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 axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 115 kNd = 12 mmL = 2.5 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø50 mm, E = 70 GPa, σ_allow = 100 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid brass shaft 0.5 m long carries a torque of 200 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 200 N·mL = 0.5 mG = 39 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A solid brass shaft of diameter 30 mm is fixed at A. A torque of 400 N·m acts at B, 0.8 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 1.5 m along. Find the maximum shear stress τ_max in the shaft.

d = 30 mmT_B = 400 N·mT_C = 100 N·mAB = 0.8 mBC = 1.5 m

τ_max = MPa

5.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 2 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 = 6 mP₁ = 70 kN at 1 mP₂ = 80 kN at 6 msupports at x = 0 and x = 4 m

M_x = kN·m

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.

20 kN @ 0 m50 kN @ 6 m20 kN @ 2 m15 kN/m, 1–5 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 150 mm wide and 16 mm thick, an overall depth of 460 mm and a web 6 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 150 mmH = 460 mmt_f = 16 mmt_w = 6 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 150 × 20 mm, a web 330 mm high and 8 mm thick, and a bottom flange 210 × 14 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 150 × 20 mmweb: 8 × 330 mmbottom flange: 210 × 14 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 0 MPa, σᵧ = 120 MPa and τₓᵧ = 30 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 = 120 MPaτ_xy = 30 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries an axial compressive force of 130 kN together with a torque of 750 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 30 mmF = −130 kNT = 750 N·m

σ₁ = 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 = 6 kN/mL = 3 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 8 mP = 40 kN at 4 m

R_B = kN

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 = 4 mE = 70 GPa (aluminium)b × h = 40 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 35 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 35 kNn = 3L = 4 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 8691 µε

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·12²/4 = 113.1 mm²

σ = P/A = 115·1000/113.1 = 1017 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1017/117000 = 0.008691 = 8691 µε

Answer

ε = 8691 µε

2.Axial stress & strain

Answer: P_max = 829.6 kN

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·40²/4 = 1257 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·5890 = 6.637 × 10⁸ N

Governing strain

ε = min(0.00125, 0.001429) = 0.00125 (steel)

Largest load

P_max = 0.00125·6.637 × 10⁸ = 829577 N = 829.6 kN

Answer

P_max = 829.6 kN

3.Torsion of shafts

Answer: d = 41.59 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.

Required polar moment

T = 200 N·m = 200000 N·mm

L = 0.5 m = 500 mm

G = 39 GPa = 39000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 200000·500/(39000·0.008727) = 293825 mm⁴

Diameter

d = (32J/π)^1/4 = 41.59 mm

Answer

d = 41.59 mm

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

4.Torsion of shafts

Answer: τ_max = 94.31 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 100 N·m

T_AB = T_C + T_B = 100 + 400 = 500 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 500 N·m

τ_max = 16·500000/(π·30³) = 94.31 MPa

Answer

τ_max = 94.31 MPa

5.Shear force & bending moment

Answer: M_x = 12.5 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.

Support reactions

Equilibrium of the whole beam. The overhang load also acts about A.

ΣM_A = 0: R_B·4 = 70·1 + 80·6

R_B = 550/4 = 137.5 kN

ΣF_y = 0: R_A = 150 − 137.5 = 12.5 kN

Cut under P₁ and keep the left part

M_x = R_A·1 = 12.5·1 = 12.5 kN·m

Answer

M_x = 12.5 kN·m

6.Shear force & bending moment

Answer: F (x = 5 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 = 57.5 kN, R₂ = 92.5 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 = −20, M_C = 10, M_D = 5, M_E = −15, M_F = −50 kN·m, M_G = 0

Answer

Mark F: M_F = −50 kN·m

7.Centroid & second moment of area

Answer: I_x = 275.9 × 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 = 460 − 2·16 = 428 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (150·460³ − 144·428³)/12

I_x = 2.759 × 10⁸ mm⁴ = 275.9 × 10⁶ mm⁴

Answer

I_x = 275.9 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 202.9 × 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 = 3000 mm², y = 354 mm

web: A = 2640 mm², y = 179 mm

bottom flange: A = 2940 mm², y = 7 mm

Centroid

ȳ = (3000·354 + 2640·179 + 2940·7)/8580 = 181.3 mm

Parallel-axis theorem

top flange: 150·20³/12 + 3000·172.7² = 8.963 × 10⁷ mm⁴

web: 8·330³/12 + 2640·2.252² = 2.397 × 10⁷ mm⁴

bottom flange: 210·14³/12 + 2940·174.3² = 8.932 × 10⁷ mm⁴

I_x = 8.963 × 10⁷ + 2.397 × 10⁷ + 8.932 × 10⁷

I_x = 2.029 × 10⁸ mm⁴ = 202.9 × 10⁶ mm⁴

Answer

I_x = 202.9 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = 31.72 °

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 = 30/67.08 = 0.4472, sin 2θ_s = −(−60)/67.08 = 0.8944

2θ_s = atan2(60, 30) = 63.43°

Angle on the element

θ_s = 31.72°

check: θ_s − θ_p = 31.72 − 76.72 = −45°

Answer

θ_s = 31.72°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ₁ = 76.77 MPa

Stresses at the surface

σ = F/A = −130000/(π·30²/4) = −130000/706.9 = −183.9 MPa

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

Mohr’s circle

C = σ/2 = −91.96 MPa

R = √((−91.96)² + 141.5²) = 168.7 MPa

σ₁ = C + R = 76.77 MPa

σ₂ = C − R = −260.7 MPa

Answer

σ₁ = 76.77 MPa

11.Beam deflection

Answer: δ = 10.33 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 = 6 N/mm, L = 3000 mm

EI = 5.88 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 6·3000⁴/(8·5.88 × 10¹²) = 10.33 mm

Answer

δ = 10.33 mm

12.Beam deflection

Answer: R_B = 12.5 kN

Remove the prop

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

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → R_B = 5P/16 = 5·40/16 = 12.5 kN

Answer

R_B = 12.5 kN

13.Column buckling

Answer: P_cr = 37.6 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 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·426667/2800² = 37598 N = 37.6 kN

Answer

P_cr = 37.6 kN

14.Column buckling

Answer: d = 59.32 mm

Required critical load

P_cr = n·P = 3·35000 = 105000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·105000·2000²/(π³·70000))^(1/4) = 59.32 mm

Answer

d = 59.32 mm

Check: the slenderness KL/r is 134.9, so the column is slender and Euler applies.

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

Stress

Strain (Hooke’s law)

Answer

Areas

Combined axial stiffness

Governing strain

Largest load

Answer

Required polar moment

Diameter

Answer

Internal torques

Maximum shear stress

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Bending moment at every mark

Answer

Outer rectangle minus the two side cut-outs

Answer

Split into three rectangles

Centroid

Parallel-axis theorem

Answer

Angle on the circle

Angle on the element

Answer

Stresses at the surface

Mohr’s circle

Answer

Consistent units

Deflection

Answer

Remove the prop

Compatibility at the prop

Answer

Smallest second moment of area

Effective length

Critical load

Answer

Required critical load

Solve for the diameter

Answer

Cut under P₁ and keep the left part