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 aluminium bar, in microstrain (µε = 10⁻⁶).

P = 80 kNd = 10 mmL = 1.5 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

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

P = 15 kNsegment 1: Ø30 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 50 mm and length 2.5 m; segment BC has diameter 25 mm and length 2 m. A torque of 200 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 200 N·md₁ = 50 mm, L₁ = 2.5 md₂ = 25 mm, L₂ = 2 mG = 39 GPa

φ_C = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 1000 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 30 kWN = 1000 rpmτ_allow = 80 MPa

d = mm

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_A at the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 40 kN at 1.5 mw = 2 kN/m from 0 to 5 m

M_A = 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 = 5 mP = 60 kN at 1.5 mw = 12 kN/m from 0 to 5 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 = 200 mmt_f = 20 mmt_w = 10 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 5 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 100 mmt_f = 30 mmt_w = 14 mmh_w = 100 mmM = 5 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 80 MPa, σᵧ = 30 MPa and τₓᵧ = 60 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 80 MPaσ_y = 30 MPaτ_xy = 60 MPaθ = -30°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 80 MPa, σᵧ = 10 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σ_x = 80 MPaσ_y = 10 MPaτ_xy = −20 MPaθ = 35°

σ_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 = 4 kN/mL = 5 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_max = 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 = 4 mw = 8 kN/m

R_B = kN

13.Column bucklingMedium

A column of length L made of steel 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 = 200 GPa (steel)d = 40 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 = 2.5 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 14551 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·10²/4 = 78.54 mm²

σ = P/A = 80·1000/78.54 = 1019 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 1019/70000 = 0.01455 = 14551 µε

Answer

ε = 14551 µε

2.Axial stress & strain

Answer: δ = 3.842 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·30²/4 = 706.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 21.22 MPa

σ₂ = 132.6 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 21.22·500/200000 = 0.05305 mm

δ₂ = σ₂L₂/E₂ = 132.6·2000/70000 = 3.789 mm

δ = δ₁ + δ₂ = 3.842 mm

Answer

δ = 3.842 mm

3.Torsion of shafts

Answer: φ_C = 16.52 °

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 moments

J₁ = π·50⁴/32 = 613592 mm⁴

J₂ = π·25⁴/32 = 38350 mm⁴

Twist of each segment

T = 200 N·m = 200000 N·mm

G = 39 GPa = 39000 MPa

φ_AB = 200000·2500/(39000·613592) = 0.02089 rad

φ_BC = 200000·2000/(39000·38350) = 0.2674 rad

Total twist

φ_C = 0.02089 + 0.2674 = 0.2883 rad = 16.52°

Answer

φ_C = 16.52°

4.Torsion of shafts

Answer: d = 26.32 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 30000/104.7 = 286.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·286479/(π·80)) = 26.32 mm

Answer

d = 26.32 mm

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

5.Shear force & bending moment

Answer: M_A = −85 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: −40·1.5 = −60 kN·m

w: −2·5²/2 = −25 kN·m

M_A = −60 − 25 = −85 kN·m

Answer

M_A = −85 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_x = 66 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·5 = 60 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 60·1.5 + 12·5·2.5

R_B = 240/5 = 48 kN

ΣF_y = 0: R_A = 120 − 48 = 72 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 = 72·1 − 12·1·0.5 = 66 kN·m

Answer

M_x = 66 kN·m

7.Centroid & second moment of area

Answer: I_x = 6.096 × 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 = 200·20 = 4000 mm², y_f = 120 + 20/2 = 130 mm

web: A_w = 10·120 = 1200 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4000·130 + 1200·60)/5200 = 113.8 mm

Parallel-axis theorem

flange: 200·20³/12 + 4000·(130 − 113.8)² = 1.177 × 10⁶ mm⁴

web: 10·120³/12 + 1200·(60 − 113.8)² = 4.919 × 10⁶ mm⁴

I_x = 1.177 × 10⁶ + 4.919 × 10⁶ = 6.096 × 10⁶ mm⁴ = 6.096 × 10⁶ mm⁴

Answer

I_x = 6.096 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 86.94 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 100·30 = 3000 mm², y_f = 100 + 30/2 = 115 mm

web: A_w = 14·100 = 1400 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3000·115 + 1400·50)/4400 = 94.32 mm

Parallel-axis theorem

flange: 100·30³/12 + 3000·(115 − 94.32)² = 1.508 × 10⁶ mm⁴

web: 14·100³/12 + 1400·(50 − 94.32)² = 3.916 × 10⁶ mm⁴

I_x = 1.508 × 10⁶ + 3.916 × 10⁶ = 5.425 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 94.32 mm

σ = M·y/I = 5 × 10⁶·94.32/5.425 × 10⁶ = 86.94 MPa

Answer

σ_bottom = 86.94 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 94.46 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.

Transformation equation for the y′ face

σ_y′ = 55 − 25·0.5 − 60·(−0.866) = 94.46 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 15.54 MPa, σ_x′ + σ_y′ = 110 = σ_x + σ_y

Answer

σ_y′ = 94.46 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 38.18 MPa

Transformation equation

σ_θ = 45 + 35·0.342 + −20·0.9397 = 38.18 MPa

Answer

σ_θ = 38.18 MPa

11.Beam deflection

Answer: δ_max = 18.5 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 = 4 N/mm, L = 5000 mm

EI = 1.76 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·4·5000⁴/(384·1.76 × 10¹²) = 18.5 mm

Answer

δ_max = 18.5 mm

12.Beam deflection

Answer: R_B = 12 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 = 3·8·4/8 = 12 kN

Answer

R_B = 12 kN

13.Column buckling

Answer: P_cr = 9.922 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 = π²·200000·125664/5000² = 9922 N = 9.922 kN

Answer

P_cr = 9.922 kN

14.Column buckling

Answer: σ_cr = 126.3 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 0.5·2500 = 1250 mm

KL/r = 1250/10 = 125

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/125² = 126.3 MPa

Answer

σ_cr = 126.3 MPa

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