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 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 50 mm, length 400 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 400 mm, E = 70 GPaδ = 0.2 mm

P = kN

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø100/Ø60 mm, E = 110 GPa, σ_allow = 80 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid aluminium 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 = 30 mmL = 3 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 1450 rpmd_o = 50 mmd_i = 40 mm

τ_max = MPa

5.Shear force & bending momentMedium

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 1.5 m from A. Find the bending moment M_x at the section x = 1 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 2 mP = 80 kN at 1.5 mw = 10 kN/m from 0 to 2 mx = 1 m

M_x = kN·m

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.

25 kN @ 0 m15 kN @ 12 m15 kN @ 4 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 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 = 20 mmt_w = 16 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 40 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 280 mmt_f = 30 mmt_w = 10 mmh_w = 260 mmM = 40 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa and σᵧ = 70 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 120 MPa?

σ_x = −30 MPaσ_y = 70 MPaσ₁ = 120 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −20 MPa and τₓᵧ = 20 MPa (σ_z = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 70 MPaσ_y = −20 MPaτ_xy = 20 MPaσ_Y = 275 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 11 kNa = 1 mL = 6 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 50 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/300. Find the smallest depth h_req of the section.

w = 1 kN/mL = 3.5 mE = 11 GPa (timber)b = 50 mm

h_req = 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 = 5 mE = 70 GPa (aluminium)b × h = 30 × 60 mm

P_cr = kN

14.Column bucklingHard

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

L = 2 mE = 200 GPa (steel)d = 25 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 31.02 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 = 706.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·706.9) = 3.537 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(70000·1963) = 2.91 × 10⁻⁶ mm/N

Load

P = 0.2/6.447 × 10⁻⁶ = 31022 N = 31.02 kN

Answer

P = 31.02 kN

2.Axial stress & strain

Answer: P_max = 584.9 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·5027 = 8.042 × 10⁸ N

Governing strain

ε = min(0.001, 0.0007273) = 0.0007273 (bronze)

Largest load

P_max = 0.0007273·8.042 × 10⁸ = 584907 N = 584.9 kN

Answer

P_max = 584.9 kN

3.Torsion of shafts

Answer: φ = 16.63 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·3000/(26000·79522) = 0.2902 rad

φ = 0.2902·180/π = 16.63°

Answer

φ = 16.63°

4.Torsion of shafts

Answer: τ_max = 9.09 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 20000/151.8 = 131.7 N·m

Polar second moment of area

J = π(50⁴ − 40⁴)/32 = 362265 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 131714·25/362265 = 9.09 MPa

Answer

τ_max = 9.09 MPa

5.Shear force & bending moment

Answer: M_x = −45 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.

Keep the free-end part

P: −80·0.5 = −40 kN·m

w: −10·1²/2 = −5 kN·m

M_x = −40 − 5 = −45 kN·m

Answer

M_x = −45 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: E (x = 8 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 = 33.33 kN, R₂ = 21.67 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, M_C = −33.33, M_D = −46.67, M_E = −60 kN·m, M_F = −30, M_G = 0

Answer

Mark E: M_E = −60 kN·m

7.Centroid & second moment of area

Answer: I_x = 34.83 × 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 = 120·20 = 2400 mm², y_f = 220 + 20/2 = 230 mm

web: A_w = 16·220 = 3520 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2400·230 + 3520·110)/5920 = 158.6 mm

Parallel-axis theorem

flange: 120·20³/12 + 2400·(230 − 158.6)² = 1.23 × 10⁷ mm⁴

web: 16·220³/12 + 3520·(110 − 158.6)² = 2.253 × 10⁷ mm⁴

I_x = 1.23 × 10⁷ + 2.253 × 10⁷ = 3.483 × 10⁷ mm⁴ = 34.83 × 10⁶ mm⁴

Answer

I_x = 34.83 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 34.56 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 280·30 = 8400 mm², y_f = 260 + 30/2 = 275 mm

web: A_w = 10·260 = 2600 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (8400·275 + 2600·130)/11000 = 240.7 mm

Parallel-axis theorem

flange: 280·30³/12 + 8400·(275 − 240.7)² = 1.05 × 10⁷ mm⁴

web: 10·260³/12 + 2600·(130 − 240.7)² = 4.652 × 10⁷ mm⁴

I_x = 1.05 × 10⁷ + 4.652 × 10⁷ = 5.702 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 290 − 240.7 = 49.27 mm

σ = M·y/I = 4 × 10⁷·49.27/5.702 × 10⁷ = 34.56 MPa

Answer

σ_top = 34.56 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 86.6 MPa

Required radius

C = (−30 + 70)/2 = 20 MPa

R = σ₁ − C = 120 − 20 = 100 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(100² − 50²) = 86.6 MPa

Answer

τ_xy = 86.6 MPa

10.Mohr’s circle & stress transformation

Answer: n = 3.094

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 = 25 MPa, R = 49.24 MPa

σ₁ = 74.24 MPa, σ₂ = −24.24 MPa, σ₃ = 0

Von Mises

σ_vM = √(74.24² − 74.24·(−24.24) + (−24.24)²) = 88.88 MPa

n = 275/88.88 = 3.094

Answer

n = 3.094

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

11.Beam deflection

Answer: δ_P = 2.122 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 = 11000 N, a = 1000 mm, b = 5000 mm, L = 6000 mm

EI = 7.2 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 11000·1000²·5000²/(3·7.2 × 10¹²·6000) = 2.122 mm

Answer

δ_P = 2.122 mm

12.Beam deflection

Answer: h_req = 154 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.

w = 1 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/300 = 3500/300 = 11.67 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1·3500⁴/(384·11000·11.67) = 1.523 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.523 × 10⁷/50)^1/3 = 154 mm

Answer

h_req = 154 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 7.614 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·135000/3500² = 7614 N = 7.614 kN

Answer

P_cr = 7.614 kN

14.Column buckling

Answer: σ_cr = 19.28 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 1·2000 = 2000 mm

KL/r = 2000/6.25 = 320

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/320² = 19.28 MPa

Answer

σ_cr = 19.28 MPa

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