Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 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

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

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)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Beam deflection

Point load at midspan: δ = PL³/(48EI)
Uniform load, simply supported: δ = 5wL⁴/(384EI)
Cantilever, end load: δ = PL³/(3EI)
Cantilever, uniform load: δ = wL⁴/(8EI)
Superposition: δ = δ₁ + δ₂

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

Find the elongation δ of the aluminium bar under the axial load.

P = 50 kNd = 25 mmL = 3 mE = 70 GPa

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 20 kNsegment 1: Ø35 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø16 mm, 2 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 50 mm and length 1 m; segment BC has diameter 25 mm and length 2 m. A torque of 200 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 200 N·md₁ = 50 mm, L₁ = 1 md₂ = 25 mm, L₂ = 2 m

τ_max = MPa

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1000 rpmd_o = 60 mmd_i = 36 mm

τ_max = MPa

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 7.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 2 kN/m from 0 to 10 mx = 7.5 m

M_x = kN·m

6.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 m55 kN @ 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)

7.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 = 300 mmt_f = 20 mmt_w = 14 mmh_w = 260 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 200 mm wide and 14 mm thick, an overall depth of 500 mm and a web 6 mm thick. It carries a bending moment of 80 kN·m about the axis x. Find the size of the maximum bending stress σ_max in the section.

B = 200 mmH = 500 mmt_f = 14 mmt_w = 6 mmM = 80 kN·m

σ_max = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 110 MPa and σᵧ = 30 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 120 MPa?

σ_x = 110 MPaσ_y = 30 MPaσ₁ = 120 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 50 mm carries a bending moment of 500 N·m together with a torque of 2000 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ₁.

d = 50 mmM = 500 N·mT = 2000 N·m

σ₁ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 2.5 m from A. Find the deflection δ_B at the free end B.

P = 30 kNa = 2.5 mL = 4 mE = 200 GPa (steel)I = 120 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1 m from support B. Find the deflection δ_mid at midspan.

P = 12 kNb = 1 mL = 6 mE = 200 GPa (steel)I = 120 × 10⁶ mm⁴

δ_mid = mm

13.Column bucklingMedium

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 200 GPa (steel)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 4 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 4.365 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·25²/4 = 490.9 mm²

σ = P/A = 50·1000/490.9 = 101.9 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 101.9/70000 = 0.001455 = 1455 µε

Elongation

δ = ε·L = 0.001455·3000 mm = 4.365 mm

Answer

δ = 4.365 mm

2.Axial stress & strain

Answer: σ_max = 99.47 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·35²/4 = 962.1 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 20.79 MPa

σ₂ = 99.47 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 99.47 MPa

Answer

σ_max = 99.47 MPa

3.Torsion of shafts

Answer: τ_max = 65.19 MPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Stress in each segment

T = 200 N·m = 200000 N·mm

τ_AB = 16·200000/(π·50³) = 8.149 MPa

τ_BC = 16·200000/(π·25³) = 65.19 MPa

Answer

τ_max = 65.19 MPa

4.Torsion of shafts

Answer: τ_max = 25.87 MPa

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 100000/104.7 = 954.9 N·m

Polar second moment of area

J = π(60⁴ − 36⁴)/32 = 1.107 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 954930·30/1.107 × 10⁶ = 25.87 MPa

Answer

τ_max = 25.87 MPa

5.Shear force & bending moment

Answer: M_x = 18.75 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 = 2·10 = 20 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 2·10·5

R_B = 100/10 = 10 kN

ΣF_y = 0: R_A = 20 − 10 = 10 kN

Cut at x = 7.5 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 = 10·7.5 − 2·7.5·3.75 = 18.75 kN·m

Answer

M_x = 18.75 kN·m

6.Shear force & bending moment

Answer: E (x = 8 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 = 8 kN, R₂ = 62 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 = 16, M_C = 32, M_D = 48, M_E = 64 kN·m, M_F = −30, M_G = 0

Answer

Mark E: M_E = 64 kN·m

7.Centroid & second moment of area

Answer: I_x = 65.11 × 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 = 300·20 = 6000 mm², y_f = 260 + 20/2 = 270 mm

web: A_w = 14·260 = 3640 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·270 + 3640·130)/9640 = 217.1 mm

Parallel-axis theorem

flange: 300·20³/12 + 6000·(270 − 217.1)² = 1.697 × 10⁷ mm⁴

web: 14·260³/12 + 3640·(130 − 217.1)² = 4.814 × 10⁷ mm⁴

I_x = 1.697 × 10⁷ + 4.814 × 10⁷ = 6.511 × 10⁷ mm⁴ = 65.11 × 10⁶ mm⁴

Answer

I_x = 65.11 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_max = 52.17 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 = 500 − 2·14 = 472 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (200·500³ − 194·472³)/12

I_x = 3.833 × 10⁸ mm⁴

Bending stress

y = 250 mm

σ = M·y/I = 8 × 10⁷·250/3.833 × 10⁸ = 52.17 MPa

Answer

σ_max = 52.17 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 30 MPa

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Required radius

C = (110 + 30)/2 = 70 MPa

R = σ₁ − C = 120 − 70 = 50 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(50² − 40²) = 30 MPa

Answer

τ_xy = 30 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 104.4 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·500000/(π·50³) = 40.74 MPa

τ = 16T/(πd³) = 16·2000000/(π·50³) = 81.49 MPa

Mohr’s circle

C = σ/2 = 20.37 MPa

R = √(20.37² + 81.49²) = 84 MPa

σ₁ = C + R = 104.4 MPa

σ₂ = C − R = −63.62 MPa

Answer

σ₁ = 104.4 MPa

11.Beam deflection

Answer: δ_B = 12.37 mm

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 30000 N, a = 2500 mm, L = 4000 mm

EI = 2.4 × 10¹³ N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 6.51 mm

θ_P = Pa²/(2EI) = 0.003906 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 12.37 mm

Answer

δ_B = 12.37 mm

12.Beam deflection

Answer: δ_mid = 1.083 mm

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 12000 N, b = 1000 mm, L = 6000 mm

EI = 2.4 × 10¹³ N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 12000·1000·(3·6000² − 4·1000²)/(48·2.4 × 10¹³) = 1.083 mm

Answer

δ_mid = 1.083 mm

13.Column buckling

Answer: P_cr = 22.21 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·180000/4000² = 22207 N = 22.21 kN

Answer

P_cr = 22.21 kN

14.Column buckling

Answer: σ_cr = 12.34 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 1·4000 = 4000 mm

KL/r = 4000/10 = 400

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa

Answer

σ_cr = 12.34 MPa

ExamProblems · examproblems.com/courses/mechanics-of-materials