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 = 15 kNd = 10 mmL = 2 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 500 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, Ø25 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 500 mm, E = 110 GPaP = 40 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 50 mm and length 1 m; segment BC has diameter 20 mm and length 1 m. A torque of 100 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 100 N·md₁ = 50 mm, L₁ = 1 md₂ = 20 mm, L₂ = 1 mG = 26 GPa

φ_C = °

4.Torsion of shaftsHard

A solid brass shaft of diameter 25 mm is fixed at A. A torque of 250 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 2 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 25 mmT_B = 250 N·mT_C = 100 N·mAB = 1.2 mBC = 2 mG = 39 GPa

φ_C = °

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 = 20 kN at 3 mw = 8 kN/m from 0 to 3 m

M_A = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment M_B at support B. Use sagging moments positive and upward shear on the left face positive.

L = 7.5 mw = 8 kN/m from 0 to 7.5 msupports at x = 0 and x = 5 m

M_B = kN·m

7.Centroid & second moment of areaMedium

A hollow rectangular (box) section has outer width B = 300 mm, outer height H = 300 mm and a uniform wall thickness t = 20 mm. Find its radius of gyration r about the horizontal centroidal axis.

B = 300 mmH = 300 mmt = 20 mm

r = mm

8.Centroid & second moment of areaHard

A welded I-section has a top flange 180 × 24 mm, a web 330 mm high and 14 mm thick, and a bottom flange 210 × 20 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 180 × 24 mmweb: 14 × 330 mmbottom flange: 210 × 20 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 0 MPa, σᵧ = −30 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 = 0 MPaσ_y = −30 MPaτ_xy = 20 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = −50 MPa and τₓᵧ = −50 MPa (σ_z = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 120 MPaσ_y = −50 MPaτ_xy = −50 MPaσ_Y = 355 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 5 m from support A. Find the deflection δ_P under the load.

P = 19 kNa = 5 mL = 6 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 150 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 2.5 kN/mL = 5 mE = 11 GPa (timber)b = 150 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 = 3 mE = 70 GPa (aluminium)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 55 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 55 kNn = 2L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 1.91 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·10²/4 = 78.54 mm²

σ = P/A = 15·1000/78.54 = 191 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 191/200000 = 0.0009549 = 954.9 µε

Elongation

δ = ε·L = 0.0009549·2000 mm = 1.91 mm

Answer

δ = 1.91 mm

2.Axial stress & strain

Answer: σ₂ = 27.57 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 = 200000·490.9/600 = 163625 N/mm

k₂ = 110000·706.9/500 = 155509 N/mm

Share of segment 2

F₂ = 40000·155509/319133 = 19491 N

Stress

σ₂ = 19491/706.9 = 27.57 MPa (compression)

Answer

σ₂ = 27.57 MPa

3.Torsion of shafts

Answer: φ_C = 14.39 °

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 moments

J₁ = π·50⁴/32 = 613592 mm⁴

J₂ = π·20⁴/32 = 15708 mm⁴

Twist of each segment

T = 100 N·m = 100000 N·mm

G = 26 GPa = 26000 MPa

φ_AB = 100000·1000/(26000·613592) = 0.006268 rad

φ_BC = 100000·1000/(26000·15708) = 0.2449 rad

Total twist

φ_C = 0.006268 + 0.2449 = 0.2511 rad = 14.39°

Answer

φ_C = 14.39°

4.Torsion of shafts

Answer: φ_C = 0.7662 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 100 N·m

T_AB = T_C − T_B = 100 − 250 = −150 N·m

Twist of each segment

J = π·25⁴/32 = 38350 mm⁴

G = 39 GPa = 39000 MPa

φ_AB = −150000·1200/(39000·38350) = −0.1204 rad

φ_BC = 100000·2000/(39000·38350) = 0.1337 rad

Twist of end C

φ_C = −0.1204 + 0.1337 = 0.01337 rad → |φ_C| = 0.7662°

Answer

φ_C = 0.7662°

5.Shear force & bending moment

Answer: M_A = −96 kN·m

Keep the free-end part

P: −20·3 = −60 kN·m

w: −8·3²/2 = −36 kN·m

M_A = −60 − 36 = −96 kN·m

Answer

M_A = −96 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_B = −25 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 = 8·7.5 = 60 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·5 = 8·7.5·3.75

R_B = 225/5 = 45 kN

ΣF_y = 0: R_A = 60 − 45 = 15 kN

Cut at B and keep the overhang

The load on the overhang, w·a, acts at a/2 from B and makes the beam hog over the support.

M_B = −w·a²/2 = −8·2.5²/2 = −25 kN·m

Answer

M_B = −25 kN·m

7.Centroid & second moment of area

Answer: r = 114.6 mm

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. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

b = 260 mm, h = 260 mm

A = BH − bh = 300·300 − 260·260 = 22400 mm²

I = (BH³ − bh³)/12 = 2.942 × 10⁸ mm⁴

Radius of gyration

r = √(I/A) = √(2.942 × 10⁸/22400) = 114.6 mm

Answer

r = 114.6 mm

8.Centroid & second moment of area

Answer: I_x = 306.2 × 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 = 4320 mm², y = 362 mm

web: A = 4620 mm², y = 185 mm

bottom flange: A = 4200 mm², y = 10 mm

Centroid

ȳ = (4320·362 + 4620·185 + 4200·10)/13140 = 187.3 mm

Parallel-axis theorem

top flange: 180·24³/12 + 4320·174.7² = 1.321 × 10⁸ mm⁴

web: 14·330³/12 + 4620·2.256² = 4.195 × 10⁷ mm⁴

bottom flange: 210·20³/12 + 4200·177.3² = 1.321 × 10⁸ mm⁴

I_x = 1.321 × 10⁸ + 4.195 × 10⁷ + 1.321 × 10⁸

I_x = 3.062 × 10⁸ mm⁴ = 306.2 × 10⁶ mm⁴

Answer

I_x = 306.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = −18.43 °

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 = 20/25 = 0.8, sin 2θ_s = −15/25 = −0.6

2θ_s = atan2(−15, 20) = −36.87°

Angle on the element

θ_s = −18.43°

check: θ_s − θ_p = −18.43 − 26.57 = −45°

Answer

θ_s = −18.43°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 1.8

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 = 35 MPa, R = 98.62 MPa

σ₁ = 133.6 MPa, σ₂ = −63.62 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 197.2 MPa

n = 355/197.2 = 1.8

Answer

n = 1.8

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_P = 15.52 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 = 19000 N, a = 5000 mm, b = 1000 mm, L = 6000 mm

EI = 1.7 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 19000·5000²·1000²/(3·1.7 × 10¹²·6000) = 15.52 mm

Answer

δ_P = 15.52 mm

12.Beam deflection

Answer: h_req = 220 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 = 2.5 N/mm, L = 5000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2.5·5000⁴/(384·11000·13.89) = 1.332 × 10⁸ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.332 × 10⁸/150)^1/3 = 220 mm

Answer

h_req = 220 mm

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

13.Column buckling

Answer: P_cr = 55.27 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 = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·180000/1500² = 55270 N = 55.27 kN

Answer

P_cr = 55.27 kN

14.Column buckling

Answer: d = 46.16 mm

Required critical load

P_cr = n·P = 2·55000 = 110000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·110000·2000²/(π³·200000))^(1/4) = 46.16 mm

Answer

d = 46.16 mm

Check: the slenderness KL/r is 173.3, so the column is slender and Euler applies.

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