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 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 80 kN. Find the shortening δ.

P = 80 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 70 GPaL = 300 mm

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 95 kNsegment 1: Ø40 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 0.5 m, brass (E = 100 GPa)

δ = mm

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 = 500 N·md = 50 mmL = 0.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 3000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 3000 rpmd_o = 40 mmd_i = 20 mm

τ_max = MPa

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 3 kN/m from 0 to 10 mx = 2 m

V_x = kN

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.

50 kN @ 0 m30 kN @ 6 m20 kN @ 2 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)

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 = 120 mmt_f = 15 mmt_w = 18 mmh_w = 220 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 20 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 100 mmt_f = 10 mmt_w = 20 mmh_w = 120 mmM = 20 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 160 MPa, σᵧ = −50 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 = 160 MPaσ_y = −50 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 = 20 MPa and τₓᵧ = 30 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = −50 MPaσ_y = 20 MPaτ_xy = 30 MPa

σ_vM = MPa

11.Beam deflectionMedium

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

w = 20 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 6 mw = 12 kN/m

R_B = kN

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 70 GPa (aluminium)d = 30 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 5 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 5 kNn = 2L = 5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.05787 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 = 80000·300/4.147 × 10⁸ = 0.05787 mm

Answer

δ = 0.05787 mm

2.Axial stress & strain

Answer: δ = 3.009 mm

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·40²/4 = 1257 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 75.6 MPa

σ₂ = 472.5 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 75.6·1000/117000 = 0.6461 mm

δ₂ = σ₂L₂/E₂ = 472.5·500/100000 = 2.362 mm

δ = δ₁ + δ₂ = 3.009 mm

Answer

δ = 3.009 mm

3.Torsion of shafts

Answer: φ = 0.5986 °

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) = 500·1000·500/(39000·613592) = 0.01045 rad

φ = 0.01045·180/π = 0.5986°

Answer

φ = 0.5986°

4.Torsion of shafts

Answer: τ_max = 27.02 MPa

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 100000/314.2 = 318.3 N·m

Polar second moment of area

J = π(40⁴ − 20⁴)/32 = 235619 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 318310·20/235619 = 27.02 MPa

Answer

τ_max = 27.02 MPa

5.Shear force & bending moment

Answer: V_x = 9 kN

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 = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 3·10·5

R_B = 150/10 = 15 kN

ΣF_y = 0: R_A = 30 − 15 = 15 kN

Cut at x = 2 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.

V = R_A − (all downward loads left of the cut)

V_x = 15 − 3·2 = 9 kN

Answer

V_x = 9 kN

6.Shear force & bending moment

Answer: B (x = 1 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 = 70 kN, R₂ = 30 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 = −50 kN·m, M_C = −30, M_D = −30, M_E = −30, M_F = −30, M_G = 0

Answer

Mark B: M_B = −50 kN·m

7.Centroid & second moment of area

Answer: I_x = 33.09 × 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 = 120·15 = 1800 mm², y_f = 220 + 15/2 = 227.5 mm

web: A_w = 18·220 = 3960 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1800·227.5 + 3960·110)/5760 = 146.7 mm

Parallel-axis theorem

flange: 120·15³/12 + 1800·(227.5 − 146.7)² = 1.178 × 10⁷ mm⁴

web: 18·220³/12 + 3960·(110 − 146.7)² = 2.131 × 10⁷ mm⁴

I_x = 1.178 × 10⁷ + 2.131 × 10⁷ = 3.309 × 10⁷ mm⁴ = 33.09 × 10⁶ mm⁴

Answer

I_x = 33.09 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 269.5 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 100·10 = 1000 mm², y_f = 120 + 10/2 = 125 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) = (1000·125 + 2400·60)/3400 = 79.12 mm

Parallel-axis theorem

flange: 100·10³/12 + 1000·(125 − 79.12)² = 2.114 × 10⁶ mm⁴

web: 20·120³/12 + 2400·(60 − 79.12)² = 3.757 × 10⁶ mm⁴

I_x = 2.114 × 10⁶ + 3.757 × 10⁶ = 5.871 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 79.12 mm

σ = M·y/I = 2 × 10⁷·79.12/5.871 × 10⁶ = 269.5 MPa

Answer

σ_bottom = 269.5 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = −55.43 °

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 = −40/112.4 = −0.356, sin 2θ_s = −105/112.4 = −0.9345

2θ_s = atan2(−105, −40) = −110.9°

Angle on the element

θ_s = −55.43°

check: θ_s − θ_p = −55.43 − (−10.43) = −45°

Answer

θ_s = −55.43°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_vM = 81.24 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. 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 = −15 MPa, R = 46.1 MPa

σ₁ = 31.1 MPa, σ₂ = −61.1 MPa, σ₃ = 0

Von Mises

σ_vM = √(31.1² − 31.1·(−61.1) + (−61.1)²) = 81.24 MPa

Answer

σ_vM = 81.24 MPa

11.Beam deflection

Answer: δ = 11.63 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 = 20 N/mm, L = 2500 mm

EI = 8.4 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 20·2500⁴/(8·8.4 × 10¹²) = 11.63 mm

Answer

δ = 11.63 mm

12.Beam deflection

Answer: R_B = 27 kN

Remove the prop

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

δ_load = wL⁴/(8EI)

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) = wL⁴/(8EI) → R_B = 3wL/8 = 3·12·6/8 = 27 kN

Answer

R_B = 27 kN

13.Column buckling

Answer: P_cr = 56.06 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·39761/700² = 56060 N = 56.06 kN

Answer

P_cr = 56.06 kN

14.Column buckling

Answer: d = 43.6 mm

Required critical load

P_cr = n·P = 2·5000 = 10000 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·10000·3500²/(π³·70000))^(1/4) = 43.6 mm

Answer

d = 43.6 mm

Check: the slenderness KL/r is 321.1, so the column is slender and Euler applies.

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