Solid mechanics, machines & thermal cycles.

To compute the slope and deflection of a loaded beam by twice integrating the bending-moment / flexural-rigidity relation EI·y'' = M(x), and to locate the point of maximum deflection and check it against the span/deflection serviceability ratio.

The curvature of a beam is proportional to the bending moment: the governing differential equation of the elastic curve is EI·d2y/dx2 = M(x). Integrating once gives the slope theta(x) = dy/dx; integrating again gives the deflection y(x). Two constants of integration are fixed by the boundary conditions. For a simply-supported beam the deflection is zero at both supports; for a cantilever the deflection and slope are both zero at the fixed end.

Here the integration is performed numerically (trapezoidal rule) over 240 stations, which reproduces the closed-form results for standard load cases.

1. Choose the support type and set the loads as in Experiment 1.
2. Set the flexural rigidity by choosing the Young's modulus E and second moment of area I.
3. The bending-moment diagram is integrated twice; read the maximum deflection and its location.
4. Increase I (a stiffer section) and confirm the deflection falls in proportion to 1/I.
5. Compare the deflection/span ratio with a typical serviceability limit of L/360.

• R. C. Hibbeler, Mechanics of Materials, 10th ed., Pearson, Ch. 12 (Deflection of Beams).
• E. P. Popov, Engineering Mechanics of Solids, Prentice Hall.
• IIT Virtual Labs — Strength of Materials, vlabs.ac.in.

To study the engineering stress-strain behaviour of common structural metals under uniaxial tension — identifying the elastic region, yield point, ultimate tensile strength (UTS) and fracture — and to evaluate the factor of safety for a chosen working stress.

In the elastic region stress is proportional to strain (Hooke's law) and the slope of the line is the Young's modulus E. At the yield point the material begins to deform permanently. Beyond yield the metal strain-hardens, the stress rising to a maximum — the ultimate tensile strength — after which a neck forms, the engineering stress drops, and the specimen fractures. The factor of safety relates a strength limit to the actual working stress.

Ductile metals (mild steel) show a large plastic plateau before fracture; aluminium and high-strength steels are stronger but less ductile.

1. Select a material from the dropdown; note its tabulated E, yield, UTS and fracture strain.
2. Increase the applied strain slider and watch the marker climb the curve.
3. Identify the strain at which the response leaves the straight elastic line (yield).
4. Continue past the UTS into the necking region and observe the load drop.
5. Set a working stress and read the factor of safety; aim for FoS of 2 or more.

• W. D. Callister, Materials Science and Engineering: An Introduction, Wiley.
• G. E. Dieter, Mechanical Metallurgy, McGraw-Hill.
• IIT Virtual Labs — Strength of Materials (Tension test), vlabs.ac.in.

To determine the axial force in every member of a simple, statically-determinate planar truss using the method of joints, and to classify each member as being in tension or compression.

A pin-jointed truss carries load only as axial forces in its members. At every joint the forces are concurrent, so two equilibrium equations apply: the sum of horizontal forces and the sum of vertical forces are each zero. Starting at a joint with no more than two unknown members, these equations are solved for the member forces, then the solver moves to the next joint. A positive result (assuming members pull on the joint) means the member is in tension; a negative result means compression.

Here a symmetric king-post truss is solved: reactions split the apex load equally, then joints A and B give the rafter, chord and vertical-tie forces directly.

1. Set the span and height; the apex point D moves to keep the truss symmetric.
2. Apply a downward point load at the apex joint D.
3. The support reactions at the pin (A) and roller (C) are computed from global equilibrium.
4. Each member force is solved by the method of joints and shown with its sign.
5. Note which members carry tension (chords) and which carry compression (rafters).
6. Raise the height and observe the rafter forces fall as the truss becomes deeper.

• R. C. Hibbeler, Structural Analysis, 10th ed., Pearson, Ch. 3 (Trusses).
• F. P. Beer & E. R. Johnston, Vector Mechanics for Engineers: Statics, McGraw-Hill.
• IIT Virtual Labs — Engineering Mechanics, vlabs.ac.in.

To construct Mohr's circle for a given plane-stress state (sigma_x, sigma_y, tau_xy), and from it to determine the principal stresses, the maximum in-plane shear stress, the orientation of the principal planes, and the stress components on any rotated plane.

Under plane stress the normal and shear stress on a plane inclined at theta vary as the plane rotates. Plotting normal stress sigma horizontally and shear stress tau vertically, every plane orientation maps to a point on a circle — Mohr's circle. Its centre lies on the sigma axis at the average normal stress; its radius equals the maximum in-plane shear. The two points where the circle crosses the sigma axis are the principal stresses, where shear vanishes. A rotation of theta on the element corresponds to a rotation of 2θ on the circle.

