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 aluminium bar under the axial load.

P = 30 kNd = 25 mmL = 1.5 mE = 70 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 200 mm, E = 110 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, Ø20 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 200 mm, E = 110 GPaP = 50 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 400 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

T = 400 N·md_i/d_o = 0.75τ_allow = 40 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 500 rpmd_o = 50 mmd_i = 30 mm

τ_max = MPa

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 5 kN/m from 0 to 6 mx = 2 m

V_x = kN

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.

60 kN @ 0 m40 kN @ 6 m15 kN/m, 1–4 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 200 mm wide and 10 mm thick, an overall depth of 500 mm and a web 10 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 200 mmH = 500 mmt_f = 10 mmt_w = 10 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 5 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 120 mmt_f = 10 mmt_w = 20 mmh_w = 100 mmM = 5 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 140 MPa, σᵧ = −20 MPa and τₓᵧ = −40 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 140 MPaσ_y = −20 MPaτ_xy = −40 MPaθ = -30°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −10 MPa and τₓᵧ = −40 MPa (σ_z = 0). The material has a yield strength of 300 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −10 MPaσ_y = −10 MPaτ_xy = −40 MPaσ_Y = 300 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 8 kN/mL = 2 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

θ_A = rad

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/300. Find the smallest depth h_req of the section.

w = 2 kN/mL = 4.5 mE = 11 GPa (timber)b = 100 mm

h_req = mm

13.Column bucklingMedium

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

L = 1 mE = 70 GPa (aluminium)d = 25 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 60 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 60 kNn = 2.5L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 1.31 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·25²/4 = 490.9 mm²

σ = P/A = 30·1000/490.9 = 61.12 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 61.12/70000 = 0.0008731 = 873.1 µε

Elongation

δ = ε·L = 0.0008731·1500 mm = 1.31 mm

Answer

δ = 1.31 mm

2.Axial stress & strain

Answer: σ₂ = 45.97 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 = 200000·314.2/300 = 209440 N/mm

k₂ = 110000·706.9/200 = 388772 N/mm

Share of segment 2

F₂ = 50000·388772/598212 = 32495 N

Stress

σ₂ = 32495/706.9 = 45.97 MPa (compression)

Answer

σ₂ = 45.97 MPa

3.Torsion of shafts

Answer: d_o = 42.08 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 = 400 N·m = 400000 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·400000/(π·40·0.6836)) = 42.08 mm

d_i = 0.75·42.08 = 31.56 mm

Answer

d_o = 42.08 mm

A solid shaft would need d = 37.07 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: τ_max = 17.88 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 20000/52.36 = 382 N·m

Polar second moment of area

J = π(50⁴ − 30⁴)/32 = 534071 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 381972·25/534071 = 17.88 MPa

Answer

τ_max = 17.88 MPa

5.Shear force & bending moment

Answer: V_x = 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 = 5·6 = 30 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 5·6·3

R_B = 90/6 = 15 kN

ΣF_y = 0: R_A = 30 − 15 = 15 kN

Cut at x = 2 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 − 5·2 = 5 kN

Answer

V_x = 5 kN

6.Shear force & bending moment

Answer: E (x = 4 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 = 75.83 kN, R₂ = 69.17 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 = −60, M_C = −51.67, M_D = −58.33, M_E = −80 kN·m, M_F = −40, M_G = 0

Answer

Mark E: M_E = −80 kN·m

7.Centroid & second moment of area

Answer: I_x = 332.3 × 10⁶ mm⁴

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 = 500 − 2·10 = 480 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (200·500³ − 190·480³)/12

I_x = 3.323 × 10⁸ mm⁴ = 332.3 × 10⁶ mm⁴

Answer

I_x = 332.3 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 49.9 MPa

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 = 120·10 = 1200 mm², y_f = 100 + 10/2 = 105 mm

web: A_w = 20·100 = 2000 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1200·105 + 2000·50)/3200 = 70.63 mm

Parallel-axis theorem

flange: 120·10³/12 + 1200·(105 − 70.63)² = 1.428 × 10⁶ mm⁴

web: 20·100³/12 + 2000·(50 − 70.63)² = 2.517 × 10⁶ mm⁴

I_x = 1.428 × 10⁶ + 2.517 × 10⁶ = 3.945 × 10⁶ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 110 − 70.63 = 39.38 mm

σ = M·y/I = 5 × 10⁶·39.38/3.945 × 10⁶ = 49.9 MPa

Answer

σ_top = 49.9 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = −14.64 MPa

Transformation equation for the y′ face

σ_y′ = 60 − 80·0.5 − (−40)·(−0.866) = −14.64 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 134.6 MPa, σ_x′ + σ_y′ = 120 = σ_x + σ_y

Answer

σ_y′ = −14.64 MPa

10.Mohr’s circle & stress transformation

Answer: n = 4.286

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 = −10 MPa, R = 40 MPa

σ₁ = 30 MPa, σ₂ = −50 MPa, σ₃ = 0

Von Mises

σ_vM = √(30² − 30·(−50) + (−50)²) = 70 MPa

n = 300/70 = 4.286

Answer

n = 4.286

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

11.Beam deflection

Answer: θ_A = 0.00202 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 2000 mm

EI = 1.32 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 8·2000³/(24·1.32 × 10¹²) = 0.00202 rad

Answer

θ_A = 0.00202 rad

That is 0.1157°.

12.Beam deflection

Answer: h_req = 198 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 N/mm, L = 4500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/300 = 4500/300 = 15 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2·4500⁴/(384·11000·15) = 6.472 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·6.472 × 10⁷/100)^1/3 = 198 mm

Answer

h_req = 198 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 = 2·1000 = 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: d = 70.55 mm

Required critical load

P_cr = n·P = 2.5·60000 = 150000 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·150000·4000²/(π³·200000))^(1/4) = 70.55 mm

Answer

d = 70.55 mm

Check: the slenderness KL/r is 226.8, so the column is slender and Euler applies.

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