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 brass (outer diameter 120 mm, inner diameter 60 mm, E = 100 GPa). Both are 600 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 100 GPaL = 600 mm

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 60 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø60 mm, 200 mm, E = 120 GPaP = 40 kN

σ₂ = MPa

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 = 400 N·md = 50 mmL = 3 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 1500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1500 rpmτ_allow = 80 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 6 kN/m from 0 to 4 mx = 2.5 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 = 24 kN/m from 2 to 6 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 190 mm and inside diameter d = 162 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 190 mmd = 162 mm

r = mm

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 25 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 140 mmt_f = 10 mmt_w = 12 mmh_w = 280 mmM = 25 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa, σᵧ = 30 MPa and τₓᵧ = −60 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 = −30 MPaσ_y = 30 MPaτ_xy = −60 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 20 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 20 MPaσ_y = 110 MPaτ_xy = 30 MPaθ = 40°

τ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_B at the free end B.

P = 8 kNa = 1 mL = 3 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_B = 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 = 18 kNw = 4 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 12 × 10⁶ mm⁴

δ_max = mm

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 = 4 mE = 70 GPa (aluminium)b × h = 25 × 50 mm

P_cr = kN

14.Column bucklingHard

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

L = 4 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.09095 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(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 100·10³·8482 = 9.896 × 10⁸ N

Shortening

δ = 150000·600/9.896 × 10⁸ = 0.09095 mm

Answer

δ = 0.09095 mm

2.Axial stress & strain

Answer: σ₂ = 11.35 MPa

Stiffnesses

k₁ = 200000·1257/600 = 418879 N/mm

k₂ = 120000·2827/200 = 1.696 × 10⁶ N/mm

Share of segment 2

F₂ = 40000·1.696 × 10⁶/2.115 × 10⁶ = 32079 N

Stress

σ₂ = 32079/2827 = 11.35 MPa (compression)

Answer

σ₂ = 11.35 MPa

3.Torsion of shafts

Answer: φ = 2.873 °

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) = 400·1000·3000/(39000·613592) = 0.05015 rad

φ = 0.05015·180/π = 2.873°

Answer

φ = 2.873°

4.Torsion of shafts

Answer: d = 27.26 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 50000/157.1 = 318.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·318310/(π·80)) = 27.26 mm

Answer

d = 27.26 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: M_x = 11.25 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·4 = 24 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 6·4·2

R_B = 48/4 = 12 kN

ΣF_y = 0: R_A = 24 − 12 = 12 kN

Cut at x = 2.5 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 = 12·2.5 − 6·2.5·1.25 = 11.25 kN·m

Answer

M_x = 11.25 kN·m

6.Shear force & bending moment

Answer: M_max = 85.33 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 = 24·4 = 96 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·6 = 24·4·4

R_B = 384/6 = 64 kN

ΣF_y = 0: R_A = 96 − 64 = 32 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 − 2) = 0

x₀ = 2 + 32/24 = 3.333 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 2)²/2 = 32·3.333 − 24·1.333²/2 = 85.33 kN·m

Answer

M_max = 85.33 kN·m

7.Centroid & second moment of area

Answer: r = 62.42 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 = 7741 mm²

I = π(D⁴ − d⁴)/64 = 3.016 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(3.016 × 10⁷/7741) = 62.42 mm

Answer

r = 62.42 mm

8.Centroid & second moment of area

Answer: σ_bottom = 106.8 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.

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 140·10 = 1400 mm², y_f = 280 + 10/2 = 285 mm

web: A_w = 12·280 = 3360 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1400·285 + 3360·140)/4760 = 182.6 mm

Parallel-axis theorem

flange: 140·10³/12 + 1400·(285 − 182.6)² = 1.468 × 10⁷ mm⁴

web: 12·280³/12 + 3360·(140 − 182.6)² = 2.806 × 10⁷ mm⁴

I_x = 1.468 × 10⁷ + 2.806 × 10⁷ = 4.274 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 182.6 mm

σ = M·y/I = 2.5 × 10⁷·182.6/4.274 × 10⁷ = 106.8 MPa

Answer

σ_bottom = 106.8 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = 76.72 °

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.

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 = −60/67.08 = −0.8944, sin 2θ_s = −(−30)/67.08 = 0.4472

2θ_s = atan2(30, −60) = 153.4°

Angle on the element

θ_s = 76.72°

check: θ_s − θ_p = 76.72 − (−58.28) = 135°

Answer

θ_s = 76.72°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 49.53 MPa

Transformation equation

τ_θ = −−45·0.9848 + 30·0.1736 = 49.53 MPa

Answer

τ_θ = 49.53 MPa

11.Beam deflection

Answer: δ_B = 2.673 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 = 8000 N, a = 1000 mm, L = 3000 mm

EI = 3.99 × 10¹² 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) = 0.6683 mm

θ_P = Pa²/(2EI) = 0.001003 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 2.673 mm

Answer

δ_B = 2.673 mm

12.Beam deflection

Answer: δ_max = 9.397 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 = 18000 N, w = 4 N/mm, L = 2500 mm, EI = 8.4 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 18000·2500³/(48·8.4 × 10¹¹) = 6.975 mm

δ_w = 5wL⁴/(384EI) = 5·4·2500⁴/(384·8.4 × 10¹¹) = 2.422 mm

Add them

δ_max = 6.975 + 2.422 = 9.397 mm

Answer

δ_max = 9.397 mm

13.Column buckling

Answer: P_cr = 5.737 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 = 50·25³/12 = 65104 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·65104/2800² = 5737 N = 5.737 kN

Answer

P_cr = 5.737 kN

14.Column buckling

Answer: σ_cr = 77.11 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 0.5·4000 = 2000 mm

KL/r = 2000/12.5 = 160

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/160² = 77.11 MPa

Answer

σ_cr = 77.11 MPa

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