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 bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaδ = 0.4 mm

P = kN

2.Axial stress & strainHard

A steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaP = 30 kN

σ₂ = MPa

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 = 200 N·md = 50 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 1000 N·m is applied at C, 2 m from A and 3 m from B. Find the angle of rotation φ_C of section C, in degrees.

d = 50 mmT₀ = 1000 N·ma = 2 mb = 3 mG = 26 GPa

φ_C = °

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 9.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 10 kN/m from 0 to 10 mx = 9.5 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 7 mw = 4 kN/m from 0 to 7 msupports at x = 0 and x = 6 m

M_max = kN·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 = 260 mmt_f = 15 mmt_w = 12 mmh_w = 140 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 60 MPa in tension and 100 MPa in compression. Find the largest hogging bending moment M_allow the beam can carry (give its size).

b_f = 120 mmt_f = 20 mmt_w = 10 mmh_w = 100 mmσ_t,allow = 60 MPaσ_c,allow = 100 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 40 MPa, σᵧ = 30 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 40 MPaσ_y = 30 MPaτ_xy = −20 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 80 MPa, σᵧ = 40 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 80 MPaσ_y = 40 MPaτ_xy = 30 MPaθ = 20°

τ_θ = MPa

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 = 15 kN/mL = 1.5 mE = 11 GPa (timber)I = 8.5 × 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 = 6 kNw = 8 kN/mL = 6 mE = 200 GPa (steel)I = 120 × 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 = 6 mE = 70 GPa (aluminium)b × h = 40 × 50 mm

P_cr = kN

14.Column bucklingHard

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

L = 6 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 30.39 kN

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

A₁ = 490.9 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(70000·706.9) = 1.011 × 10⁻⁵ mm/N

Load

P = 0.4/1.316 × 10⁻⁵ = 30393 N = 30.39 kN

Answer

P = 30.39 kN

2.Axial stress & strain

Answer: σ₂ = 16.32 MPa

Stiffnesses

k₁ = 200000·706.9/500 = 282743 N/mm

k₂ = 100000·706.9/400 = 176715 N/mm

Share of segment 2

F₂ = 30000·176715/459458 = 11538 N

Stress

σ₂ = 11538/706.9 = 16.32 MPa (compression)

Answer

σ₂ = 16.32 MPa

3.Torsion of shafts

Answer: φ = 0.4669 °

Polar second moment of area

J = π·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2000/(80000·613592) = 0.008149 rad

φ = 0.008149·180/π = 0.4669°

Answer

φ = 0.4669°

4.Torsion of shafts

Answer: φ_C = 4.31 °

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 = 1000

T_A·a = T_B·b

T_A = 1000·3/5 = 600 N·m

T_B = 1000·2/5 = 400 N·m

Rotation of C

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

J = π·50⁴/32 = 613592 mm⁴

G = 26 GPa = 26000 MPa

φ_C = T_A·a/(GJ) = 600000·2000/(26000·613592) = 0.07522 rad = 4.31°

Answer

φ_C = 4.31°

5.Shear force & bending moment

Answer: M_x = 23.75 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·10 = 100 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 10·10·5

R_B = 500/10 = 50 kN

ΣF_y = 0: R_A = 100 − 50 = 50 kN

Cut at x = 9.5 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 = 50·9.5 − 10·9.5·4.75 = 23.75 kN·m

Answer

M_x = 23.75 kN·m

6.Shear force & bending moment

Answer: M_max = 17.01 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 4·7 = 28 kN, acting at x̄ = 3.5 m

ΣM_A = 0: R_B·6 = 4·7·3.5

R_B = 98/6 = 16.33 kN

ΣF_y = 0: R_A = 28 − 16.33 = 11.67 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 11.67/4 = 2.917 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 11.67²/(2·4) = 17.01 kN·m

Answer

M_max = 17.01 kN·m

7.Centroid & second moment of area

Answer: I_x = 9.87 × 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 = 260·15 = 3900 mm², y_f = 140 + 15/2 = 147.5 mm

web: A_w = 12·140 = 1680 mm², y_w = 140/2 = 70 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3900·147.5 + 1680·70)/5580 = 124.2 mm

Parallel-axis theorem

flange: 260·15³/12 + 3900·(147.5 − 124.2)² = 2.196 × 10⁶ mm⁴

web: 12·140³/12 + 1680·(70 − 124.2)² = 7.673 × 10⁶ mm⁴

I_x = 2.196 × 10⁶ + 7.673 × 10⁶ = 9.87 × 10⁶ mm⁴ = 9.87 × 10⁶ mm⁴

Answer

I_x = 9.87 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 3.741 kN·m

Centroid

ȳ = (2400·110 + 1000·50)/3400 = 92.35 mm

top edge: c = 120 − 92.35 = 27.65 mm

Second moment of area

I = 120·20³/12 + 2400·17.65² + 10·100³/12 + 1000·42.35² = 3.455 × 10⁶ mm⁴

Moment allowed by each edge

Hogging: the top is in tension, the bottom in compression.

top edge: M = 60·3.455 × 10⁶/27.65 = 7.497 kN·m

bottom edge: M = 100·3.455 × 10⁶/92.35 = 3.741 kN·m

Governing value

M_allow = min = 3.741 kN·m (bottom edge)

Answer

M_allow = 3.741 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_max = 20.62 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.

Radius of Mohr’s circle

τ_max = R = √(((40 − 30)/2)² + −20²) = 20.62 MPa

Answer

τ_max = 20.62 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 10.13 MPa

Transformation equation

τ_θ = −20·0.6428 + 30·0.766 = 10.13 MPa

Answer

τ_θ = 10.13 MPa

11.Beam deflection

Answer: θ_A = 0.02256 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 = 15 N/mm, L = 1500 mm

EI = 9.35 × 10¹⁰ N·mm²

End slope

θ_A = wL³/(24EI) = 15·1500³/(24·9.35 × 10¹⁰) = 0.02256 rad

Answer

θ_A = 0.02256 rad

That is 1.293°.

12.Beam deflection

Answer: δ_max = 6.75 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 = 6000 N, w = 8 N/mm, L = 6000 mm, EI = 2.4 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 6000·6000³/(48·2.4 × 10¹³) = 1.125 mm

δ_w = 5wL⁴/(384EI) = 5·8·6000⁴/(384·2.4 × 10¹³) = 5.625 mm

Add them

δ_max = 1.125 + 5.625 = 6.75 mm

Answer

δ_max = 6.75 mm

13.Column buckling

Answer: P_cr = 20.47 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 = 50·40³/12 = 266667 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·266667/3000² = 20470 N = 20.47 kN

Answer

P_cr = 20.47 kN

14.Column buckling

Answer: σ_cr = 2.142 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 = 2·6000 = 12000 mm

KL/r = 12000/12.5 = 960

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/960² = 2.142 MPa

Answer

σ_cr = 2.142 MPa

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