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 bronze (diameter 40 mm, length 500 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaδ = 0.25 mm

P = kN

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 60 kNsegment 1: Ø35 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 0.5 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 250 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

T = 250 N·md_i/d_o = 0.7τ_allow = 40 MPa

d_o = mm

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 750 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 20 kWN = 750 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 8 kN/m from 0 to 4 mx = 3.5 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 = 5 mw = 12 kN/m from 0 to 5 msupports at x = 0 and x = 4 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 = 200 mmt_f = 20 mmt_w = 12 mmh_w = 160 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 100 mm × 240 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 144 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 240 mmcut-out: 50 × 40 mmhole centre: y = 144 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 140 MPa, σᵧ = 50 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 = 140 MPaσ_y = 50 MPaτ_xy = −30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −50 MPa, σᵧ = 60 MPa and τₓᵧ = 40 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = −50 MPaσ_y = 60 MPaτ_xy = 40 MPa

σ_vM = 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 = 8 kN/mL = 6 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 2 m from support B. Find the maximum deflection δ_max of the beam.

P = 4 kNb = 2 mL = 6 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_max = mm

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 = 1 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 105 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 105 kNn = 2.5L = 2.5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 34.95 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 = 1257 mm²

Flexibilities

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

L₂/(E₂A₂) = 500/(110000·1257) = 3.617 × 10⁻⁶ mm/N

Load

P = 0.25/7.154 × 10⁻⁶ = 34946 N = 34.95 kN

Answer

P = 34.95 kN

2.Axial stress & strain

Answer: σ_max = 191 MPa

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·35²/4 = 962.1 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 62.36 MPa

σ₂ = 191 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 191 MPa

Answer

σ_max = 191 MPa

3.Torsion of shafts

Answer: d_o = 34.73 mm

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.

Outer diameter from the allowable stress

T = 250 N·m = 250000 N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·250000/(π·40·0.7599)) = 34.73 mm

d_i = 0.7·34.73 = 24.31 mm

Answer

d_o = 34.73 mm

A solid shaft would need d = 31.69 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 31.89 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 20000/78.54 = 254.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·254648/(π·40)) = 31.89 mm

Answer

d = 31.89 mm

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

5.Shear force & bending moment

Answer: M_x = 7 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·4 = 32 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 8·4·2

R_B = 64/4 = 16 kN

ΣF_y = 0: R_A = 32 − 16 = 16 kN

Cut at x = 3.5 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 = 16·3.5 − 8·3.5·1.75 = 7 kN·m

Answer

M_x = 7 kN·m

6.Shear force & bending moment

Answer: M_max = 21.09 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 = 12·5 = 60 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·4 = 12·5·2.5

R_B = 150/4 = 37.5 kN

ΣF_y = 0: R_A = 60 − 37.5 = 22.5 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 22.5/12 = 1.875 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 22.5²/(2·12) = 21.09 kN·m

Answer

M_max = 21.09 kN·m

7.Centroid & second moment of area

Answer: I_x = 14.74 × 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 = 160 + 20/2 = 170 mm

web: A_w = 12·160 = 1920 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4000·170 + 1920·80)/5920 = 140.8 mm

Parallel-axis theorem

flange: 200·20³/12 + 4000·(170 − 140.8)² = 3.541 × 10⁶ mm⁴

web: 12·160³/12 + 1920·(80 − 140.8)² = 1.12 × 10⁷ mm⁴

I_x = 3.541 × 10⁶ + 1.12 × 10⁷ = 1.474 × 10⁷ mm⁴ = 14.74 × 10⁶ mm⁴

Answer

I_x = 14.74 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 113.7 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (24000·120 − 2000·144)/(24000 − 2000) = 117.8 mm

Full plate about ȳ

100·240³/12 + 24000·2.182² = 115.3 × 10⁶ mm⁴

Hole about ȳ

50·40³/12 + 2000·26.18² = 1.638 × 10⁶ mm⁴

Subtract

I_x = 113.7 × 10⁶ mm⁴

Answer

I_x = 113.7 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = −16.85 °

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, 140 − 50) = −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: σ_vM = 117.9 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. 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 = 5 MPa, R = 68.01 MPa

σ₁ = 73.01 MPa, σ₂ = −63.01 MPa, σ₃ = 0

Von Mises

σ_vM = √(73.01² − 73.01·(−63.01) + (−63.01)²) = 117.9 MPa

Answer

σ_vM = 117.9 MPa

11.Beam deflection

Answer: δ_max = 79.41 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 = 8 N/mm, L = 6000 mm

EI = 1.7 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·6000⁴/(384·1.7 × 10¹²) = 79.41 mm

Answer

δ_max = 79.41 mm

12.Beam deflection

Answer: δ_max = 3.88 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 = 4000 N, b = 2000 mm, L = 6000 mm

EI = 3.99 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((6000² − 2000²)/3) = 3266 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 3.88 mm

Answer

δ_max = 3.88 mm

13.Column buckling

Answer: P_cr = 992.2 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/2000² = 992201 N = 992.2 kN

Answer

P_cr = 992.2 kN

14.Column buckling

Answer: d = 69.78 mm

Required critical load

P_cr = n·P = 2.5·105000 = 262500 N

KL = 0.7·2500 = 1750 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·262500·1750²/(π³·70000))^(1/4) = 69.78 mm

Answer

d = 69.78 mm

Check: the slenderness KL/r is 100.3, so the column is slender and Euler applies.

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