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 = (−1, 2, 5) and b = (−2, 4, 5).

a = (−1, 2, 5)b = (−2, 4, 5)

a·b =

2.Vectors: dot & cross productHard

Find the area A of the triangle with corners A(4, 1, 0), B(−1, 2, −1) and C(−1, −1, 2). Coordinates are in metres.

A(4, 1, 0)B(−1, 2, −1)C(−1, −1, 2)

A = m²

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 2.5 m from A and a clockwise couple M at 3.5 m from A. Find the vertical reaction R_A, taking upward as positive.

L = 6 mP = 50 kN at 2.5 mM = 40 kN·m (clockwise) at 3.5 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 = 8 mP = 45 kN at 7 mw = 4 kN/m from 1 to 5 m

R_A = kN

5.Truss: method of jointsMedium

Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.

D: 10 kN ↓G: 25 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

6.Truss: method of jointsHard

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 = 3 mB is 3 m from Avertical load = 30 kNhorizontal load = 30 kN →

F_AB = kN

7.Frames & machinesMedium

The A-frame shown stands on a pin at A and a roller at B, 2.4 m apart. The legs meet at the pin C, 3 m high, and the horizontal tie DE is 0.8 m above the ground. A load of 5 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.

P = 5 kNAB = 2.4 mheight of C = 3 mheight of DE = 0.8 m

C = kN

8.Frames & machinesHard

The frame shown has pins at A, B and C; A and C are fixed to the ground 4 m apart, and B is 4 m high and 1.2 m to the right of A. Member AB carries a vertical load of 10 kN at K, 0.6 m (horizontally) from A, and a horizontal load of 8 kN to the right acts at B on member AB. Find the size of the horizontal reaction A_x at A.

P = 10 kNQ = 8 kNL = 4 mH = 4 mhorizontal distance A to B = 1.2 mhorizontal distance A to K = 0.6 m

A_x = kN

9.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 1.5 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 = 6.5 mP₁ = 45 kN at 3 mP₂ = 15 kN at 6.5 msupports at x = 0 and x = 5 m

M_x = kN·m

10.Shear force & bending momentHard

Find the bending moment M_x at the section x = 6 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 15 kN at 6 mw = 6 kN/m from 0 to 8 mx = 6 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 = 10 mmt_w = 16 mmh_w = 100 mm

I_x = × 10⁶ mm⁴

12.Centroid & second moment of areaHard

A symmetric I-section has flanges 100 mm wide and 24 mm thick, an overall depth of 320 mm and a web 12 mm thick. It carries a bending moment of 40 kN·m about the axis x. Find the size of the maximum bending stress σ_max in the section.

B = 100 mmH = 320 mmt_f = 24 mmt_w = 12 mmM = 40 kN·m

σ_max = MPa

3Solutions

1.Vectors: dot & cross product

Answer: a·b = 35

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 = (−1)·(−2) + 2·4 + 5·5 = 2 + 8 + 25 = 35

Answer

a·b = 35

Positive: the angle between the vectors is less than 90°.

2.Vectors: dot & cross product

Answer: A = 10.61 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, 1, −1)

AC = C − A = (−5, −2, 2)

Cross product

(AB × AC)_x = AB_yAC_z − AB_zAC_y = 1·2 − (−1)·(−2) = 0

(AB × AC)_y = AB_zAC_x − AB_xAC_z = (−1)·(−5) − (−5)·2 = 15

(AB × AC)_z = AB_xAC_y − AB_yAC_x = (−5)·(−2) − 1·(−5) = 15

|AB × AC| = √(0² + 15² + 15²) = 21.21

Area of the triangle

A = ½·21.21 = 10.61 m²

Answer

A = 10.61 m²

3.Beam support reactions

Answer: R_A = 22.5 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 − 50·2.5 − 40 = 0

R_B = 165/6 = 27.5 kN

Vertical equilibrium

R_A = 50 − 27.5 = 22.5 kN

Answer

R_A = 22.5 kN

A negative reaction acts downward.

4.Beam support reactions

Answer: R_A = 15.63 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 = 4·4 = 16 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·8 = 45·7 + 4·4·3

R_B = 363/8 = 45.38 kN

ΣF_y = 0: R_A = 61 − 45.38 = 15.63 kN

Answer

R_A = 15.63 kN

5.Truss: method of joints

Answer: member BF

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 B

AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.

