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 elongation δ of the brass bar under the axial load.

P = 90 kNd = 30 mmL = 0.5 mE = 100 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaP = 20 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid steel 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 = 100 N·md = 20 mmL = 1.2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft of diameter 25 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 3 m from A and 1.5 m from B. Find the angle of rotation φ_C of section C, in degrees.

d = 25 mmT₀ = 250 N·ma = 3 mb = 1.5 mG = 80 GPa

φ_C = °

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 3 kN/m from 0 to 5 mx = 0.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 4 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 20 kN at 3.5 mw = 4 kN/m from 0 to 5 mx = 4 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 = 160 mmt_f = 15 mmt_w = 18 mmh_w = 200 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 10 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 100 mmt_f = 10 mmt_w = 10 mmh_w = 220 mmM = 10 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 50 MPa, σᵧ = 70 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 = 50 MPaσ_y = 70 MPaτ_xy = −10 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries a bending moment of 1500 N·m together with a torque of 200 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τ_max.

d = 40 mmM = 1500 N·mT = 200 N·m

τ_max = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_P under the load.

P = 26 kNa = 1 mL = 4 mE = 11 GPa (timber)I = 12 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth h_req of the section.

w = 4 kN/mL = 5 mE = 11 GPa (timber)b = 100 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 70 GPa (aluminium)d = 25 mm

P_cr = kN

14.Column bucklingHard

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

L = 4 mE = 200 GPa (steel)d = 60 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.6366 mm

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 = 90·1000/706.9 = 127.3 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 127.3/100000 = 0.001273 = 1273 µε

Elongation

δ = ε·L = 0.001273·500 mm = 0.6366 mm

Answer

δ = 0.6366 mm

2.Axial stress & strain

Answer: σ₂ = 6.211 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 = 200000·1257/400 = 628319 N/mm

k₂ = 100000·706.9/400 = 176715 N/mm

Share of segment 2

F₂ = 20000·176715/805033 = 4390 N

Stress

σ₂ = 4390/706.9 = 6.211 MPa (compression)

Answer

σ₂ = 6.211 MPa

3.Torsion of shafts

Answer: φ = 5.471 °

Polar second moment of area

J = π·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1200/(80000·15708) = 0.09549 rad

φ = 0.09549·180/π = 5.471°

Answer

φ = 5.471°

4.Torsion of shafts

Answer: φ_C = 4.669 °

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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 250

T_A·a = T_B·b

T_A = 250·1.5/4.5 = 83.33 N·m

T_B = 250·3/4.5 = 166.7 N·m

Rotation of C

Use segment AC: it is fixed at A and carries T_A.

J = π·25⁴/32 = 38350 mm⁴

G = 80 GPa = 80000 MPa

φ_C = T_A·a/(GJ) = 83333·3000/(80000·38350) = 0.08149 rad = 4.669°

Answer

φ_C = 4.669°

5.Shear force & bending moment

Answer: M_x = 3.375 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.

UDL resultant: W = w·ℓ = 3·5 = 15 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 3·5·2.5

R_B = 37.5/5 = 7.5 kN

ΣF_y = 0: R_A = 15 − 7.5 = 7.5 kN

Cut at x = 0.5 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 = 7.5·0.5 − 3·0.5·0.25 = 3.375 kN·m

Answer

M_x = 3.375 kN·m

6.Shear force & bending moment

Answer: M_x = 22 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 4·5 = 20 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 20·3.5 + 4·5·2.5

R_B = 120/5 = 24 kN

ΣF_y = 0: R_A = 40 − 24 = 16 kN

Cut at x = 4 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 = 16·4 − 20·(4 − 3.5) − 4·4·2 = 22 kN·m

Answer

M_x = 22 kN·m

7.Centroid & second moment of area

Answer: I_x = 28.69 × 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 = 160·15 = 2400 mm², y_f = 200 + 15/2 = 207.5 mm

web: A_w = 18·200 = 3600 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2400·207.5 + 3600·100)/6000 = 143 mm

Parallel-axis theorem

flange: 160·15³/12 + 2400·(207.5 − 143)² = 1.003 × 10⁷ mm⁴

web: 18·200³/12 + 3600·(100 − 143)² = 1.866 × 10⁷ mm⁴

I_x = 1.003 × 10⁷ + 1.866 × 10⁷ = 2.869 × 10⁷ mm⁴ = 28.69 × 10⁶ mm⁴

Answer

I_x = 28.69 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 46.77 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 100·10 = 1000 mm², y_f = 220 + 10/2 = 225 mm

web: A_w = 10·220 = 2200 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1000·225 + 2200·110)/3200 = 145.9 mm

Parallel-axis theorem

flange: 100·10³/12 + 1000·(225 − 145.9)² = 6.259 × 10⁶ mm⁴

web: 10·220³/12 + 2200·(110 − 145.9)² = 1.171 × 10⁷ mm⁴

I_x = 6.259 × 10⁶ + 1.171 × 10⁷ = 1.797 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 230 − 145.9 = 84.06 mm

σ = M·y/I = 1 × 10⁷·84.06/1.797 × 10⁷ = 46.77 MPa

Answer

σ_top = 46.77 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −67.5 °

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, 50 − 70) = −135°

Angle on the element

θ_p = −67.5°

Answer

θ_p = −67.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_max = 120.4 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·1500000/(π·40³) = 238.7 MPa

τ = 16T/(πd³) = 16·200000/(π·40³) = 15.92 MPa

Mohr’s circle

C = σ/2 = 119.4 MPa

R = √(119.4² + 15.92²) = 120.4 MPa

Answer

τ_max = 120.4 MPa

11.Beam deflection

Answer: δ_P = 65.66 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.

P = 26000 N, a = 1000 mm, L = 4000 mm

EI = 1.32 × 10¹¹ N·mm²

Deflection under the load

δ_P = Pa³/(3EI) = 26000·1000³/(3·1.32 × 10¹¹) = 65.66 mm

Answer

δ_P = 65.66 mm

12.Beam deflection

Answer: h_req = 260.9 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.

w = 4 N/mm, L = 5000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/250 = 5000/250 = 20 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·4·5000⁴/(384·11000·20) = 1.48 × 10⁸ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.48 × 10⁸/100)^1/3 = 260.9 mm

Answer

h_req = 260.9 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 3.312 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·19175/2000² = 3312 N = 3.312 kN

Answer

P_cr = 3.312 kN

14.Column buckling

Answer: σ_cr = 27.76 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 = 1·4000 = 4000 mm

KL/r = 4000/15 = 266.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/266.7² = 27.76 MPa

Answer

σ_cr = 27.76 MPa

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