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 steel bar under the axial load.

P = 50 kNd = 30 mmL = 2.5 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 50 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 300 mm, E = 110 GPaP = 40 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 = 200 N·md = 30 mmL = 3 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft of diameter 25 mm is fixed at A. A torque of 400 N·m acts at B, 1.2 m from A, and a torque of 100 N·m in the opposite direction acts at the free end C, a further 0.8 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 25 mmT_B = 400 N·mT_C = 100 N·mAB = 1.2 mBC = 0.8 mG = 39 GPa

φ_C = °

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 3 kN/m from 0 to 10 mx = 2.5 m

V_x = kN

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 35 kN at 6 mw = 6 kN/m from 0 to 8 mx = 7.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 220 mm wide and 22 mm thick, an overall depth of 380 mm and a web 6 mm thick. Find its elastic section modulus S_x for bending about the horizontal centroidal axis x.

B = 220 mmH = 380 mmt_f = 22 mmt_w = 6 mm

S_x = × 10³ mm³

8.Centroid & second moment of areaHard

A welded I-section has a top flange 160 × 16 mm, a web 330 mm high and 14 mm thick, and a bottom flange 230 × 22 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 160 × 16 mmweb: 14 × 330 mmbottom flange: 230 × 22 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 110 MPa, σᵧ = 10 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 110 MPaσ_y = 10 MPaτ_xy = −20 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 10 MPa, σᵧ = 100 MPa and τₓᵧ = 30 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 10 MPaσ_y = 100 MPaτ_xy = 30 MPa

σ_vM = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 8 kN/mL = 4 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ = 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 = 19 kNw = 6 kN/mL = 6 mE = 200 GPa (steel)I = 57 × 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 = 1.5 mE = 70 GPa (aluminium)b × h = 40 × 60 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 3 against buckling. Find the smallest diameter d.

P = 105 kNn = 3L = 3 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.8842 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 = 50·1000/706.9 = 70.74 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 70.74/200000 = 0.0003537 = 353.7 µε

Elongation

δ = ε·L = 0.0003537·2500 mm = 0.8842 mm

Answer

δ = 0.8842 mm

2.Axial stress & strain

Answer: σ₂ = 12.31 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 = 200000·706.9/300 = 471239 N/mm

k₂ = 110000·1963/300 = 719948 N/mm

Share of segment 2

F₂ = 40000·719948/1.191 × 10⁶ = 24176 N

Stress

σ₂ = 24176/1963 = 12.31 MPa (compression)

Answer

σ₂ = 12.31 MPa

3.Torsion of shafts

Answer: φ = 5.404 °

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/(80000·79522) = 0.09431 rad

φ = 0.09431·180/π = 5.404°

Answer

φ = 5.404°

4.Torsion of shafts

Answer: φ_C = 10.73 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 100 N·m

T_AB = T_C − T_B = 100 − 400 = −300 N·m

Twist of each segment

J = π·25⁴/32 = 38350 mm⁴

G = 39 GPa = 39000 MPa

φ_AB = −300000·1200/(39000·38350) = −0.2407 rad

φ_BC = 100000·800/(39000·38350) = 0.05349 rad

Twist of end C

φ_C = −0.2407 + 0.05349 = −0.1872 rad → |φ_C| = 10.73°

Answer

φ_C = 10.73°

5.Shear force & bending moment

Answer: V_x = 7.5 kN

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 = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 3·10·5

R_B = 150/10 = 15 kN

ΣF_y = 0: R_A = 30 − 15 = 15 kN

Cut at x = 2.5 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.

V = R_A − (all downward loads left of the cut)

V_x = 15 − 3·2.5 = 7.5 kN

Answer

V_x = 7.5 kN

6.Shear force & bending moment

Answer: M_x = 24.38 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·8 = 48 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 35·6 + 6·8·4

R_B = 402/8 = 50.25 kN

ΣF_y = 0: R_A = 83 − 50.25 = 32.75 kN

Cut at x = 7.5 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 = 32.75·7.5 − 35·(7.5 − 6) − 6·7.5·3.75 = 24.38 kN·m

Answer

M_x = 24.38 kN·m

7.Centroid & second moment of area

Answer: S_x = 1734 × 10³ mm³

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 380 − 2·22 = 336 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (220·380³ − 214·336³)/12

I_x = 3.295 × 10⁸ mm⁴

Section modulus

c = H/2 = 190 mm

S_x = I/c = 3.295 × 10⁸/190

S_x = 1.734 × 10⁶ mm³ = 1734 × 10³ mm³

Answer

S_x = 1734 × 10³ mm³

8.Centroid & second moment of area

Answer: I_x = 259.2 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 2560 mm², y = 360 mm

web: A = 4620 mm², y = 187 mm

bottom flange: A = 5060 mm², y = 11 mm

Centroid

ȳ = (2560·360 + 4620·187 + 5060·11)/12240 = 150.4 mm

Parallel-axis theorem

top flange: 160·16³/12 + 2560·209.6² = 1.125 × 10⁸ mm⁴

web: 14·330³/12 + 4620·36.58² = 4.811 × 10⁷ mm⁴

bottom flange: 230·22³/12 + 5060·139.4² = 9.857 × 10⁷ mm⁴

I_x = 1.125 × 10⁸ + 4.811 × 10⁷ + 9.857 × 10⁷

I_x = 2.592 × 10⁸ mm⁴ = 259.2 × 10⁶ mm⁴

Answer

I_x = 259.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = −55.9 °

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −20/53.85 = −0.3714, sin 2θ_s = −50/53.85 = −0.9285

2θ_s = atan2(−50, −20) = −111.8°

Angle on the element

θ_s = −55.9°

check: θ_s − θ_p = −55.9 − (−10.9) = −45°

Answer

θ_s = −55.9°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_vM = 108.6 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 = 55 MPa, R = 54.08 MPa

σ₁ = 109.1 MPa, σ₂ = 0.9167 MPa, σ₃ = 0

Von Mises

σ_vM = √(109.1² − 109.1·0.9167 + 0.9167²) = 108.6 MPa

Answer

σ_vM = 108.6 MPa

11.Beam deflection

Answer: δ = 64.16 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 = 4000 mm

EI = 3.99 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 8·4000⁴/(8·3.99 × 10¹²) = 64.16 mm

Answer

δ = 64.16 mm

12.Beam deflection

Answer: δ_max = 16.38 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 = 19000 N, w = 6 N/mm, L = 6000 mm, EI = 1.14 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 19000·6000³/(48·1.14 × 10¹³) = 7.5 mm

δ_w = 5wL⁴/(384EI) = 5·6·6000⁴/(384·1.14 × 10¹³) = 8.882 mm

Add them

δ_max = 7.5 + 8.882 = 16.38 mm

Answer

δ_max = 16.38 mm

13.Column buckling

Answer: P_cr = 200.5 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 = 60·40³/12 = 320000 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·320000/1050² = 200525 N = 200.5 kN

Answer

P_cr = 200.5 kN

14.Column buckling

Answer: d = 80 mm

Required critical load

P_cr = n·P = 3·105000 = 315000 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·315000·2100²/(π³·70000))^(1/4) = 80 mm

Answer

d = 80 mm

Check: the slenderness KL/r is 105, so the column is slender and Euler applies.

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