Agricultural Engineering · Foundations · also Mechanics of Materials, Mechanics of Solids
How members carry load: stress, strain, bending, torsion, deflection and buckling, used in machine and structure design.
In the exams: NCEES FE 9–14 questions; NCEES PE strength of materials; ICAR AIEEA-PG Unit IV
Strength of materials, also called mechanics of materials or mechanics of solids, answers one question: when a member carries a load, how hard is the material working inside it, and how much does the member change shape? Engineering mechanics treats bodies as rigid and finds the forces on them. Strength of materials takes those forces and asks whether a pin will shear, a shaft will twist too far, a beam will sag or a post will buckle. Every machine element and every farm structure is sized with these methods.
For an agricultural engineer the members are specific: the drawbar pin and the three-point linkage of a tractor, the power take-off (PTO) shaft that drives a rotary tiller or thresher, the tool bar of a cultivator, the boom of a field sprayer, the purlins and trusses of a farm shed, the legs of an elevated grain bin, the shell of a spray tank or an air receiver, and the steel pipes of an irrigation main that expand in the sun. The same twelve ideas cover all of them: stress and strain, elastic constants, thermal stress, shear force and bending moment, bending and shear stresses, torsion, deflection, columns, thin shells, combined stress and Mohr's circle, strain energy, and theories of failure.
Worldwide the subject is examined as short numerical problems. The NCEES FE Other Disciplines exam in the USA gives it 9–14 of 110 questions, answered with the NCEES Reference Handbook. India's ICAR AIEEA-PG lists it in Unit IV with soil mechanics and engineering mechanics. Licensing exams for agricultural and biosystems engineers in the Philippines, the NCEES PE Agricultural and Biological exam and national registration exams elsewhere use it inside structures and machine design questions. The skills that score are the same everywhere: draw the free-body diagram, pick the right formula, keep N, mm and MPa consistent, and check whether the answer is sensible.
This manual follows the standard textbook order. Each chapter defines its terms, derives or states the working formula, and solves a farm example. All numbers in worked examples are illustrative: material properties such as modulus of elasticity, yield strength and allowable stress vary with grade, heat treatment, timber species, moisture content and the design code of each country, so a real design must take them from the relevant national code or the supplier's certified data.
Stress is internal force per unit area. When a bar of cross-sectional area A carries an axial force P, the normal stress is σ = P/A, tensile if the bar is pulled and compressive if it is pushed. In SI the unit is the pascal (N/m²); engineers use the megapascal, and 1 MPa equals 1 N/mm², which is why working in newtons and millimetres avoids most unit slips. Shear stress τ acts parallel to a surface: a pin or rivet that resists a force sliding one plate over another carries τ = P/A in single shear, and τ = P/(2A) in double shear, where the pin is cut at two planes. Bearing stress is the contact pressure between a pin and the hole it sits in, taken as P/(d t) on the projected area of the pin diameter d times plate thickness t.
Farm example. A tractor drawbar pulls 20 kN through a clevis, and the hitch pin of 25 mm diameter is in double shear. The pin area is π(25)²/4 = 490.9 mm², so τ = 20,000/(2 × 490.9) = 20.37 MPa. The common mistake is to forget the second shear plane and report twice this value. Designers then compare the working stress with an allowable stress, the yield or ultimate strength divided by a factor of safety. The factor chosen depends on the code, the certainty of the load and the consequence of failure; shock-loaded farm machinery usually needs a larger factor than a static building member.
Normal strain ε is change in length divided by original length, δ/L, and has no unit. Shear strain γ is the change in a right angle, in radians. In a tension test on mild steel the stress–strain curve starts as a straight line up to the limit of proportionality; the elastic limit, beyond which the bar does not return fully to its length, lies just above it. Mild steel then shows an upper and a lower yield point where the bar stretches with little rise in load, followed by strain hardening up to the ultimate tensile stress, then necking and fracture. Ductile materials such as mild steel and aluminium show large plastic strain before fracture; brittle materials such as cast iron and concrete break with little. Materials without a clear yield point are given a proof stress, usually the 0.2 percent offset stress.
