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 25 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 50 mm, E = 100 GPa). Both are 600 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 100 GPaL = 600 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 aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 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/Ø70 mm, E = 70 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid steel shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 250 N·md = 50 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft transmits 5 kW at 300 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 5 kWN = 300 rpmτ_allow = 60 MPaθ_allow = 0.75°/mG = 26 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 12 kN/m from 0 to 4 mx = 1.5 m

M_x = kN·m

6.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 = 10 mP = 50 kN at 1 mw = 3 kN/m from 0 to 10 mx = 1 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 240 mm wide with two upright legs, 160 mm high overall. Base and legs are all 16 mm thick. Find its second moment of area I_x about the horizontal axis x through its centroid.

b = 240 mmh = 160 mmt = 16 mm

I_x = × 10⁶ mm⁴

8.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 30 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 300 mmt_f = 25 mmt_w = 20 mmh_w = 120 mmM = 30 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 130 MPa, σᵧ = 80 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 15° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 130 MPaσ_y = 80 MPaτ_xy = −10 MPaθ = 15°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 100 MPa, σᵧ = 30 MPa and τₓᵧ = 60 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = 100 MPaσ_y = 30 MPaτ_xy = 60 MPaθ = 50°

σ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 8 kN/mL = 5 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ_max = mm

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 = 7 kNw = 10 kN/mL = 2 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 4 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 75 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 75 kNn = 3L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.08731 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·25²/4 = 490.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·5890 = 6.872 × 10⁸ N

Shortening

δ = 100000·600/6.872 × 10⁸ = 0.08731 mm

Answer

δ = 0.08731 mm

2.Axial stress & strain

Answer: P_max = 378.6 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·4006 = 3.786 × 10⁸ N

Governing strain

ε = min(0.001, 0.001) = 0.001 (aluminium)

Largest load

P_max = 0.001·3.786 × 10⁸ = 378562 N = 378.6 kN

Answer

P_max = 378.6 kN

3.Torsion of shafts

Answer: φ = 0.4377 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·1500/(80000·613592) = 0.007639 rad

φ = 0.007639·180/π = 0.4377°

Answer

φ = 0.4377°

4.Torsion of shafts

Answer: d = 46.72 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 5000/31.42 = 159.2 N·m

Diameter for strength

T = 159.2 N·m = 159155 N·mm

d_τ = ∛(16·159155/(π·60)) = 23.82 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·159155/(π·26000·1.309 × 10⁻⁵))^1/4 = 46.72 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(23.82, 46.72) = 46.72 mm

Answer

d = 46.72 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 22.5 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.

UDL resultant: W = w·ℓ = 12·4 = 48 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 12·4·2

R_B = 96/4 = 24 kN

ΣF_y = 0: R_A = 48 − 24 = 24 kN

Cut at x = 1.5 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 = 24·1.5 − 12·1.5·0.75 = 22.5 kN·m

Answer

M_x = 22.5 kN·m

6.Shear force & bending moment

Answer: M_x = 58.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 50·1 + 3·10·5

R_B = 200/10 = 20 kN

ΣF_y = 0: R_A = 80 − 20 = 60 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 = 60·1 − 3·1·0.5 = 58.5 kN·m

Answer

M_x = 58.5 kN·m

7.Centroid & second moment of area

Answer: I_x = 21.45 × 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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 240·16 = 3840 mm², y₁ = 8 mm

two legs: A₂ = 2·16·144 = 4608 mm², y₂ = 16 + 144/2 = 88 mm

Centroid

ȳ = (3840·8 + 4608·88)/8448 = 51.64 mm

Parallel-axis theorem

base: 240·16³/12 + 3840·(8 − 51.64)² = 7.394 × 10⁶ mm⁴

two legs: 2·16·144³/12 + 4608·(88 − 51.64)² = 1.406 × 10⁷ mm⁴

I_x = 7.394 × 10⁶ + 1.406 × 10⁷

I_x = 2.145 × 10⁷ mm⁴ = 21.45 × 10⁶ mm⁴

Answer

I_x = 21.45 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 70.34 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 300·25 = 7500 mm², y_f = 120 + 25/2 = 132.5 mm

web: A_w = 20·120 = 2400 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (7500·132.5 + 2400·60)/9900 = 114.9 mm

Parallel-axis theorem

flange: 300·25³/12 + 7500·(132.5 − 114.9)² = 2.707 × 10⁶ mm⁴

web: 20·120³/12 + 2400·(60 − 114.9)² = 1.012 × 10⁷ mm⁴

I_x = 2.707 × 10⁶ + 1.012 × 10⁷ = 1.283 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 145 − 114.9 = 30.08 mm

σ = M·y/I = 3 × 10⁷·30.08/1.283 × 10⁷ = 70.34 MPa

Answer

σ_top = 70.34 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 88.35 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.

Transformation equation for the y′ face

σ_y′ = 105 − 25·0.866 − (−10)·0.5 = 88.35 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 121.7 MPa, σ_x′ + σ_y′ = 210 = σ_x + σ_y

Answer

σ_y′ = 88.35 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 118 MPa

Transformation equation

σ_θ = 65 + 35·−0.1736 + 60·0.9848 = 118 MPa

Answer

σ_θ = 118 MPa

11.Beam deflection

Answer: δ_max = 2.035 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.

w = 8 N/mm, L = 5000 mm

EI = 3.2 × 10¹³ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·5000⁴/(384·3.2 × 10¹³) = 2.035 mm

Answer

δ_max = 2.035 mm

12.Beam deflection

Answer: δ_max = 1.912 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 = 7000 N, w = 10 N/mm, L = 2000 mm, EI = 1.7 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 7000·2000³/(48·1.7 × 10¹²) = 0.6863 mm

δ_w = 5wL⁴/(384EI) = 5·10·2000⁴/(384·1.7 × 10¹²) = 1.225 mm

Add them

δ_max = 0.6863 + 1.225 = 1.912 mm

Answer

δ_max = 1.912 mm

13.Column buckling

Answer: P_cr = 86.82 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/4000² = 86818 N = 86.82 kN

Answer

P_cr = 86.82 kN

14.Column buckling

Answer: d = 47.81 mm

Required critical load

P_cr = n·P = 3·75000 = 225000 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·225000·1500²/(π³·200000))^(1/4) = 47.81 mm

Answer

d = 47.81 mm

Check: the slenderness KL/r is 125.5, so the column is slender and Euler applies.

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