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 copper (outer diameter 100 mm, inner diameter 70 mm, E = 120 GPa). Both are 600 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 120 GPaL = 600 mm

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 60 kNsegment 1: Ø30 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 50 mm and length 0.8 m; segment BC has diameter 40 mm and length 0.8 m. A torque of 200 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 200 N·md₁ = 50 mm, L₁ = 0.8 md₂ = 40 mm, L₂ = 0.8 m

τ_max = MPa

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 100 kWN = 500 rpmτ_allow = 75 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 30 kN·m at x = 3.5 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 5 mM₀ = 30 kN·m clockwise, at 3.5 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1.5 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 = 5.5 mw = 4 kN/m from 0 to 5.5 msupports at x = 0 and x = 4 m

M_B = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 210 mm wide and 18 mm thick, an overall depth of 460 mm and a web 14 mm thick. Find its elastic section modulus S_x for bending about the horizontal centroidal axis x.

B = 210 mmH = 460 mmt_f = 18 mmt_w = 14 mm

S_x = × 10³ mm³

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 bending stress σ_bottom at the bottom edge.

b_f = 140 mmt_f = 25 mmt_w = 8 mmh_w = 200 mmM = 10 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −40 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 = −40 MPaσ_y = −30 MPaτ_xy = 20 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries a bending moment of 1000 N·m together with a torque of 750 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τ_max.

d = 30 mmM = 1000 N·mT = 750 N·m

τ_max = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 20 kNa = 1 mL = 8 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the size of the bending moment M_A at the fixed end.

L = 4 mw = 5 kN/m

M_A = kN·m

13.Column bucklingMedium

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

L = 6 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 5 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 5 kNn = 3L = 2.5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.05183 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(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 120·10³·4006 = 5.788 × 10⁸ N

Shortening

δ = 50000·600/5.788 × 10⁸ = 0.05183 mm

Answer

δ = 0.05183 mm

2.Axial stress & strain

Answer: σ_max = 298.4 MPa

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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 84.88 MPa

σ₂ = 298.4 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 298.4 MPa

Answer

σ_max = 298.4 MPa

3.Torsion of shafts

Answer: τ_max = 15.92 MPa

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.

Stress in each segment

T = 200 N·m = 200000 N·mm

τ_AB = 16·200000/(π·50³) = 8.149 MPa

τ_BC = 16·200000/(π·40³) = 15.92 MPa

Answer

τ_max = 15.92 MPa

4.Torsion of shafts

Answer: d = 50.62 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 100000/52.36 = 1910 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·1.91 × 10⁶/(π·75)) = 50.62 mm

Answer

d = 50.62 mm

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

5.Shear force & bending moment

Answer: M_x = 9 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 30/5 = 6 kN (upward)

R_A = −R_B = −6 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −6·3.5 + 30 = 9 kN·m

Answer

M_x = 9 kN·m

6.Shear force & bending moment

Answer: M_B = −4.5 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 = 4·5.5 = 22 kN, acting at x̄ = 2.75 m

ΣM_A = 0: R_B·4 = 4·5.5·2.75

R_B = 60.5/4 = 15.13 kN

ΣF_y = 0: R_A = 22 − 15.13 = 6.875 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 = −4·1.5²/2 = −4.5 kN·m

Answer

M_B = −4.5 kN·m

7.Centroid & second moment of area

Answer: S_x = 1993 × 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 = 460 − 2·18 = 424 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (210·460³ − 196·424³)/12

I_x = 4.584 × 10⁸ mm⁴

Section modulus

c = H/2 = 230 mm

S_x = I/c = 4.584 × 10⁸/230

S_x = 1.993 × 10⁶ mm³ = 1993 × 10³ mm³

Answer

S_x = 1993 × 10³ mm³

8.Centroid & second moment of area

Answer: σ_bottom = 91.28 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 = 140·25 = 3500 mm², y_f = 200 + 25/2 = 212.5 mm

web: A_w = 8·200 = 1600 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3500·212.5 + 1600·100)/5100 = 177.2 mm

Parallel-axis theorem

flange: 140·25³/12 + 3500·(212.5 − 177.2)² = 4.542 × 10⁶ mm⁴

web: 8·200³/12 + 1600·(100 − 177.2)² = 1.487 × 10⁷ mm⁴

I_x = 4.542 × 10⁶ + 1.487 × 10⁷ = 1.941 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 177.2 mm

σ = M·y/I = 1 × 10⁷·177.2/1.941 × 10⁷ = 91.28 MPa

Answer

σ_bottom = 91.28 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = 7.018 °

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

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

cos 2θ_s = 20/20.62 = 0.9701, sin 2θ_s = −(−5)/20.62 = 0.2425

2θ_s = atan2(5, 20) = 14.04°

Angle on the element

θ_s = 7.018°

check: θ_s − θ_p = 7.018 − 52.02 = −45°

Answer

θ_s = 7.018°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_max = 235.8 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·1000000/(π·30³) = 377.3 MPa

τ = 16T/(πd³) = 16·750000/(π·30³) = 141.5 MPa

Mohr’s circle

C = σ/2 = 188.6 MPa

R = √(188.6² + 141.5²) = 235.8 MPa

Answer

τ_max = 235.8 MPa

11.Beam deflection

Answer: δ_P = 23.2 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 = 20000 N, a = 1000 mm, b = 7000 mm, L = 8000 mm

EI = 1.76 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 20000·1000²·7000²/(3·1.76 × 10¹²·8000) = 23.2 mm

Answer

δ_P = 23.2 mm

12.Beam deflection

Answer: M_A = 10 kN·m

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·5·4/8 = 7.5 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = wL²/2 − R_B·L = 40 − 30 = 10 kN·m

Answer

M_A = 10 kN·m

13.Column buckling

Answer: P_cr = 139.5 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 round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/3000² = 139528 N = 139.5 kN

Answer

P_cr = 139.5 kN

14.Column buckling

Answer: d = 26.24 mm

Required critical load

P_cr = n·P = 3·5000 = 15000 N

KL = 0.7·2500 = 1750 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·15000·1750²/(π³·200000))^(1/4) = 26.24 mm

Answer

d = 26.24 mm

Check: the slenderness KL/r is 266.8, so the column is slender and Euler applies.

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