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

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa). Both are 300 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 70 GPaL = 300 mm

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø100/Ø50 mm, E = 120 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 40 mm, inner diameter 24 mm, length 2.5 m) is fixed at one end and twisted by a torque of 100 N·m at the other. Find the angle of twist φ in degrees.

T = 100 N·md_o = 40 mmd_i = 24 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 15 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 15 kWN = 300 rpmd_o = 75 mmd_i = 56 mm

τ_max = MPa

5.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 m45 kN @ 5 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)

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment M_max in the beam.

L = 4 mw = 20 kN/m from 0 to 3 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 70 mm and inside diameter d = 42 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 70 mmd = 42 mm

r = mm

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 210 mm wide and 20 mm thick, an overall depth of 320 mm and a web 12 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 = 210 mmH = 320 mmt_f = 20 mmt_w = 12 mmM = 80 kN·m

σ_max = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −10 MPa, σᵧ = −40 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = −10 MPaσ_y = −40 MPaτ_xy = −40 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −50 MPa, σᵧ = 110 MPa and τₓᵧ = −40 MPa (σ_z = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −50 MPaσ_y = 110 MPaτ_xy = −40 MPaσ_Y = 275 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 5 kN/mL = 5 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

θ_A = rad

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 22 kNw = 2 kN/mL = 4 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 5 mE = 70 GPa (aluminium)d = 60 mm

P_cr = kN

14.Column bucklingHard

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

L = 3 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.1085 mm

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·5027 = 4.147 × 10⁸ N

Shortening

δ = 150000·300/4.147 × 10⁸ = 0.1085 mm

Answer

δ = 0.1085 mm

2.Axial stress & strain

Answer: P_max = 469.6 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 120·10³·5890 = 8.05 × 10⁸ N

Governing strain

ε = min(0.001, 0.0005833) = 0.0005833 (copper)

Largest load

P_max = 0.0005833·8.05 × 10⁸ = 469603 N = 469.6 kN

Answer

P_max = 469.6 kN

3.Torsion of shafts

Answer: φ = 2.518 °

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.

Polar second moment of area

J = π(40⁴ − 24⁴)/32 = 218755 mm⁴

Angle of twist

T = 100 N·m = 100000 N·mm

L = 2.5 m = 2500 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 100000·2500/(26000·218755) = 0.04395 rad

φ = 0.04395·180/π = 2.518°

Answer

φ = 2.518°

4.Torsion of shafts

Answer: τ_max = 8.364 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 15000/31.42 = 477.5 N·m

Polar second moment of area

J = π(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 477465·37.5/2.141 × 10⁶ = 8.364 MPa

Answer

τ_max = 8.364 MPa

5.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 = 47.5 kN, R₂ = 57.5 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 = 47.5, M_C = 95 kN·m, M_D = 82.5, M_E = 70, M_F = 57.5, M_G = 0

Answer

Mark C: M_C = 95 kN·m

6.Shear force & bending moment

Answer: M_max = 35.16 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 20·3 = 60 kN, acting at x̄ = 1.5 m

ΣM_A = 0: R_B·4 = 20·3·1.5

R_B = 90/4 = 22.5 kN

ΣF_y = 0: R_A = 60 − 22.5 = 37.5 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 0) = 0

x₀ = 0 + 37.5/20 = 1.875 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 0)²/2 = 37.5·1.875 − 20·1.875²/2 = 35.16 kN·m

Answer

M_max = 35.16 kN·m

7.Centroid & second moment of area

Answer: r = 20.41 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 2463 mm²

I = π(D⁴ − d⁴)/64 = 1.026 × 10⁶ mm⁴

Radius of gyration

r = √(I/A) = √(1.026 × 10⁶/2463) = 20.41 mm

Answer

r = 20.41 mm

8.Centroid & second moment of area

Answer: σ_max = 60.6 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.

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 − 2·20 = 280 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (210·320³ − 198·280³)/12

I_x = 2.112 × 10⁸ mm⁴

Bending stress

y = 160 mm

σ = M·y/I = 8 × 10⁷·160/2.112 × 10⁸ = 60.6 MPa

Answer

σ_max = 60.6 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = −79.72 °

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −40/42.72 = −0.9363, sin 2θ_s = −15/42.72 = −0.3511

2θ_s = atan2(−15, −40) = −159.4°

Angle on the element

θ_s = −79.72°

check: θ_s − θ_p = −79.72 − (−34.72) = −45°

Answer

θ_s = −79.72°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 1.743

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 30 MPa, R = 89.44 MPa

σ₁ = 119.4 MPa, σ₂ = −59.44 MPa, σ₃ = 0

Von Mises

σ_vM = √(119.4² − 119.4·(−59.44) + (−59.44)²) = 157.8 MPa

n = 275/157.8 = 1.743

Answer

n = 1.743

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: θ_A = 0.003617 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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.

w = 5 N/mm, L = 5000 mm

EI = 7.2 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 5·5000³/(24·7.2 × 10¹²) = 0.003617 rad

Answer

θ_A = 0.003617 rad

That is 0.2072°.

12.Beam deflection

Answer: δ_max = 2.143 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 = 22000 N, w = 2 N/mm, L = 4000 mm, EI = 1.68 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 22000·4000³/(48·1.68 × 10¹³) = 1.746 mm

δ_w = 5wL⁴/(384EI) = 5·2·4000⁴/(384·1.68 × 10¹³) = 0.3968 mm

Add them

δ_max = 1.746 + 0.3968 = 2.143 mm

Answer

δ_max = 2.143 mm

13.Column buckling

Answer: P_cr = 70.32 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 round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·636173/2500² = 70322 N = 70.32 kN

Answer

P_cr = 70.32 kN

14.Column buckling

Answer: σ_cr = 7.676 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 = 1·3000 = 3000 mm

KL/r = 3000/10 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa

Answer

σ_cr = 7.676 MPa

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