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

Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).

P = 15 kNd = 30 mmL = 2.5 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø100/Ø60 mm, E = 110 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 75 mm, inner diameter 56 mm, length 0.8 m) is fixed at one end and twisted by a torque of 400 N·m at the other. Find the angle of twist φ in degrees.

T = 400 N·md_o = 75 mmd_i = 56 mmL = 0.8 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 300 rpmd_o = 100 mmd_i = 80 mm

τ_max = MPa

5.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. It carries a point load P₁ between the supports and a point load P₂ at C. Find the bending moment M_x under the load P₁. Use sagging moments positive and upward shear on the left face positive.

L = 7 mP₁ = 45 kN at 2 mP₂ = 10 kN at 7 msupports at x = 0 and x = 6 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 10 kN at 2.5 mw = 3 kN/m from 0 to 4 mx = 1 m

M_x = 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 = 280 mmt_f = 10 mmt_w = 16 mmh_w = 120 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 = 140 mmt_f = 15 mmt_w = 20 mmh_w = 160 mmM = 40 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 160 MPa, σᵧ = 70 MPa and τₓᵧ = 30 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 = 30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 160 MPaσ_y = −30 MPaτ_xy = 40 MPaθ = -45°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 12 kN/mL = 2 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴

δ_max = 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 = 16 kNw = 2 kN/mL = 2.5 mE = 11 GPa (timber)I = 8.5 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 2.5 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is pinned at both ends. It must carry 45 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 45 kNn = 2.5L = 2.5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 106.1 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·30²/4 = 706.9 mm²

σ = P/A = 15·1000/706.9 = 21.22 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 21.22/200000 = 0.0001061 = 106.1 µε

Answer

ε = 106.1 µε

2.Axial stress & strain

Answer: P_max = 390.7 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

A_r = π·25²/4 = 490.9 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·5027 = 6.511 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0006364) = 0.0006 (steel)

Largest load

P_max = 0.0006·6.511 × 10⁸ = 390657 N = 390.7 kN

Answer

P_max = 390.7 kN

3.Torsion of shafts

Answer: φ = 0.3294 °

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(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴

Angle of twist

T = 400 N·m = 400000 N·mm

L = 0.8 m = 800 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 400000·800/(26000·2.141 × 10⁶) = 0.005749 rad

φ = 0.005749·180/π = 0.3294°

Answer

φ = 0.3294°

4.Torsion of shafts

Answer: τ_max = 27.46 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 100000/31.42 = 3183 N·m

Polar second moment of area

J = π(100⁴ − 80⁴)/32 = 5.796 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 3.183 × 10⁶·50/5.796 × 10⁶ = 27.46 MPa

Answer

τ_max = 27.46 MPa

5.Shear force & bending moment

Answer: M_x = 56.67 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.

Support reactions

Equilibrium of the whole beam. The overhang load also acts about A.

ΣM_A = 0: R_B·6 = 45·2 + 10·7

R_B = 160/6 = 26.67 kN

ΣF_y = 0: R_A = 55 − 26.67 = 28.33 kN

Cut under P₁ and keep the left part

M_x = R_A·2 = 28.33·2 = 56.67 kN·m

Answer

M_x = 56.67 kN·m

6.Shear force & bending moment

Answer: M_x = 8.25 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·4 = 12 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 10·2.5 + 3·4·2

R_B = 49/4 = 12.25 kN

ΣF_y = 0: R_A = 22 − 12.25 = 9.75 kN

Cut at x = 1 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 = 9.75·1 − 3·1·0.5 = 8.25 kN·m

Answer

M_x = 8.25 kN·m

7.Centroid & second moment of area

Answer: I_x = 7.14 × 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 = 280·10 = 2800 mm², y_f = 120 + 10/2 = 125 mm

web: A_w = 16·120 = 1920 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2800·125 + 1920·60)/4720 = 98.56 mm