1. Enter the in-plane stress components sigma_x, sigma_y and the shear tau_xy.
2. Read the circle centre and radius, plotted directly on the sigma-tau axes.
3. Identify the principal stresses sigma_1 and sigma_2 where the circle meets the sigma axis.
4. Read tau_max (the top of the circle) and the principal-plane angle theta_p.
5. Sweep the plane angle theta and watch the X-face point travel 2θ around the circle.

• R. C. Hibbeler, Mechanics of Materials, 10th ed., Pearson, Ch. 9 (Stress Transformation).
• J. M. Gere & B. J. Goodno, Mechanics of Materials, Cengage.
• IIT Virtual Labs — Strength of Materials, vlabs.ac.in.

To analyse the kinematics of simple and compound gear trains — computing the overall velocity ratio, the output speed, the multiplied output torque and the direction of rotation — and to observe live meshing of the gears.

Two meshing gears have the same pitch-line velocity, so their speeds are inversely proportional to their numbers of teeth. For a compound train the overall ratio is the product of the individual stage ratios. Each external mesh reverses the direction of rotation, so an even number of meshes returns the output to the input direction. Neglecting losses, power is conserved, so the torque is multiplied by the same ratio that the speed is divided by; a small efficiency factor accounts for friction at each mesh.

1. Choose a single-stage or two-stage (compound) train.
2. Set the tooth counts; a larger driven gear gives a speed reduction and torque gain.
3. Set the input speed and torque and read the multiplied output values.
4. Check the output direction: one mesh reverses it, two meshes restore it.
5. Watch the gears mesh; smaller gears spin proportionally faster.

• R. L. Norton, Design of Machinery, McGraw-Hill, Ch. 9 (Gear Trains).
• J. E. Shigley, Theory of Machines and Mechanisms, Oxford.
• IIT Virtual Labs — Theory of Machines, vlabs.ac.in.

To plot the P-V diagram of the air-standard Otto cycle and the Carnot cycle, and to compute the thermal efficiency, heat input and rejection, net work and mean effective pressure, comparing the Otto efficiency with the Carnot upper bound for the same temperature limits.

The air-standard Otto cycle (the ideal spark-ignition engine) consists of isentropic compression, constant-volume heat addition, isentropic expansion and constant-volume heat rejection. Its efficiency depends only on the compression ratio and the specific-heat ratio gamma. The Carnot cycle, two isotherms joined by two adiabats, sets the maximum possible efficiency for any cycle operating between the same hot and cold temperatures. The net work equals the area enclosed by the cycle on the P-V plane.

For the same temperature limits the Otto efficiency is always below the Carnot limit, since heat is added and rejected across finite temperature differences.

1. Select the Otto or Carnot cycle.
2. Set the compression ratio and the ratio of specific heats gamma.
3. Set the cold (intake) and hot (peak) temperatures.
4. Read the thermal efficiency and compare it with the Carnot limit shown alongside.
5. Increase the compression ratio and watch the Otto efficiency rise toward, but never reach, the Carnot value.

• Y. A. Cengel & M. A. Boles, Thermodynamics: An Engineering Approach, McGraw-Hill.
• M. J. Moran et al., Fundamentals of Engineering Thermodynamics, Wiley.
• IIT Virtual Labs — Thermodynamics, vlabs.ac.in.

To study the free and forced vibration of a single-degree-of-freedom mass-spring-damper system: to find its natural frequency and damping ratio, to classify the damping regime (under-, critically- or over-damped), and to observe resonance as the forcing frequency approaches the natural frequency.

A single mass on a spring and damper obeys the second-order equation of motion m·x'' + c·x' + k·x = F(t). Its undamped natural frequency is set by the stiffness-to-mass ratio. The damping ratio zeta compares the actual damping to the critical value: below 1 the system oscillates with a decaying envelope (under-damped); at 1 it returns to rest as fast as possible without overshoot (critical); above 1 it crawls back slowly (over-damped). Under harmonic forcing the steady-state amplitude is largest when the forcing frequency nears the natural frequency — resonance.

1. Choose free vibration (an initial 20 mm displacement) or harmonic forcing.
2. Set the mass, stiffness and damping; read the natural frequency and damping ratio.
3. Note the regime: under-damped oscillates and decays, over-damped does not oscillate.
4. For forced vibration, sweep the forcing frequency toward the natural frequency and watch the amplitude grow at resonance.
5. Increase the damping and confirm the resonant peak shrinks and broadens.

• S. S. Rao, Mechanical Vibrations, 6th ed., Pearson.
• D. J. Inman, Engineering Vibration, Pearson.
• IIT Virtual Labs — Dynamics of Machines, vlabs.ac.in.

To discretise a 2D cantilever plate into constant-strain triangular (CST) finite elements, assemble the global stiffness matrix, apply a clamped support and a tip shear load, solve the equilibrium system K·U = F with a hand-written Gauss-elimination solver, recover the element stresses, and visualise the von Mises stress field over the deformed shape.