Within the proportional limit, Hooke's law holds: σ = Eε, where E is Young's modulus. Combining the definitions gives the elongation of a prismatic bar, δ = PL/(AE). A steel tie rod of a farm-shed truss, 16 mm in diameter and 3 m long, carries 25 kN. Its area is 201.06 mm², stress 124.3 MPa, and with E = 200 GPa the elongation is 25,000 × 3000/(201.06 × 200,000) = 1.87 mm. For a stepped bar, add PL/AE for each segment. For a bar tapering from diameter d₁ to d₂, δ = 4PL/(πE d₁ d₂). A bar hanging under its own weight W stretches WL/(2AE), half what the same load applied at the end would give.
When a load is shared by two materials, such as a concrete post with steel bars or a steel tube around a brass core, equilibrium alone does not give the forces. Add a compatibility condition: both materials shorten by the same amount, so P₁L/(A₁E₁) = P₂L/(A₂E₂), and P₁ + P₂ = P. The stiffer material takes the larger share in proportion to AE. The ratio E₁/E₂ is called the modular ratio, and the same idea is used to convert a composite section into an equivalent section of one material. The method of equilibrium plus compatibility plus the force–deformation law solves every indeterminate problem in this subject, including thermal stress and propped beams.
Example. A steel core of area 500 mm² (E = 200 GPa) sits inside a copper tube of area 1000 mm² (E = 100 GPa), both the same length, and together carry 100 kN. The stiffness ratio is (500 × 200)/(1000 × 100) = 1, so each carries 50 kN: steel stress 100 MPa and copper stress 50 MPa. Equal strain, unequal stress: the stiffer material is always the more highly stressed.
An isotropic linear-elastic material, one whose properties are the same in every direction, is described by four elastic constants. Young's modulus E relates normal stress to normal strain. The modulus of rigidity or shear modulus G relates shear stress to shear strain, τ = Gγ. The bulk modulus K relates a uniform hydrostatic pressure to the volumetric strain it causes, K = p/(ΔV/V). Poisson's ratio ν is the ratio of lateral strain to longitudinal strain under uniaxial stress, taken positive: a bar that stretches along its length gets thinner. Only two of the four are independent for an isotropic material. The relations are E = 2G(1 + ν), E = 3K(1 − 2ν), and, eliminating ν, E = 9KG/(3K + G).
Worked example: steel with E = 200 GPa and ν = 0.3 gives G = 200/(2 × 1.3) = 76.9 GPa and K = 200/(3 × 0.4) = 166.7 GPa. Because K must be positive, ν cannot exceed 0.5; a material with ν = 0.5, like rubber approximately, is incompressible and has infinite K. Common metals have ν between about 0.25 and 0.35. Cork has a Poisson's ratio near zero, which is why it seals a bottle neck without bulging. Timber is not isotropic: its stiffness along the grain is many times that across the grain, and values vary with species and moisture, so the isotropic relations do not apply to it.
Volumetric strain is change in volume over original volume. For small strains it is the sum of the three normal strains, ε_v = ε_x + ε_y + ε_z. For a bar under uniaxial stress σ, the lateral strains are each −νε, so ε_v = ε(1 − 2ν). Under three normal stresses, ε_x = [σ_x − ν(σ_y + σ_z)]/E and similarly for the others, and ε_v = (σ_x + σ_y + σ_z)(1 − 2ν)/E. For a sphere, whose diameter strain is ε_d, ε_v = 3ε_d. For a cylinder, ε_v = 2ε_d + ε_L, where ε_L is the longitudinal strain; this is the basis for finding the extra fluid that must be pumped into a pressurised tank.
Poisson's effect matters on the farm in places students do not expect. A bolt tightened in tension gets slightly thinner. A rubber mounting under an operator's seat is almost incompressible, so it deflects by bulging sideways and must be given room to do so. In the tension test, measuring both longitudinal and lateral strain with strain gauges gives E and ν directly, and G then follows from the relation above.
A bar heated by ΔT expands freely by δ = αLΔT, where α is the coefficient of linear thermal expansion. If the bar is free, there is no stress. If the ends are fixed, the wall pushes back by exactly the amount needed to squeeze the bar back to its original length, so the strain caused by stress is −αΔT and the stress is σ = EαΔT, compressive on heating and tensile on cooling. The length cancels: a short restrained bar and a long one reach the same thermal stress. If the support yields or a gap of size g is left, only the part of the expansion that is blocked causes stress: σ = E(αLΔT − g)/L, and if αLΔT ≤ g the stress is zero.
