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 value of k for which the vectors a = (6, 4, −4) and b = (6, −6, k) are perpendicular.

a = (6, 4, −4)b = (6, −6, k)

k =

2.Vectors: dot & cross productHard

A force F = (−40, 20, −60) N acts at point A(−1, 3, 1). Coordinates are in metres. Find the y-component M_y of the moment of the force about the origin O(0, 0, 0).

F = (−40, 20, −60) NA(−1, 3, 1) mO(0, 0, 0) m

M_y = N·m

3.Beam support reactionsMedium

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

L = 8 mP₁ = 15 kN at 4 mP₂ = 25 kN at 5 m

R_A = kN

4.Beam support reactionsHard

The simply supported beam has a pin at A and a roller at B. A load P acts at 2.5 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 1.5 m from A. Find the magnitude of the total reaction force R_A at the pin A.

L = 4 mP = 55 kN at 2.5 m, θ = 40°Q = 40 kN at 1.5 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.

F: 25 kN ↓H: 40 kN ↓

1member AB
2member BC
3member CD
4member DE
5member FG
6member GH
7member AF
8member HE
9member BF
10member CG
11member DH
12member BG
13member DG

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 = 5 mheight = 2 mB is 2 m from Avertical load = 40 kNhorizontal load = 5 kN →

F_AC = kN

7.Frames & machinesMedium

The A-frame shown stands on a pin at A and a roller at B, 6 m apart, with its apex pin C 2 m high and the horizontal tie DE 1 m above the ground. A vertical load of 4 kN acts on leg AC at K, 2.25 m (horizontally) from A. Find the tension T in the tie.

P = 4 kNAB = 6 mheight of C = 2 mheight of DE = 1 mhorizontal distance A to K = 2.25 m

T = kN

8.Frames & machinesHard

The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 0.8 m below A). The beam carries a uniform load of 1.5 kN/m over its full length and a point load of 3 kN at its free end B. Find the force F_CD in the strut.

w = 1.5 kN/mP = 3 kNAD = 1.6 mAB = 2.5 mAC = 0.8 m

F_CD = kN

9.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 10 kN/m from 0 to 5 mx = 2.5 m

M_x = kN·m

10.Shear force & bending momentHard

Find the bending moment M_x at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 40 kN at 1 mw = 2 kN/m from 0 to 5 mx = 1 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 = 180 mmt_f = 20 mmt_w = 16 mmh_w = 120 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 bending stress σ_bottom at the bottom edge.

b_f = 280 mmt_f = 25 mmt_w = 10 mmh_w = 200 mmM = 5 kN·m

σ_bottom = MPa

3Solutions

1.Vectors: dot & cross product

Answer: k = 3

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 = 6·6 + 4·(−6) + (−4)·k = 0

12 + (−4)·k = 0

Solve for k

k = −12/(−4) = 3

Answer

k = 3

2.Vectors: dot & cross product

Answer: M_y = −100 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 = (−1, 3, 1) − (0, 0, 0) = (−1, 3, 1) m

Moment M = r × F

(r × F)_x = r_yF_z − r_zF_y = 3·(−60) − 1·20 = −200 N·m

(r × F)_y = r_zF_x − r_xF_z = 1·(−40) − (−1)·(−60) = −100 N·m

(r × F)_z = r_xF_y − r_yF_x = (−1)·20 − 3·(−40) = 100 N·m

Requested component

M_y = −100 N·m

Answer

M_y = −100 N·m

3.Beam support reactions

Answer: R_A = 16.88 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.

ΣM_A = 0: R_B·8 = 15·4 + 25·5

R_B = 185/8 = 23.13 kN

ΣF_y = 0: R_A = 40 − 23.13 = 16.88 kN

Answer

R_A = 16.88 kN

4.Beam support reactions

Answer: R_A = 56.91 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.

