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 = 45 kNd = 16 mmL = 1.5 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 300 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 200 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 200 mm, E = 70 GPaP = 50 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 40 mm, inner diameter 24 mm, length 0.5 m) is fixed at one end and twisted by a torque of 200 N·m at the other. Find the angle of twist φ in degrees.

T = 200 N·md_o = 40 mmd_i = 24 mmL = 0.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 3000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 10 kWN = 3000 rpmτ_allow = 60 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 3 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 = 3 mP = 75 kN at 3 mw = 6 kN/m from 0 to 3 m

M_A = 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.

50 kN @ 12 m25 kN @ 4 m6 kN/m, 0–10 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 = 140 mmt_f = 25 mmt_w = 8 mmh_w = 240 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 170 × 14 mm, a web 370 mm high and 14 mm thick, and a bottom flange 230 × 24 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 170 × 14 mmweb: 14 × 370 mmbottom flange: 230 × 24 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −80 MPa, σᵧ = 0 MPa and τₓᵧ = 10 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 = −80 MPaσ_y = 0 MPaτ_xy = 10 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 25 mm carries an axial tensile force of 150 kN together with a torque of 200 N·m. At the surface of the shaft, find the maximum in-plane shear stress τ_max.

d = 25 mmF = 150 kNT = 200 N·m

τ_max = 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 = 3 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 6 kN/mL = 1.5 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

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

L = 6 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 1119 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·16²/4 = 201.1 mm²

σ = P/A = 45·1000/201.1 = 223.8 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 223.8/200000 = 0.001119 = 1119 µε

Answer

ε = 1119 µε

2.Axial stress & strain

Answer: σ₂ = 16.13 MPa

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.

Stiffnesses

k₁ = 200000·1257/300 = 837758 N/mm

k₂ = 70000·706.9/200 = 247400 N/mm

Share of segment 2

F₂ = 50000·247400/1.085 × 10⁶ = 11399 N

Stress

σ₂ = 11399/706.9 = 16.13 MPa (compression)

Answer

σ₂ = 16.13 MPa

3.Torsion of shafts

Answer: φ = 0.6716 °

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(40⁴ − 24⁴)/32 = 218755 mm⁴

Angle of twist

T = 200 N·m = 200000 N·mm

L = 0.5 m = 500 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 200000·500/(39000·218755) = 0.01172 rad

φ = 0.01172·180/π = 0.6716°

Answer

φ = 0.6716°

4.Torsion of shafts

Answer: d = 13.93 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 10000/314.2 = 31.83 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·31831/(π·60)) = 13.93 mm

Answer

d = 13.93 mm

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

5.Shear force & bending moment

Answer: M_A = −252 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: −75·3 = −225 kN·m

w: −6·3²/2 = −27 kN·m

M_A = −225 − 27 = −252 kN·m

Answer

M_A = −252 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: F (x = 10 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 = 35 kN, R₂ = 100 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 = 58, M_C = 92, M_D = 52, M_E = −12, M_F = −100 kN·m, M_G = 0

Answer

Mark F: M_F = −100 kN·m

7.Centroid & second moment of area

Answer: I_x = 31.17 × 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 = 140·25 = 3500 mm², y_f = 240 + 25/2 = 252.5 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) = (3500·252.5 + 1920·120)/5420 = 205.6 mm

Parallel-axis theorem

flange: 140·25³/12 + 3500·(252.5 − 205.6)² = 7.893 × 10⁶ mm⁴

web: 8·240³/12 + 1920·(120 − 205.6)² = 2.327 × 10⁷ mm⁴

I_x = 7.893 × 10⁶ + 2.327 × 10⁷ = 3.117 × 10⁷ mm⁴ = 31.17 × 10⁶ mm⁴

Answer

I_x = 31.17 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 331 × 10⁶ mm⁴

Split into three rectangles

Measure every y from the bottom edge.

top flange: A = 2380 mm², y = 401 mm

web: A = 5180 mm², y = 209 mm

bottom flange: A = 5520 mm², y = 12 mm

Centroid

ȳ = (2380·401 + 5180·209 + 5520·12)/13080 = 160.8 mm

Parallel-axis theorem

top flange: 170·14³/12 + 2380·240.2² = 1.374 × 10⁸ mm⁴

web: 14·370³/12 + 5180·48.2² = 7.113 × 10⁷ mm⁴

bottom flange: 230·24³/12 + 5520·148.8² = 1.225 × 10⁸ mm⁴

I_x = 1.374 × 10⁸ + 7.113 × 10⁷ + 1.225 × 10⁸

I_x = 3.31 × 10⁸ mm⁴ = 331 × 10⁶ mm⁴

Answer

I_x = 331 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 82.98 °

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(2·10, −80 − 0) = 166°

Angle on the element

θ_p = 82.98°

Answer

θ_p = 82.98°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_max = 166.1 MPa

Stresses at the surface

σ = F/A = 150000/(π·25²/4) = 150000/490.9 = 305.6 MPa

τ = 16T/(πd³) = 16·200000/(π·25³) = 65.19 MPa

Mohr’s circle

C = σ/2 = 152.8 MPa

R = √(152.8² + 65.19²) = 166.1 MPa

Answer

τ_max = 166.1 MPa

11.Beam deflection

Answer: δ_max = 5.273 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 = 3000 mm

EI = 2.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·12·3000⁴/(384·2.4 × 10¹²) = 5.273 mm

Answer

δ_max = 5.273 mm

12.Beam deflection

Answer: I_req = 1.356 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 1500/360 = 4.167 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·6·1500⁴/(384·70000·4.167) = 1.356 × 10⁶ mm⁴

I_req = 1.356 × 10⁶ mm⁴

Answer

I_req = 1.356 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 45.32 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·180000/2800² = 45320 N = 45.32 kN

Answer

P_cr = 45.32 kN

14.Column buckling

Answer: σ_cr = 30.71 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 0.5·6000 = 3000 mm

KL/r = 3000/20 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/150² = 30.71 MPa

Answer

σ_cr = 30.71 MPa

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