Farm example. A steel irrigation main 12 m long, anchored at both ends, heats by 30 °C in the sun. With α = 12 × 10⁻⁶ per °C and E = 200 GPa, the fully restrained stress is 200,000 × 12 × 10⁻⁶ × 30 = 72 MPa. Free expansion would be 4.32 mm. If a 0.5 mm gap is left at a joint, the blocked expansion is 3.82 mm and the stress falls to 63.7 MPa. The value of α varies with the grade of steel, and values for aluminium (about twice steel) and plastics (much higher) matter when pipe materials are mixed.
When two materials with different α are joined, as in a steel bolt through a brass or aluminium sleeve, and the assembly is heated, the material with larger α wants to expand more. Neither can move freely, so the one with larger α is put in compression and the one with smaller α in tension. The forces are equal and opposite (no external load), and compatibility says the final length is the same for both: α₁ΔT − σ₁/E₁ = α₂ΔT + σ₂/E₂. Solving these two equations gives both stresses.
Shrink fitting uses thermal expansion deliberately. A steel tyre or ring slightly smaller than the wheel it fits is heated, slipped on and allowed to cool; it tries to return to its smaller diameter and grips the wheel. For a thin ring on a rigid wheel, the hoop strain is (D − d)/d and the hoop stress is E(D − d)/d, where d is the ring's internal diameter and D the wheel's diameter. Steel cartwheel tyres were fitted this way, and bearing races and gears are still shrunk onto shafts. Thermal stress also governs large steel grain bins in hot climates, greenhouse frames, long steel roof sheeting, and the joints between steel and concrete in farm structures, which is why expansion joints and slotted holes are provided.
A beam carries loads perpendicular to its axis. At any section, the shear force V is the algebraic sum of the transverse forces on one side of the section, and the bending moment M is the algebraic sum of the moments of those forces about the section. The usual convention takes a bending moment that makes the beam sag (concave upward) as positive and one that makes it hog as negative. A shear force diagram (SFD) and bending moment diagram (BMD) plot V and M along the span. Two relations make drawing them quick: the slope of the SFD equals minus the load intensity, dV/dx = −w, and the slope of the BMD equals the shear force, dM/dx = V. Hence the maximum bending moment occurs where the shear force is zero or changes sign, and the change in moment between two sections equals the area of the SFD between them.
Consequences worth memorising: under a point load the SFD jumps and the BMD has a kink; over a uniformly distributed load (UDL) the SFD is a sloping straight line and the BMD a parabola; under a uniformly varying load the SFD is a parabola and the BMD a cubic. A concentrated couple makes the BMD jump without changing the SFD.
Cantilever with end load W: shear W throughout, moment WL at the fixed end, zero at the free end. Cantilever with UDL w over length L: shear wL and moment wL²/2 at the fixed end. Simply supported beam with central load W: reactions W/2, maximum moment WL/4 at midspan. With a load W at distances a and b from the supports, the maximum moment is Wab/L under the load. Simply supported beam with UDL: reactions wL/2, maximum moment wL²/8 at midspan, where the shear is zero. A beam with overhangs can have both sagging and hogging moments; the point where the moment changes sign is a point of contraflexure, a good place for a splice.
Farm examples. One half of a field sprayer boom, 6 m long, acts as a cantilever from the centre frame. Treating pipes, nozzles and the frame as a UDL of 200 N/m, the root shear is 1200 N and the root moment 1200 × 3 = 3600 N·m; any bounce over rough ground multiplies this. A timber floor joist in a feed store spans 4 m, simply supported, and carries 2 kN/m. The reactions are 4 kN and the midspan moment is 2 × 4²/8 = 4 kN·m. A cultivator tool bar with tines spaced along it is a beam with several point loads, the draft forces, and its critical section is at the mounting brackets.
Simple bending theory assumes the material is homogeneous, isotropic and linear elastic with the same E in tension and compression, the beam is initially straight, plane sections remain plane after bending, and the radius of curvature is large compared with the depth. Under these assumptions the strain varies linearly from the neutral axis, the fibre that neither stretches nor shortens, which passes through the centroid of the section. The result is the flexure formula M/I = σ/y = E/R: stress is proportional to distance y from the neutral axis, zero at the axis and largest at the extreme fibre. I is the second moment of area about the neutral axis and R the radius of curvature.