Parallel-axis theorem

flange: 280·10³/12 + 2800·(125 − 98.56)² = 1.981 × 10⁶ mm⁴

web: 16·120³/12 + 1920·(60 − 98.56)² = 5.159 × 10⁶ mm⁴

I_x = 1.981 × 10⁶ + 5.159 × 10⁶ = 7.14 × 10⁶ mm⁴ = 7.14 × 10⁶ mm⁴

Answer

I_x = 7.14 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 145.6 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 140·15 = 2100 mm², y_f = 160 + 15/2 = 167.5 mm

web: A_w = 20·160 = 3200 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2100·167.5 + 3200·80)/5300 = 114.7 mm

Parallel-axis theorem

flange: 140·15³/12 + 2100·(167.5 − 114.7)² = 5.901 × 10⁶ mm⁴

web: 20·160³/12 + 3200·(80 − 114.7)² = 1.067 × 10⁷ mm⁴

I_x = 5.901 × 10⁶ + 1.067 × 10⁷ = 1.657 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 175 − 114.7 = 60.33 mm

σ = M·y/I = 4 × 10⁷·60.33/1.657 × 10⁷ = 145.6 MPa

Answer

σ_top = 145.6 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 16.85 °

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

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(2·30, 160 − 70) = 33.69°

Angle on the element

θ_p = 16.85°

Answer

θ_p = 16.85°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 95 MPa

Transformation equation

τ_θ = −95·−1 + 40·0 = 95 MPa

Answer

τ_θ = 95 MPa

11.Beam deflection

Answer: δ_max = 2.706 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 = 12 N/mm, L = 2000 mm

EI = 9.24 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·12·2000⁴/(384·9.24 × 10¹¹) = 2.706 mm

Answer

δ_max = 2.706 mm

12.Beam deflection

Answer: δ_max = 66.58 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 = 16000 N, w = 2 N/mm, L = 2500 mm, EI = 9.35 × 10¹⁰ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 16000·2500³/(48·9.35 × 10¹⁰) = 55.7 mm

δ_w = 5wL⁴/(384EI) = 5·2·2500⁴/(384·9.35 × 10¹⁰) = 10.88 mm

Add them

δ_max = 55.7 + 10.88 = 66.58 mm

Answer

δ_max = 66.58 mm

13.Column buckling

Answer: P_cr = 3.473 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 2·2500 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/5000² = 3473 N = 3.473 kN

Answer

P_cr = 3.473 kN

14.Column buckling

Answer: d = 67.48 mm

Required critical load

P_cr = n·P = 2.5·45000 = 112500 N

KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·112500·2500²/(π³·70000))^(1/4) = 67.48 mm

Answer

d = 67.48 mm

Check: the slenderness KL/r is 148.2, so the column is slender and Euler applies.

ExamProblems · examproblems.com/courses/mechanics-of-materials

Mechanics of Materials: cram pack

1Formulas and methods

Axial stress & strain

Torsion of shafts

Method

Shear force & bending moment

Method

Centroid & second moment of area

Mohr’s circle & stress transformation

Method

Beam deflection

Column buckling

2Practice problems

3Solutions

Stress

Strain (Hooke’s law)

Answer

Areas

Combined axial stiffness

Governing strain

Largest load

Answer

Polar second moment of area

Angle of twist

Answer

Torque

Polar second moment of area

Maximum shear stress

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Cut at x = 1 m and keep the left part

Answer

Split into two rectangles

Centroid

Parallel-axis theorem

Answer

Split into two rectangles

Centroid

Parallel-axis theorem

Bending stress

Answer

Angle on the circle

Angle on the element

Answer

Transformation equation

Answer

Consistent units

Deflection

Answer

Consistent units

Each load on its own

Add them

Answer

Smallest second moment of area

Effective length

Critical load

Answer

Required critical load

Solve for the diameter

Answer

Cut under P₁ and keep the left part