The finite-element method replaces a continuous body by an assembly of simple elements. For a constant-strain triangle the displacement varies linearly, so strain (and therefore stress) is constant within each element. The strain-displacement matrix B is built from the nodal coordinates; the constitutive matrix D is the plane-stress elasticity matrix. The element stiffness is the volume integral of B-transpose D B, which for a CST reduces to a closed form because B and D are constant. Elements are summed into a global stiffness matrix K; constrained degrees of freedom are removed, and the remaining linear system is solved for the nodal displacements.

The computed tip deflection is compared with the Euler-Bernoulli beam estimate P·L^3/(3·E·I); the two converge as the mesh is refined.

1. Set the plate length, height and thickness, and the mesh density (elements along each side).
2. Each rectangular cell is split into two CST elements; nodes are numbered automatically.
3. Choose the material modulus E and Poisson ratio nu; the plane-stress D matrix is formed.
4. Apply the downward tip shear load P; the left edge is fully clamped.
5. The global stiffness matrix is assembled and K·U = F is solved by Gauss elimination.
6. Read the tip deflection and the maximum von Mises stress; compare with beam theory.
7. Refine the mesh and watch the FE tip deflection approach the beam-theory value.

• O. C. Zienkiewicz & R. L. Taylor, The Finite Element Method, Vol. 1, Butterworth-Heinemann.
• T. R. Chandrupatla & A. D. Belegundu, Introduction to Finite Elements in Engineering, Pearson.
• R. D. Cook et al., Concepts and Applications of Finite Element Analysis, Wiley.

To compute the Euler critical buckling load of a slender column for different end conditions, to relate the critical stress to the slenderness ratio K·L/r through the Euler curve, to locate the operating point and the transition to material yielding, and to draw the fundamental buckled mode shape for the chosen supports.

A perfectly straight, axially loaded column remains straight until the load reaches a critical value; beyond it the straight configuration becomes unstable and the column buckles sideways. Euler showed the critical load depends on the flexural rigidity EI, the length, and how the ends are restrained, captured by an effective-length factor K. Dividing by the cross-sectional area gives the critical stress, which falls off as the square of the slenderness ratio. Very stocky columns fail instead by material yielding, so the usable curve is the lower of the Euler stress and the yield stress.

For a solid circular section I = pi·d^4/64 and A = pi·d^2/4, so the radius of gyration is simply d/4.

1. Choose the end conditions; the effective-length factor K updates accordingly.
2. Set the column length, modulus, diameter and the material yield stress.
3. Read the slenderness ratio and the Euler critical load P_cr.
4. Locate the operating point on the Euler curve and note whether it lies above or below the yield cut-off.
5. Apply a working load P and read the factor of safety against buckling.
6. Observe the buckled mode shape change with the end conditions (number of half-waves and inflection points).

• S. P. Timoshenko & J. M. Gere, Theory of Elastic Stability, McGraw-Hill.
• R. C. Hibbeler, Mechanics of Materials, 10th ed., Pearson, Ch. 13 (Buckling of Columns).
• IIT Virtual Labs — Strength of Materials, vlabs.ac.in.

To construct the S-N (stress-life) curve for a chosen metal using the Basquin relation, to plot the Goodman, Soderberg and Gerber mean-stress failure lines on a constant-life (Haigh) diagram, to evaluate the fatigue factor of safety for a given mean and alternating stress, and to estimate the cycles to failure from the equivalent fully-reversed stress.

Components subjected to fluctuating loads can fail at stresses well below the static strength — fatigue. A cyclic stress is described by its mean and alternating components. The endurance limit Se is the fully-reversed stress amplitude the material can sustain for a very large number of cycles. A non-zero mean stress is detrimental: the Goodman, Soderberg and Gerber criteria each draw a failure boundary on the mean-versus-alternating stress plane, the safe region lying below the line. The Soderberg line (to yield) is the most conservative, Goodman (to ultimate) is widely used, and the Gerber parabola best fits ductile test data. Below the infinite-life region, the finite life follows the Basquin power law fitted between the low-cycle and endurance anchor points.

The endurance limit is estimated as Se = 0.5·Sut for steels (capped near 700 MPa), the standard first approximation before Marin surface, size and reliability factors are applied.

1. Select a material; its ultimate, yield and endurance strengths are tabulated.
2. Set the mean stress sigma_m and alternating stress sigma_a of the service load.
3. Read the load point plotted on the Haigh (constant-life) diagram against the three failure lines.
4. Compare the Goodman, Soderberg and Gerber factors of safety.
5. Read the equivalent fully-reversed stress sigma_ar and the estimated cycles to failure from the S-N curve.
6. Increase the mean stress and observe the factor of safety fall even though the amplitude is unchanged.

• R. G. Budynas & J. K. Nisbett, Shigley's Mechanical Engineering Design, McGraw-Hill, Ch. 6 (Fatigue Failure).
• N. E. Dowling, Mechanical Behavior of Materials, Pearson.
• R. L. Norton, Machine Design: An Integrated Approach, Pearson.