The section modulus Z = I/y_max gives the extreme-fibre stress directly, σ_max = M/Z. For a rectangle of width b and depth d, I = bd³/12 and Z = bd²/6, so doubling depth quadruples strength while doubling width only doubles it; this is why joists are set on edge. For a solid circle, Z = πd³/32; for a hollow circle, Z = π(D⁴ − d⁴)/(32D). I-sections and channels put most material far from the neutral axis and so give high Z per kilogram. The 4 m feed-store joist of Chapter 4 with a 4 kN·m moment, if made from a 100 × 200 mm timber on edge, has Z = 100 × 200²/6 = 666667 mm³ and σ_max = 4 × 10⁶/666667 = 6.0 MPa, which is then checked against the permissible bending stress for that species and grade in the national timber code.
Where the bending moment changes along the beam, shear force exists and shear stress is distributed across the depth: τ = VAȳ/(Ib), where A ȳ is the first moment of the area above the level considered about the neutral axis, and b the width at that level. The shear stress is zero at the top and bottom faces and maximum at the neutral axis for rectangles and circles. For a rectangle τ_max = 1.5 τ_avg, where τ_avg = V/(bd); for a solid circle τ_max = (4/3)τ_avg. In an I-section the web carries most of the shear force and the flanges most of the bending moment, and the shear stress jumps where the width changes from flange to web.
For the joist above, the support shear is 4 kN, so τ_avg = 4000/(100 × 200) = 0.20 MPa and τ_max = 0.30 MPa. That looks small, but timber is weak in shear parallel to the grain, so short, deep, heavily loaded timber beams can fail in horizontal shear before bending. A beam of uniform strength, one whose section varies so that M/Z is constant, is used for leaf springs and some tapered implement beams to save weight. Composite (flitched) beams, timber with a steel plate bolted to the side, are analysed by the modular-ratio method of Chapter 1.
A circular shaft twisted by a torque T develops shear stress that varies linearly from zero at the centre to a maximum at the outer surface. Assuming plane cross-sections remain plane and radii remain straight, which is exact only for circular sections, the torsion equation is T/J = τ/r = Gθ/L. Here J is the polar second moment of area, πd⁴/32 for a solid shaft and π(D⁴ − d⁴)/32 for a hollow one, θ is the angle of twist in radians over length L, and G is the shear modulus. The maximum shear stress in a solid shaft is τ = 16T/(πd³); J/r is called the polar section modulus. GJ is the torsional rigidity and GJ/L the torsional stiffness.
Power and torque are linked by P = 2πNT/60 with P in watts, N in rev/min and T in N·m. Shafts are sized for strength (τ within allowable) and for stiffness (θ within an allowable angle per metre); the larger diameter governs. For shafts in series (joined end to end) the torque is the same and the twists add. For shafts in parallel (a compound shaft fixed at both ends and loaded at the junction) the twists are equal and the torques add.
The tractor power take-off turns at standard speeds of 540 or 1000 rev/min (ISO 500). Suppose 30 kW is taken at 540 rev/min to a rotary tiller. The torque is T = 30,000 × 60/(2π × 540) = 530.5 N·m. For an allowable shear stress of 40 MPa, d³ = 16T/(πτ) gives d = 40.7 mm; round up to 41 mm. The actual stress is then 39.2 MPa and, with G = 80 GPa, the twist over one metre is 0.0239 rad, or 1.37°. Running the same power at 1000 rev/min needs only 540/1000 of the torque, which is one reason the higher PTO speed is used for high-power implements.
A hollow shaft is stronger and stiffer than a solid shaft of the same weight because material near the centre carries little stress; for the same strength it is lighter. That is why propeller shafts and PTO drivelines use tubes. Where torque fluctuates, designers use the maximum rather than mean torque, and implement drivelines include shear bolts or friction slip clutches that limit the torque. Non-circular sections warp when twisted and need other formulas; the square and telescoping profiles of PTO drivelines transmit torque but are designed from specialised data, not the circular-shaft equation.
