

R_A = 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 = 6 mheight = 2 mB is 1 m from Avertical load = 30 kN

F_BC = kN

6.Truss: method of jointsHard

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

span = 4 mheight = 3 mB is 1 m from Avertical load = 10 kNhorizontal load = 5 kN ←

F_BC = kN

7.Frames & machinesMedium

The bell crank shown turns about the pin O. A vertical force of 300 N at A (100 mm from O) is balanced by a horizontal force at B (100 mm above O). Find the magnitude of the pin reaction F_O.

F = 300 NOA = 100 mmOB = 100 mm

F_O = N

8.Frames & machinesHard

The A-frame shown stands on a pin at A and a roller at B, 2.4 m apart, with its apex pin C 4 m high and the horizontal tie DE 1.5 m above the ground. A vertical load of 12 kN acts on leg AC at K, 0.3 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.

P = 12 kNAB = 2.4 mheight of C = 4 mheight of DE = 1.5 mhorizontal distance A to K = 0.3 m

C = 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.

60 kN @ 2 m30 kN @ 1 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

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.

15 kN @ 12 m45 kN @ 2 m10 kN/m, 0–8 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 150 mm × 240 mm plate has a circular hole of diameter 30 mm on its 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 = 150 mmh = 240 mmd = 30 mmhole centre: y = 84 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 5 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 140 mmt_f = 20 mmt_w = 20 mmh_w = 280 mmM = 5 kN·m

σ_top = MPa

3Solutions

1.Vectors: dot & cross product

Answer: k = −7.167

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 equals zero

a·b = 6·(−3) + (−6)·k + 5·(−5) = 0

−43 + (−6)·k = 0

Solve for k

k = −(−43)/(−6) = −7.167

Answer

k = −7.167

2.Vectors: dot & cross product

Answer: (a × b)_x = 5

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, component by component

(a × b)_x = a_yb_z − a_zb_y = 1·(−4) − 3·(−3) = 5

(a × b)_y = a_zb_x − a_xb_z = 3·3 − (−4)·(−4) = −7

(a × b)_z = a_xb_y − a_yb_x = (−4)·(−3) − 1·3 = 9

Requested component

(a × b)_x = 5

Answer

(a × b)_x = 5

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

3.Beam support reactions

Answer: R_B = 16.67 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·6 − 35·4 + 40 = 0

R_B = 100/6 = 16.67 kN

Vertical equilibrium

R_A = 35 − 16.67 = 18.33 kN

Answer

R_B = 16.67 kN

A negative reaction acts downward.

4.Beam support reactions

Answer: R_A = 14.06 kN

Resultant of the distributed load

W = 5·5 = 25 kN, x̄ = (5 + 10)/2 = 7.5 m

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣM_A = 0: R_B·8 = 20·3 + 25·7.5

R_B = 247.5/8 = 30.94 kN

R_A = 45 − 30.94 = 14.06 kN

Answer

R_A = 14.06 kN

5.Truss: method of joints

Answer: F_BC = −13.46 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 = 30·1 → C_y = 5 kN

ΣF_y = 0: A_y = 30 − 5 = 25 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 = 2/5.385 = 0.3714, cos γ = 5/5.385 = 0.9285

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −5/0.3714 = −13.46 kN

Answer

F_BC = −13.46 kN (compression)

6.Truss: method of joints

Answer: F_BC = 1.768 kN

Support reactions

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

ΣM_A = 0: C_y·4 = 10·1 − 5·3 → C_y = −1.25 kN

ΣF_y = 0: A_y = 10 − −1.25 = 11.25 kN

ΣF_x = 0: A_x = 5 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 = 3/4.243 = 0.7071, cos γ = 3/4.243 = 0.7071

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −−1.25/0.7071 = 1.768 kN

Answer

F_BC = 1.768 kN (tension)

7.Frames & machines

Answer: F_O = 424.3 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).

Force at B

Q = 300·100/100 = 300 N

Pin reaction

O_x = 300 N, O_y = 300 N

F_O = √(300² + 300²) = 424.3 N

Answer

F_O = 424.3 N

8.Frames & machines

Answer: C = 1.664 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.

Support reaction at B

B_y = 12·0.3/2.4 = 1.5 kN

Tie force

T = 1.5·1.2/2.5 = 0.72 kN

Pin force on BC

C_x = 0.72 kN, C_y = −1.5 kN

C = √(0.72² + 1.5²) = 1.664 kN

Answer

C = 1.664 kN

9.Shear force & bending moment

Answer: C (x = 2 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 = 65 kN, R₂ = 25 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 = 65, M_C = 100 kN·m, M_D = 75, M_E = 50, M_F = 25, M_G = 0

Answer

Mark C: M_C = 100 kN·m

10.Shear force & bending moment

Answer: B (x = 2 m)

Support reactions

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

R₁ = 66.25 kN, R₂ = 73.75 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 = 112.5 kN·m, M_C = 95, M_D = 37.5, M_E = −60, M_F = −30, M_G = 0

Answer

Mark B: M_B = 112.5 kN·m

11.Centroid & second moment of area

Answer: I_x = 171.8 × 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 = (36000·120 − 706.9·84)/(36000 − 706.9) = 120.7 mm

Full plate about ȳ

150·240³/12 + 36000·−0.721² = 172.8 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−36.72² = 0.9929 × 10⁶ mm⁴

Subtract

I_x = 171.8 × 10⁶ mm⁴

Answer

I_x = 171.8 × 10⁶ mm⁴

12.Centroid & second moment of area

Answer: σ_top = 6.99 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 = 140·20 = 2800 mm², y_f = 280 + 20/2 = 290 mm

web: A_w = 20·280 = 5600 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2800·290 + 5600·140)/8400 = 190 mm

Parallel-axis theorem

flange: 140·20³/12 + 2800·(290 − 190)² = 2.809 × 10⁷ mm⁴

web: 20·280³/12 + 5600·(140 − 190)² = 5.059 × 10⁷ mm⁴

I_x = 2.809 × 10⁷ + 5.059 × 10⁷ = 7.868 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 300 − 190 = 110 mm

σ = M·y/I = 5 × 10⁶·110/7.868 × 10⁷ = 6.99 MPa

Answer

σ_top = 6.99 MPa

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