ΣF⊥ = 0 → F_BF = 0

Check with the method of joints

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

AB: 18.75 kN (T), BC: 18.75 kN (T), CD: 31.25 kN (T), DE: 31.25 kN (T), FG: 37.5 kN (C), GH: 37.5 kN (C), AF: 26.52 kN (C), HE: 44.19 kN (C), BF: 0, CG: 25 kN (C), DH: 10 kN (T), FC: 26.52 kN (T), HC: 8.839 kN (T)

Answer

Zero-force member: BF

6.Truss: method of joints

Answer: F_AB = 21.21 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 = 30·3 + 30·3 → C_y = 45 kN

ΣF_y = 0: A_y = 30 − 45 = −15 kN

ΣF_x = 0: A_x = −30 kN

Joint A

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

sin α = 3/4.243 = 0.7071

ΣF_y = 0: A_y + F_AB·sin α = 0

F_AB = −−15/0.7071 = 21.21 kN

Answer

F_AB = 21.21 kN (tension)

7.Frames & machines

Answer: C = 2.848 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 = 2.5 kN

T = 2.5·1.2/2.2 = 1.364 kN

Components of the pin force on BC

C_x = T = 1.364 kN, C_y = −B_y = −2.5 kN

Magnitude

C = √(1.364² + 2.5²) = 2.848 kN

Answer

C = 2.848 kN

8.Frames & machines

Answer: A_x = 1.35 kN

Direction of BC

BC = 4.883 m, cos = 0.5735, sin = 0.8192

Whole frame: moments about A

F_BC·0.8192·4 = 10·0.6 + 8·4

F_BC = 11.6 kN

Horizontal balance

A_x + 8 − 11.6·0.5735 = 0

A_x = −1.35 kN → |A_x| = 1.35 kN

Answer

A_x = 1.35 kN

The minus sign means A_x points to the left.

9.Shear force & bending moment

Answer: M_x = 40.5 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·5 = 45·3 + 15·6.5

R_B = 232.5/5 = 46.5 kN

ΣF_y = 0: R_A = 60 − 46.5 = 13.5 kN

Cut under P₁ and keep the left part

M_x = R_A·3 = 13.5·3 = 40.5 kN·m

Answer

M_x = 40.5 kN·m

10.Shear force & bending moment

Answer: M_x = 58.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·8 = 48 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 15·6 + 6·8·4

R_B = 282/8 = 35.25 kN

ΣF_y = 0: R_A = 63 − 35.25 = 27.75 kN

Cut at x = 6 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 = 27.75·6 − 6·6·3 = 58.5 kN·m

Answer

M_x = 58.5 kN·m

11.Centroid & second moment of area

Answer: I_x = 3.203 × 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 = 100·10 = 1000 mm², y_f = 100 + 10/2 = 105 mm

web: A_w = 16·100 = 1600 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1000·105 + 1600·50)/2600 = 71.15 mm

Parallel-axis theorem

flange: 100·10³/12 + 1000·(105 − 71.15)² = 1.154 × 10⁶ mm⁴

web: 16·100³/12 + 1600·(50 − 71.15)² = 2.049 × 10⁶ mm⁴

I_x = 1.154 × 10⁶ + 2.049 × 10⁶ = 3.203 × 10⁶ mm⁴ = 3.203 × 10⁶ mm⁴

Answer

I_x = 3.203 × 10⁶ mm⁴

12.Centroid & second moment of area

Answer: σ_max = 51 MPa

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 320 − 2·24 = 272 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (100·320³ − 88·272³)/12

I_x = 1.255 × 10⁸ mm⁴

Bending stress

y = 160 mm

σ = M·y/I = 4 × 10⁷·160/1.255 × 10⁸ = 51 MPa

Answer

σ_max = 51 MPa

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

Multiply matching components and add

Answer

Edge vectors

Cross product

Area of the triangle

Answer

Moments about A

Vertical equilibrium

Answer

Free-body diagram

Equilibrium

Answer

Joint B

Check with the method of joints

Answer

Support reactions

Joint A

Answer

Tie force from leg BC

Components of the pin force on BC

Magnitude

Answer

Direction of BC

Whole frame: moments about A

Horizontal balance

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Cut at x = 6 m and keep the left part

Answer

Split into two rectangles

Centroid

Parallel-axis theorem

Answer

Outer rectangle minus the two side cut-outs

Bending stress

Answer

Cut under P₁ and keep the left part