The deflected centreline of a beam is the elastic curve. For small slopes, its curvature is d²y/dx² = M/(EI), so integrating the bending moment twice gives slope and deflection, with constants fixed by support conditions: zero deflection at a simple support, zero slope and zero deflection at a fixed end. This is the double-integration method. Macaulay's method writes a single moment expression valid across the span using bracket terms that are ignored when negative, so one pair of constants covers beams with several loads. The moment-area method uses two theorems: the change in slope between two points equals the area of the M/EI diagram between them, and the deflection of one point relative to the tangent at another equals the first moment of that area. The conjugate-beam method turns the same ideas into a loading problem. Energy methods (Chapter 11) give deflections too.
Standard results to know: cantilever with end load, slope WL²/(2EI) and deflection WL³/(3EI) at the tip; cantilever with UDL, slope wL³/(6EI) and deflection wL⁴/(8EI); simply supported beam with central load, end slope WL²/(16EI) and central deflection WL³/(48EI); simply supported beam with UDL, end slope wL³/(24EI) and central deflection 5wL⁴/(384EI).
Many farm members are governed by deflection, not strength. Take the 4 m joist with 2 kN/m (2 N/mm), 100 × 200 mm, I = bd³/12 = 66.67 × 10⁶ mm⁴. Assuming E = 10 GPa for the timber (the real value depends on species, grade and moisture), δ = 5 × 2 × 4000⁴/(384 × 10,000 × 66.67 × 10⁶) = 10.0 mm. Codes typically limit such deflections to a fraction of the span, commonly between span/180 and span/360 depending on use and country, so this joist must be checked against the local rule. Timber also creeps under long-term load, and codes increase long-term deflection accordingly.
Deflection depends on EI and very strongly on span: doubling the span of a UDL beam multiplies the deflection by 16. A sprayer boom that sags lets end nozzles run closer to the crop than centre nozzles, spoiling spray distribution, and a tool bar that deflects changes tine depth across the implement. Increasing depth is the cheapest cure because I grows with the cube of depth. A propped cantilever or fixed-ended beam is statically indeterminate; its reactions are found by writing deflection compatibility, for example setting the deflection at a prop to zero.
A short, stocky compression member fails by crushing when P/A reaches the compressive strength. A long, slender one fails at a much lower load by buckling: it bows sideways and loses stability. Euler's formula gives the critical load of an ideal, initially straight, centrally loaded elastic column: P_cr = π²EI/L_e², where I is the least second moment of area (buckling happens about the weaker axis) and L_e the effective length. For both ends pinned L_e = L; one end fixed and one free L_e = 2L; both ends fixed L_e = L/2; one fixed and one pinned L_e = L/√2. The radius of gyration is k = √(I/A), and the slenderness ratio L_e/k controls behaviour. Dividing by area gives the critical stress π²E/(L_e/k)². Euler's formula is valid only while this stress is below the proportional limit, which for mild steel means slenderness ratios above roughly 80 to 100, depending on the steel.
Example. A solid steel rod 40 mm in diameter and 2 m long, pinned at both ends, with E = 200 GPa: I = π × 40⁴/64 = 125664 mm⁴, k = d/4 = 10 mm, L_e/k = 200. Euler gives P_cr = π² × 200,000 × 125664/2000² = 62.0 kN. Fixed at the base and free at the top, as a post of a lean-to shed may effectively be, the effective length doubles and P_cr falls to a quarter, 15.5 kN.
Real columns are not perfectly straight or centrally loaded, and intermediate columns fail by a mix of crushing and buckling. Rankine (Rankine–Gordon) combines the two: 1/P = 1/P_c + 1/P_e, which gives P = σ_c A/[1 + a(L_e/k)²], where σ_c is the crushing stress and a = σ_c/(π²E) is Rankine's constant. For short columns it approaches the crushing load and for long ones the Euler load. Indian textbooks commonly use σ_c = 320 MPa and a = 1/7500 for mild steel (pinned ends) as illustrative values. For the 40 mm rod above, these give P = 320 × 1256.6/(1 + 200²/7500) = 63.5 kN, close to Euler because the rod is slender.
Other column formulas appear in other syllabi: the Johnson parabolic formula for intermediate columns and the secant formula for eccentrically loaded columns are in the NCEES handbook tradition, and national steel and timber codes give their own buckling curves with safety factors. Farm uses: posts and legs of elevated grain bins and water tanks, the struts of a roof truss, the piston rod of a front-loader hydraulic cylinder (fully extended, it is a slender column), and the tie-and-strut frames of greenhouses. Bracing a column at mid-height halves its effective length and multiplies its Euler load by four.
