Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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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 30 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 110 GPaL = 600 mm

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø80/Ø60 mm, E = 110 GPa, σ_allow = 60 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid brass 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 = 100 N·md = 20 mmL = 2.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 200 N·m is applied at C, 2.5 m from A and 0.5 m from B. Find the angle of rotation φ_C of section C, in degrees.

d = 40 mmT₀ = 200 N·ma = 2.5 mb = 0.5 mG = 26 GPa

φ_C = °

5.Shear force & bending momentMedium

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 mw = 10 kN/m from 0 to 8 mx = 6 m

M_x = kN·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 = 6 mw = 16 kN/m from 3 to 5 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 180 mm wide with two upright legs, 190 mm high overall. Base and legs are all 14 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 180 mmh = 190 mmt = 14 mm

ȳ = mm

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 120 mm wide and 18 mm thick, an overall depth of 440 mm and a web 6 mm thick. It carries a bending moment of 30 kN·m about the axis x. Find the size of the bending stress σ_j at the junction of the web and the top flange.

B = 120 mmH = 440 mmt_f = 18 mmt_w = 6 mmM = 30 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 160 MPa, σᵧ = −70 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 160 MPaσ_y = −70 MPaτ_xy = −60 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 90 MPa, σᵧ = 0 MPa and τₓᵧ = −50 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 90 MPaσ_y = 0 MPaτ_xy = −50 MPaσ_Y = 350 MPa

n =

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 8 kN/mL = 3 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 5 mP = 55 kN at 2.5 m

R_B = kN

13.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)d = 60 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 50 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 50 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.1252 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·30²/4 = 706.9 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·2199 = 3.833 × 10⁸ N

Shortening

δ = 80000·600/3.833 × 10⁸ = 0.1252 mm

Answer

δ = 0.1252 mm

2.Axial stress & strain

Answer: P_max = 166.2 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·2199 = 3.047 × 10⁸ N

Governing strain

ε = min(0.001, 0.0005455) = 0.0005455 (bronze)

Largest load

P_max = 0.0005455·3.047 × 10⁸ = 166219 N = 166.2 kN

Answer

P_max = 166.2 kN

3.Torsion of shafts

Answer: φ = 23.38 °

Polar second moment of area

J = π·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·2500/(39000·15708) = 0.4081 rad

φ = 0.4081·180/π = 23.38°

Answer

φ = 23.38°

4.Torsion of shafts

Answer: φ_C = 0.7307 °

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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 200

T_A·a = T_B·b

T_A = 200·0.5/3 = 33.33 N·m

T_B = 200·2.5/3 = 166.7 N·m

Rotation of C

Use segment AC: it is fixed at A and carries T_A.

J = π·40⁴/32 = 251327 mm⁴

G = 26 GPa = 26000 MPa

φ_C = T_A·a/(GJ) = 33333·2500/(26000·251327) = 0.01275 rad = 0.7307°

Answer

φ_C = 0.7307°

5.Shear force & bending moment

Answer: M_x = 60 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·8 = 80 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 10·8·4

R_B = 320/8 = 40 kN

ΣF_y = 0: R_A = 80 − 40 = 40 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 = 40·6 − 10·6·3 = 60 kN·m

Answer

M_x = 60 kN·m

6.Shear force & bending moment

Answer: M_max = 35.56 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 = 16·2 = 32 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·6 = 16·2·4

R_B = 128/6 = 21.33 kN

ΣF_y = 0: R_A = 32 − 21.33 = 10.67 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 − 3) = 0

x₀ = 3 + 10.67/16 = 3.667 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 3)²/2 = 10.67·3.667 − 16·0.6667²/2 = 35.56 kN·m

Answer

M_max = 35.56 kN·m

7.Centroid & second moment of area

Answer: ȳ = 69.86 mm

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 = 180·14 = 2520 mm², y₁ = 7 mm

two legs: A₂ = 2·14·176 = 4928 mm², y₂ = 14 + 176/2 = 102 mm

Centroid

ȳ = (2520·7 + 4928·102)/7448 = 69.86 mm

Answer

ȳ = 69.86 mm

8.Centroid & second moment of area

Answer: σ_j = 26.88 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 = 440 − 2·18 = 404 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (120·440³ − 114·404³)/12

I_x = 2.254 × 10⁸ mm⁴

Bending stress

y = 220 − 18 = 202 mm

σ = M·y/I = 3 × 10⁷·202/2.254 × 10⁸ = 26.88 MPa

Answer

σ_j = 26.88 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −13.78 °

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−60, 160 − −70) = −27.55°

Angle on the element

θ_p = −13.78°

Answer

θ_p = −13.78°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 2.802

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 = 45 MPa, R = 67.27 MPa

σ₁ = 112.3 MPa, σ₂ = −22.27 MPa, σ₃ = 0

Von Mises

σ_vM = √(112.3² − 112.3·(−22.27) + (−22.27)²) = 124.9 MPa

n = 350/124.9 = 2.802

Answer

n = 2.802

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

11.Beam deflection

Answer: δ = 11.25 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 = 3000 mm

EI = 7.2 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 8·3000⁴/(8·7.2 × 10¹²) = 11.25 mm

Answer

δ = 11.25 mm

12.Beam deflection

Answer: R_B = 17.19 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = 5PL³/(48EI) → R_B = 5P/16 = 5·55/16 = 17.19 kN

Answer

R_B = 17.19 kN

13.Column buckling

Answer: P_cr = 313.9 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 = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN

Answer

P_cr = 313.9 kN

14.Column buckling

Answer: d = 41.28 mm

Required critical load

P_cr = n·P = 2.5·50000 = 125000 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·125000·1500²/(π³·200000))^(1/4) = 41.28 mm

Answer

d = 41.28 mm

Check: the slenderness KL/r is 145.4, so the column is slender and Euler applies.

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