Components of P

P_v = 55·sin 40° = 35.35 kN

P_h = 55·cos 40° = 42.13 kN

Reactions

ΣM_A = 0: R_B·4 = 35.35·2.5 + 40·1.5 → R_B = 37.1 kN

A_y = 35.35 + 40 − 37.1 = 38.26 kN

A_x = P_h = 42.13 kN

Resultant at A

R_A = √(42.13² + 38.26²) = 56.91 kN

Answer

R_A = 56.91 kN

5.Truss: method of joints

Answer: member CG

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 C

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

ΣF⊥ = 0 → F_CG = 0

Check with the method of joints

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

AB: 28.75 kN (T), BC: 32.5 kN (T), CD: 32.5 kN (T), DE: 36.25 kN (T), FG: 28.75 kN (C), GH: 36.25 kN (C), AF: 40.66 kN (C), HE: 51.27 kN (C), BF: 3.75 kN (T), CG: 0, DH: 3.75 kN (C), BG: 5.303 kN (C), DG: 5.303 kN (T)

Answer

Zero-force member: CG

6.Truss: method of joints

Answer: F_AC = 27 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·5 = 40·2 + 5·2 → C_y = 18 kN

ΣF_y = 0: A_y = 40 − 18 = 22 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 = 2/3.606 = 0.5547, cos γ = 3/3.606 = 0.8321

ΣF_y = 0: C_y + F_BC·sin γ = 0 → F_BC = −18/0.5547 = −32.45 kN

ΣF_x = 0: −F_AC − F_BC·cos γ = 0 → F_AC = −(−32.45)·0.8321 = 27 kN

Answer

F_AC = 27 kN (tension)

7.Frames & machines

Answer: T = 4.5 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.

Whole frame: moments about A

B_y = 4·2.25/6 = 1.5 kN

Leg BC: moments about C

1.5·3 = T·1

T = 4.5 kN

Answer

T = 4.5 kN

8.Frames & machines

Answer: F_CD = 17.03 kN

Resultant of the uniform load

w·L = 1.5·2.5 = 3.75 kN at 1.25 m

Moments about A for the beam

F_CD·0.4472·1.6 = 3.75·1.25 + 3·2.5 = 12.19 kN·m

F_CD = 17.03 kN

Answer

F_CD = 17.03 kN

9.Shear force & bending moment

Answer: M_x = 31.25 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 = 10·5 = 50 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 10·5·2.5

R_B = 125/5 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 2.5 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 = 25·2.5 − 10·2.5·1.25 = 31.25 kN·m

Answer

M_x = 31.25 kN·m

10.Shear force & bending moment

Answer: M_x = 36 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 2·5 = 10 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 40·1 + 2·5·2.5

R_B = 65/5 = 13 kN

ΣF_y = 0: R_A = 50 − 13 = 37 kN

Cut at x = 1 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 = 37·1 − 2·1·0.5 = 36 kN·m

Answer

M_x = 36 kN·m

11.Centroid & second moment of area

Answer: I_x = 8.56 × 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 = 180·20 = 3600 mm², y_f = 120 + 20/2 = 130 mm

web: A_w = 16·120 = 1920 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3600·130 + 1920·60)/5520 = 105.7 mm

Parallel-axis theorem

flange: 180·20³/12 + 3600·(130 − 105.7)² = 2.254 × 10⁶ mm⁴

web: 16·120³/12 + 1920·(60 − 105.7)² = 6.306 × 10⁶ mm⁴

I_x = 2.254 × 10⁶ + 6.306 × 10⁶ = 8.56 × 10⁶ mm⁴ = 8.56 × 10⁶ mm⁴

Answer

I_x = 8.56 × 10⁶ mm⁴

12.Centroid & second moment of area

Answer: σ_bottom = 35.09 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 280·25 = 7000 mm², y_f = 200 + 25/2 = 212.5 mm

web: A_w = 10·200 = 2000 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (7000·212.5 + 2000·100)/9000 = 187.5 mm

Parallel-axis theorem

flange: 280·25³/12 + 7000·(212.5 − 187.5)² = 4.74 × 10⁶ mm⁴

web: 10·200³/12 + 2000·(100 − 187.5)² = 2.198 × 10⁷ mm⁴

I_x = 4.74 × 10⁶ + 2.198 × 10⁷ = 2.672 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 187.5 mm

σ = M·y/I = 5 × 10⁶·187.5/2.672 × 10⁷ = 35.09 MPa

Answer

σ_bottom = 35.09 MPa

ExamProblems · examproblems.com/courses/statics