A shell is thin when its wall thickness is small compared with its diameter; a common textbook criterion is t less than about D/20, though some texts use D/10. The stress through the wall is then treated as uniform and the radial stress, at most equal to the internal pressure, is neglected. Cutting a closed cylinder of internal diameter D, thickness t, under internal pressure p along its length and balancing forces gives the hoop (circumferential) stress σ_h = pD/(2t). Cutting it across gives the longitudinal stress σ_l = pD/(4t), half the hoop stress. That is why a cylinder bursts along its length. A thin sphere has the same stress in every direction, σ = pD/(4t), so for the same pressure, diameter and stress a sphere needs only half the thickness of a cylinder.
Welded or riveted joints are weaker than the plate. With a longitudinal joint efficiency η_l, the hoop stress becomes pD/(2tη_l); with a circumferential joint efficiency η_c, the longitudinal stress becomes pD/(4tη_c). A tractor-mounted sprayer's pressure vessel or a small air receiver of D = 400 mm and t = 3 mm at p = 1.0 MPa (about 10 bar) has σ_h = 1 × 400/(2 × 3) = 66.7 MPa and σ_l = 33.3 MPa. Pressure-vessel design is governed by national codes with their own allowable stresses, corrosion allowances and inspection rules.
The hoop strain ε_h = (σ_h − νσ_l)/E = (pD/4tE)(2 − ν), the longitudinal strain ε_l = (pD/4tE)(1 − 2ν), and the volumetric strain of the cylinder ε_v = 2ε_h + ε_l = (pD/4tE)(5 − 4ν). For a thin sphere, ε_v = 3(pD/4tE)(1 − ν). The change in volume is ε_v multiplied by the internal volume; this is how the additional liquid needed to pressurise a tank to a test pressure is found, allowing also for the compressibility of the liquid through its bulk modulus.
The maximum in-plane shear stress in a thin cylinder is (σ_h − σ_l)/2 = pD/(8t); including the radial direction, where the stress is close to zero, the absolute maximum shear stress is σ_h/2 = pD/(4t). Farm examples include spray tanks and pressure vessels, compressed-air receivers, hydraulic accumulators and cylinders, pipes of pressurised irrigation systems, and biogas storage vessels. When the wall is thick relative to the diameter, as in hydraulic cylinders at high pressure, Lamé's thick-cylinder equations are needed and the hoop stress is largest at the inner surface.
At a point loaded by normal stresses σ_x and σ_y and shear stress τ_xy, the stresses on a plane rotated by θ are σ_θ = (σ_x + σ_y)/2 + [(σ_x − σ_y)/2]cos 2θ + τ_xy sin 2θ and τ_θ = −[(σ_x − σ_y)/2]sin 2θ + τ_xy cos 2θ. The principal planes carry no shear, and on them the normal stresses reach their extreme values, the principal stresses σ₁,₂ = (σ_x + σ_y)/2 ± √{[(σ_x − σ_y)/2]² + τ_xy²}. The principal planes are at tan 2θ_p = 2τ_xy/(σ_x − σ_y), and the planes of maximum in-plane shear lie at 45° to them, carrying τ_max = √{[(σ_x − σ_y)/2]² + τ_xy²} together with a normal stress equal to the average (σ_x + σ_y)/2.
Mohr's circle draws all this. Plot the points (σ_x, τ_xy) and (σ_y, −τ_xy); the line joining them is a diameter, the centre sits at (σ_x + σ_y)/2 on the σ axis, and the radius equals τ_max. Rotating 2θ on the circle corresponds to θ on the element. For σ_x = 80 MPa, σ_y = -20 MPa and τ_xy = 30 MPa: centre 30 MPa, radius √(50² + 30²) = 58.31 MPa, so σ₁ = 88.3 MPa, σ₂ = -28.3 MPa, τ_max = 58.3 MPa, and θ_p = ½ tan⁻¹(60/100) = 15.5°. The sum σ₁ + σ₂ always equals σ_x + σ_y. In pure shear the circle is centred at the origin and the principal stresses are ±τ at 45°, which is why a brittle chalk twisted to failure breaks on a 45° helix.
