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 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 400 mm, E = 100 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, Ø30 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaδ = 0.4 mm

P = kN

2.Axial stress & strainHard

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

P = 20 kNsegment 1: Ø30 mm, 1 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1 m, brass (E = 100 GPa)

δ = mm

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 must transmit 50 kW at 750 rpm. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter d_o.

P = 50 kWN = 750 rpmd_i/d_o = 0.8τ_allow = 80 MPa

d_o = 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 0.5 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 = 2.5 mP = 60 kN at 0.5 mw = 10 kN/m from 0 to 2.5 m

M_A = kN·m

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load over part of its span. Find the distance x₀ from A to the section where the bending moment is largest.

L = 8 mw = 10 kN/m from 0 to 6 m

x₀ = 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 = 10 mmt_w = 14 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 20 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 160 mmt_f = 30 mmt_w = 8 mmh_w = 240 mmM = 20 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 100 MPa and τₓᵧ = 60 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 120 MPaσ_y = 100 MPaτ_xy = 60 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 6 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 1500 mmt = 6 mmp = 0.8 MPaα = 60°

|τ_n| = 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 = 2 kN/mL = 3 mE = 11 GPa (timber)I = 23 × 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 = 23 kNw = 15 kN/mL = 6 mE = 70 GPa (aluminium)I = 160 × 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 = 3 mE = 70 GPa (aluminium)b × h = 25 × 40 mm

P_cr = kN

14.Column bucklingHard

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

L = 3 mE = 70 GPa (aluminium)d = 60 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 47.12 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 = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(100000·706.9) = 5.659 × 10⁻⁶ mm/N

Load

P = 0.4/8.488 × 10⁻⁶ = 47124 N = 47.12 kN

Answer

P = 47.12 kN

2.Axial stress & strain

Answer: δ = 2.01 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 = 28.29 MPa

σ₂ = 176.8 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 28.29·1000/117000 = 0.2418 mm

δ₂ = σ₂L₂/E₂ = 176.8·1000/100000 = 1.768 mm

δ = δ₁ + δ₂ = 2.01 mm

Answer

δ = 2.01 mm

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: d_o = 40.95 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 50000/78.54 = 636.6 N·m

Outer diameter from the allowable stress

T = 636.6 N·m = 636620 N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·636620/(π·80·0.5904)) = 40.95 mm

d_i = 0.8·40.95 = 32.76 mm

Answer

d_o = 40.95 mm

A solid shaft would need d = 34.35 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: M_A = −61.25 kN·m

Keep the free-end part

P: −60·0.5 = −30 kN·m

w: −10·2.5²/2 = −31.25 kN·m

M_A = −30 − 31.25 = −61.25 kN·m

Answer

M_A = −61.25 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: x₀ = 3.75 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 = 10·6 = 60 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·8 = 10·6·3

R_B = 180/8 = 22.5 kN

ΣF_y = 0: R_A = 60 − 22.5 = 37.5 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 − 0) = 0

x₀ = 0 + 37.5/10 = 3.75 m

Answer

x₀ = 3.75 m

7.Centroid & second moment of area

Answer: I_x = 5.89 × 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 = 200·10 = 2000 mm², y_f = 120 + 10/2 = 125 mm

web: A_w = 14·120 = 1680 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2000·125 + 1680·60)/3680 = 95.33 mm

Parallel-axis theorem

flange: 200·10³/12 + 2000·(125 − 95.33)² = 1.778 × 10⁶ mm⁴

web: 14·120³/12 + 1680·(60 − 95.33)² = 4.113 × 10⁶ mm⁴

I_x = 1.778 × 10⁶ + 4.113 × 10⁶ = 5.89 × 10⁶ mm⁴ = 5.89 × 10⁶ mm⁴

Answer

I_x = 5.89 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 125.2 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 160·30 = 4800 mm², y_f = 240 + 30/2 = 255 mm

web: A_w = 8·240 = 1920 mm², y_w = 240/2 = 120 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4800·255 + 1920·120)/6720 = 216.4 mm

Parallel-axis theorem

flange: 160·30³/12 + 4800·(255 − 216.4)² = 7.501 × 10⁶ mm⁴

web: 8·240³/12 + 1920·(120 − 216.4)² = 2.707 × 10⁷ mm⁴

I_x = 7.501 × 10⁶ + 2.707 × 10⁷ = 3.457 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 216.4 mm

σ = M·y/I = 2 × 10⁷·216.4/3.457 × 10⁷ = 125.2 MPa

Answer

σ_bottom = 125.2 MPa

9.Mohr’s circle & stress transformation

Answer: τ_abs = 85.41 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 110 MPa, R = √(10² + 60²) = 60.83 MPa

σ₁ = 170.8 MPa, σ₂ = 49.17 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(121.7, 170.8, 49.17)/2 = 85.41 MPa

Answer

τ_abs = 85.41 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 21.65 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. Thin-walled cylinder: hoop stress σ_h = pr/t and axial stress σ_a = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 1500/2 = 750 mm

σ_h = pr/t = 0.8·750/6 = 100 MPa

σ_a = pr/(2t) = 50 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

|τ_n| = (100 − 50)·sin 60°·cos 60° = 21.65 MPa

Answer

|τ_n| = 21.65 MPa

11.Beam deflection

Answer: δ_max = 8.337 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 = 2 N/mm, L = 3000 mm

EI = 2.53 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·2·3000⁴/(384·2.53 × 10¹¹) = 8.337 mm

Answer

δ_max = 8.337 mm

12.Beam deflection

Answer: δ_max = 31.84 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 = 23000 N, w = 15 N/mm, L = 6000 mm, EI = 1.12 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 23000·6000³/(48·1.12 × 10¹³) = 9.241 mm

δ_w = 5wL⁴/(384EI) = 5·15·6000⁴/(384·1.12 × 10¹³) = 22.6 mm

Add them

δ_max = 9.241 + 22.6 = 31.84 mm

Answer

δ_max = 31.84 mm

13.Column buckling

Answer: P_cr = 8.159 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 = 40·25³/12 = 52083 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·52083/2100² = 8159 N = 8.159 kN

Answer

P_cr = 8.159 kN

14.Column buckling

Answer: σ_cr = 35.25 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 0.7·3000 = 2100 mm

KL/r = 2100/15 = 140

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa

Answer

σ_cr = 35.25 MPa

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