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

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 60 mm, E = 110 GPa). Both are 800 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø25 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 110 GPaL = 800 mm

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 80 kNsegment 1: Ø40 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)

δ = mm

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 = 1000 N·md = 25 mmL = 0.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 3000 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 50 kWN = 3000 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. It carries a point load P₁ between the supports and a point load P₂ at C. Find the bending moment M_x under the load P₁. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP₁ = 35 kN at 2.5 mP₂ = 30 kN at 5 msupports at x = 0 and x = 4 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 10 kN at 5 mw = 6 kN/m from 0 to 8 mx = 5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 210 mm wide and 22 mm thick, an overall depth of 340 mm and a web 10 mm thick. Find its elastic section modulus S_x for bending about the horizontal centroidal axis x.

B = 210 mmH = 340 mmt_f = 22 mmt_w = 10 mm

S_x = × 10³ mm³

8.Centroid & second moment of areaHard

A 150 mm × 200 mm plate has a rectangular cut-out 80 mm wide and 50 mm high, centred on the vertical centre line with its centre 80 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 200 mmcut-out: 80 × 50 mmhole centre: y = 80 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 100 MPa and τₓᵧ = 10 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 120 MPaσ_y = 100 MPaτ_xy = 10 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 10 MPa, σᵧ = 90 MPa and τₓᵧ = 50 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° clockwise from the x-axis.

σ_x = 10 MPaσ_y = 90 MPaτ_xy = 50 MPaθ = -20°

σ_θ = 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 = 10 kN/mL = 4 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 125 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 = 1.5 kN/mL = 6 mE = 11 GPa (timber)b = 125 mm

h_req = mm

13.Column bucklingMedium

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

L = 1 mE = 70 GPa (aluminium)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1.5 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.07758 mm

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.

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·8482 = 1.031 × 10⁹ N

Shortening

δ = 100000·800/1.031 × 10⁹ = 0.07758 mm

Answer

δ = 0.07758 mm

2.Axial stress & strain

Answer: δ = 4.387 mm

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 63.66 MPa

σ₂ = 707.4 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 63.66·1500/70000 = 1.364 mm

δ₂ = σ₂L₂/E₂ = 707.4·500/117000 = 3.023 mm

δ = δ₁ + δ₂ = 4.387 mm

Answer

δ = 4.387 mm

3.Torsion of shafts

Answer: φ = 9.338 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·500/(80000·38350) = 0.163 rad

φ = 0.163·180/π = 9.338°

Answer

φ = 9.338°

4.Torsion of shafts

Answer: d = 27.26 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 50000/314.2 = 159.2 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·159155/(π·40)) = 27.26 mm

Answer

d = 27.26 mm

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

5.Shear force & bending moment

Answer: M_x = 14.06 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. The overhang load also acts about A.

ΣM_A = 0: R_B·4 = 35·2.5 + 30·5

R_B = 237.5/4 = 59.38 kN

ΣF_y = 0: R_A = 65 − 59.38 = 5.625 kN

Cut under P₁ and keep the left part

M_x = R_A·2.5 = 5.625·2.5 = 14.06 kN·m

Answer

M_x = 14.06 kN·m

6.Shear force & bending moment

Answer: M_x = 63.75 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 = 10·5 + 6·8·4

R_B = 242/8 = 30.25 kN

ΣF_y = 0: R_A = 58 − 30.25 = 27.75 kN

Cut at x = 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 = 27.75·5 − 6·5·2.5 = 63.75 kN·m

Answer

M_x = 63.75 kN·m

7.Centroid & second moment of area

Answer: S_x = 1503 × 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 = 340 − 2·22 = 296 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (210·340³ − 200·296³)/12

I_x = 2.556 × 10⁸ mm⁴

Section modulus

c = H/2 = 170 mm

S_x = I/c = 2.556 × 10⁸/170

S_x = 1.503 × 10⁶ mm³ = 1503 × 10³ mm³

Answer

S_x = 1503 × 10³ mm³

8.Centroid & second moment of area

Answer: I_x = 97.32 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (30000·100 − 4000·80)/(30000 − 4000) = 103.1 mm

Full plate about ȳ

150·200³/12 + 30000·−3.077² = 100.3 × 10⁶ mm⁴

Hole about ȳ

80·50³/12 + 4000·−23.08² = 2.964 × 10⁶ mm⁴

Subtract

I_x = 97.32 × 10⁶ mm⁴

Answer

I_x = 97.32 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 62.07 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 110 MPa, R = √(10² + 10²) = 14.14 MPa

σ₁ = 124.1 MPa, σ₂ = 95.86 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(28.28, 124.1, 95.86)/2 = 62.07 MPa

Answer

τ_abs = 62.07 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: σ_θ = −12.78 MPa

Transformation equation

σ_θ = 50 + −40·0.766 + 50·−0.6428 = −12.78 MPa

Answer

σ_θ = −12.78 MPa

11.Beam deflection

Answer: δ = 28.57 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 = 10 N/mm, L = 4000 mm

EI = 1.12 × 10¹³ N·mm²

Deflection

δ = wL⁴/(8EI) = 10·4000⁴/(8·1.12 × 10¹³) = 28.57 mm

Answer

δ = 28.57 mm

12.Beam deflection

Answer: h_req = 236.7 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 = 1.5 N/mm, L = 6000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 6000/360 = 16.67 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1.5·6000⁴/(384·11000·16.67) = 1.381 × 10⁸ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.381 × 10⁸/125)^1/3 = 236.7 mm

Answer

h_req = 236.7 mm

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

13.Column buckling

Answer: P_cr = 124.4 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 = 1·1000 = 1000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·180000/1000² = 124357 N = 124.4 kN

Answer

P_cr = 124.4 kN

14.Column buckling

Answer: σ_cr = 34.27 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 2·1500 = 3000 mm

KL/r = 3000/12.5 = 240

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/240² = 34.27 MPa

Answer

σ_cr = 34.27 MPa

ExamProblems · examproblems.com/courses/mechanics-of-materials

Mechanics of Materials: cram pack

1Formulas and methods

Axial stress & strain

Torsion of shafts

Method

Shear force & bending moment

Method

Centroid & second moment of area

Mohr’s circle & stress transformation

Method

Beam deflection

Column buckling

2Practice problems

3Solutions

Areas

Combined axial stiffness

Shortening

Answer

Same force in both segments

Add the segment elongations

Answer

Polar second moment of area

Angle of twist

Answer

Torque

Diameter from the allowable stress

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Cut at x = 5 m and keep the left part

Answer

Outer rectangle minus the two side cut-outs

Section modulus

Answer

Centroid

Full plate about ȳ

Hole about ȳ

Subtract

Answer

In-plane principal stresses

Largest of the three circles

Answer

Transformation equation

Answer

Consistent units

Deflection

Answer

Consistent units

Deflection limit

Required second moment of area

Required depth

Answer

Smallest second moment of area

Effective length

Critical load

Answer

Smallest second moment of area

Radius of gyration and slenderness

Critical stress

Answer

Cut under P₁ and keep the left part