Real members carry several actions at once. An axial load plus bending gives σ = P/A ± M/Z. A load P applied with eccentricity e on a short column adds a moment Pe; for a rectangular section b wide, no tension develops if e ≤ b/6 (the middle-third rule), and for a solid circle if e ≤ d/8. This core of the section matters for masonry and concrete, which crack in tension: farm-structure walls, masonry piers and small check dams are proportioned so the resultant stays within the middle third.
Shafts carry bending from pulley, sprocket or gear forces together with torque. The bending stress σ = 32M/(πd³) and the torsional shear τ = 16T/(πd³) act at the same surface point. Combining them gives the equivalent twisting moment T_e = √(M² + T²), used with the maximum shear stress theory, and the equivalent bending moment M_e = ½[M + √(M² + T²)], used with the maximum principal stress theory. For a thresher drum shaft with M = 1.2 kN·m and T = 0.9 kN·m, T_e = 1.50 kN·m and M_e = 1.350 kN·m. Rotavator, thresher and combine shafts are sized this way, with shock and fatigue factors from machine-design practice added.
Work done by a load in deforming an elastic body is stored as strain energy U, recoverable when the load is removed. For a bar under gradually applied load P, the load rises from zero, so U = ½Pδ = σ²V/(2E), where V is the volume. The strain energy per unit volume at a stress σ is σ²/(2E). Resilience is the strain energy stored; proof resilience is the maximum stored at the elastic limit; the modulus of resilience is proof resilience per unit volume, σ_e²/(2E). Toughness, the energy absorbed up to fracture, is the area under the whole stress–strain curve and is much larger for ductile materials.
The way a load is applied changes the stress. A gradually applied load P gives σ = P/A. A load applied suddenly, the full value at once but with no velocity, gives twice that, σ = 2P/A, because the load does work Pδ while the bar stores only ½σAδ. A falling load W dropping height h onto a collar at the end of a bar of length L gives σ = (W/A)[1 + √(1 + 2AEh/(WL))]; if h is large compared with the extension, σ ≈ √(2WhE/(AL)). Example: W = 1 kN falling 10 mm onto a bar 2 m long with A = 1000 mm² and E = 200 GPa gives σ = 1 × [1 + √(1 + 2 × 1000 × 200,000 × 10/(1000 × 2000))] = 45.7 MPa, compared with 1 MPa if the same weight were placed gently.
In shear, strain energy per unit volume is τ²/(2G). For a solid circular shaft in torsion, U = τ_max²V/(4G), and for a hollow shaft U = [τ_max²/(4G)](D² + d²)/D² × V. For bending, U = ∫M²dx/(2EI), and for torsion in general U = T²L/(2GJ). Castigliano's theorem states that the partial derivative of the strain energy with respect to a load gives the deflection in the direction of that load; it gives beam deflections and solves indeterminate problems. For a cantilever with end load W, U = W²L³/(6EI) and ∂U/∂W = WL³/(3EI), matching Chapter 7.
Strain energy explains practical design. A shock load is resisted by energy absorption, so a longer, more flexible member with a lower stiffness can survive an impact that shatters a short stiff one; bolts subject to impact are sometimes turned down in the shank to the root diameter so the energy is spread along the length. Tine springs on cultivators, shock absorbers on implement mounts and the safety shear bolts on ploughs and PTO drives all work with energy: the spring stores the energy of hitting a stone, and the shear bolt fails at a set load before costlier parts do.
A tension test gives a single failure stress, but real parts see combined stresses. A theory of failure states which quantity, computed for the combined state, is compared with its value at failure in the simple tension test. The maximum principal stress theory (Rankine) says failure occurs when the largest principal stress reaches the tensile strength; it suits brittle materials. The maximum shear stress theory (Tresca, also called Guest's theory) says yield occurs when the maximum shear stress reaches half the tensile yield stress; it suits ductile materials and is slightly conservative. The maximum principal strain theory (Saint-Venant) compares the largest principal strain with the yield strain. The total strain energy theory (Haigh) compares the total strain energy per unit volume. The distortion energy or shear strain energy theory (von Mises, also Hencky or Maxwell–Huber–Hencky–von Mises) compares the energy of shape change only and agrees best with tests on ductile metals.
In pure shear the theories predict different shear yield strengths: Tresca τ_y = 0.5σ_y, von Mises τ_y = σ_y/√3 ≈ 0.577σ_y, and maximum principal stress τ_y = σ_y. Under hydrostatic stress, distortion energy is zero, so ductile metals do not yield in pure hydrostatic compression, which the distortion-energy theory captures and the total strain energy theory does not.
For a member with one normal stress σ and one shear stress τ, as at the surface of a shaft in bending and torsion, the equivalent stresses are: Tresca, σ_eq = √(σ² + 4τ²); von Mises, σ_eq = √(σ² + 3τ²); maximum principal stress, σ₁ = σ/2 + √[(σ/2)² + τ²]. For σ = 100 MPa and τ = 50 MPa, Tresca gives 141.4 MPa and von Mises 132.3 MPa. Each is compared with the yield stress divided by the factor of safety. For a shaft, Tresca leads to d³ = 16T_e/(πτ_allow) with T_e = √(M² + T²), which is the equivalent-torque method of Chapter 10.
Choosing a theory: use maximum principal stress (or the Coulomb–Mohr theory where compressive strength exceeds tensile strength) for brittle materials such as cast iron; use Tresca or von Mises for ductile steels and aluminium, von Mises when accuracy matters and Tresca when a simple conservative check is wanted. Plotted in the σ₁–σ₂ plane for plane stress, Tresca is a hexagon inscribed in the von Mises ellipse; they coincide for uniaxial stress and differ most in pure shear, by about 15 percent. Fatigue, which governs most rotating farm-machine parts, is a separate check using endurance limits and stress concentration factors, treated in machine design.
A B.Tech student uses this subject in every design course that follows. The first habit to build is a clean free-body diagram, the second is units: newtons and millimetres give stress straight in MPa. In the laboratory the student runs tension, torsion, deflection and column tests and compares measured E, G and buckling loads with theory. For exams, memorise the standard SF/BM and deflection results, effective lengths, thin-cylinder formulas and the relation E = 2G(1 + ν), and practise Mohr's circle until it takes under three minutes. A project such as a trolley frame, a thresher shaft or a small shed truss is where these pieces first come together.
Five years into practice, an engineer in a machinery firm, testing centre or structures unit uses these methods to check rather than to derive. Typical tasks: confirm a PTO driveline or gearbox shaft for the torque of a heavier tractor, size a hitch pin or shear bolt, check whether a sprayer boom or tool bar will deflect too much, verify posts and bracing of a poly-house or grain-bin support, and read strain-gauge results from field tests. The decisions are practical: add a gusset, change section, increase depth, add a brace to halve an effective length, or reject a supplier's lighter tube. Knowing which national code sets allowable stresses and deflection limits is part of the job.
A senior engineer with 25 years of work sets design rules and signs off. Failures come back from the field: a cracked tool bar at a weld, a bent loader cylinder rod, a split spray tank. The senior engineer reads the fracture, decides whether the cause was overload, buckling, fatigue or a stress concentration, and changes the design standard or the test procedure. They choose factors of safety for new product lines, balance weight against durability, make sure calculations follow the governing code for each market, and decide when finite-element analysis and prototype strain testing are needed instead of hand calculation. They also train juniors to sanity-check computer output against simple beam and shaft formulas.
A researcher with a PhD and long experience uses strength of materials as the base for advanced work: finite-element modelling of implement frames and soil-engaging tools, fatigue life prediction from measured field load spectra, design of lightweight composite or high-strength-steel booms, buckling and wall-pressure behaviour of thin-walled grain silos, and stress analysis of biological materials such as grain kernels, fruit and stems under compression. The researcher validates models against instrumented tests, publishes load data that design standards can adopt, and supervises students who must still master the classical closed-form solutions to judge whether numerical results are plausible.
This manual covers the classical B.Tech syllabus with closed-form solutions. It does not yet cover thick cylinders beyond naming Lamé's equations, curved beams, unsymmetrical bending and shear centre, springs (treated in machine design), fatigue and stress concentration (machine design), plastic analysis, the secant and Johnson column formulas beyond naming them, or finite-element methods. It gives no design allowable stresses, deflection limits or material property tables for any specific code, because these differ between countries and grades; worked examples use illustrative values only. Timber and composite material behaviour (orthotropy, creep, moisture effects) is only mentioned.
Original practice questions written to the syllabus and pattern of the exams named in each question's examTags. They are not past papers and do not reproduce any board's questions. Material properties in the questions are illustrative; real design values come from the relevant national code and certified supplier data.
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