Questions & explanations
1. A wall has thermal resistance to conduction of 0.5 m²K/W. The inner surface is at 30°C, outer at 10°C. Convection (h=5 W/m²K) and radiation (ε=0.9, surroundings at 10°C) act on the outer surface. Find the heat flux through the wall.
Let T_o be the outer surface temperature. The conduction through the wall: q = (30 - T_o)/0.5. At the outer surface, convection: q_conv = 5*(T_o - 10), radiation: q_rad = 0.9*5.67×10⁻⁸*(T_o⁴ - 283⁴). Energy balance: q = q_conv + q_rad. Solve iteratively. Assume T_o=15°C (288 K). q_cond = (30-15)/0.5=30 W/m². q_conv=5*(15-10)=25. q_rad=0.9*5.67e-8*(288⁴-283⁴)=0.9*5.67e-8*(6.88e9-6.42e9)=0.9*5.67e-8*4.6e8=0.9*26.1=23.5. Total=48.5 >30. Try T_o=12°C (285 K). q_cond=(30-12)/0.5=36. q_conv=5*(12-10)=10. q_rad=0.9*5.67e-8*(285⁴-283⁴)=0.9*5.67e-8*(6.60e9-6.42e9)=0.9*5.67e-8*1.8e8=0.9*10.2=9.2. Total=19.2 <36. So T_o between 12 and 15. Interpolate: at T_o=13.5°C (286.5 K), q_cond=33, q_conv=17.5, q_rad≈? 286.5⁴=6.74e9, 283⁴=6.42e9, diff=0.32e9, q_rad=0.9*5.67e-8*3.2e8=0.9*18.1=16.3, total=33.8 ≈33. So heat flux ≈33 W/m².
2. What does the Reynolds transport theorem allow us to do when analyzing fluid flow?
The Reynolds transport theorem is a tool that helps us convert equations written for a moving fluid mass (a system) into equations for a fixed region in space (a control volume). This is useful because in engineering we often want to analyze flow through pipes or around objects, where it's easier to track what enters and leaves a fixed volume than to follow a specific mass of fluid. The theorem relates the rate of change of a property (like mass or momentum) in a system to the rate of change inside a control volume plus the net flow of that property across the control surface. For example, it forms the basis of the continuity equation and the momentum equation for control volumes. Without it, we would have to track individual fluid particles, which is impractical for most real-world problems.
3. How does the ASME elliptic criterion compare to the Goodman and Gerber criteria?
The ASME elliptic criterion gives results between the Goodman and Gerber criteria for many materials. At zero mean stress, all three give the endurance limit. At high mean stress near ultimate strength, the ASME elliptic criterion allows some alternating stress, while Goodman gives zero and Gerber gives a small amount. For example, with endurance limit 200 MPa and ultimate 600 MPa, at mean stress 400 MPa: Goodman gives alternating stress = 200*(1-400/600)=66.7 MPa; Gerber gives 200*sqrt(1-(400/600)^2)=149.1 MPa; ASME elliptic gives 200*sqrt(1-(400/600)^2)=149.1 MPa as well (same as Gerber because both are quadratic). But at intermediate mean stresses, the shapes differ slightly. The ASME elliptic is often preferred for its smooth curve and good agreement with test data.
4. How does the Gerber criterion differ from the Soderberg criterion?
The Gerber criterion uses a parabolic curve instead of a straight line. Its equation is alternating stress / endurance limit + (mean stress / ultimate strength)^2 = 1. The Gerber line is less conservative than the Soderberg line, meaning it allows higher alternating stress for a given mean stress. Soderberg uses yield strength and is very safe, while Gerber uses ultimate strength and is more realistic for many ductile materials. For example, at a mean stress of 200 MPa with yield 400 MPa and ultimate 600 MPa, Soderberg gives alternating stress = endurance limit * (1 - 200/400) = half the endurance limit, while Gerber gives alternating stress = endurance limit * (1 - (200/600)^2) = about 0.89 times endurance limit. So Gerber allows more alternating stress.
5. What are Marin factors in fatigue design?
Marin factors are correction factors used to adjust the endurance limit of a material for real-world conditions. The endurance limit from a standard test (rotating beam test) applies only to a smooth, polished specimen in a laboratory. Marin factors account for differences like surface finish, size, loading type, temperature, and reliability. Each factor is a number between 0 and 1 (or sometimes >1) that multiplies the endurance limit. For example, a rough surface reduces the endurance limit, so the surface factor is less than 1. The modified endurance limit = endurance limit * surface factor * size factor * load factor * temperature factor * reliability factor. This gives a more realistic fatigue strength for the actual part.
6. A point in a structure has σx = -50 MPa, σy = 20 MPa, and τxy = -40 MPa. Using Mohr's circle, find the orientation of the principal planes.
First, compute σ_avg = (-50+20)/2 = -15 MPa. Center at (-15,0). Radius R = sqrt(((-50-20)/2)^2 + (-40)^2) = sqrt((-35)^2 + 1600) = sqrt(1225+1600)= sqrt(2825)≈53.15 MPa. Plot point A (σx, τxy)=(-50, -40). The angle 2θp from the x-axis to the principal plane is given by tan(2θp) = (2τxy)/(σx-σy) = (2*(-40))/(-50-20) = -80/(-70)=1.1429, so 2θp ≈ 48.8° (since tan positive, angle in first quadrant). But note: the sign convention: if τxy is negative, the angle is measured clockwise from the x-axis. So θp ≈ 24.4° clockwise from the x-axis. Alternatively, from Mohr's circle, the angle from point A to the principal stress point (σ1) is 2θp measured counterclockwise from A to the rightmost point. So the orientation is 24.4° clockwise.
7. How does spin softening affect the natural frequencies of a rotating beam?
Spin softening is the reduction in apparent stiffness of a rotating structure due to centrifugal forces. When a beam rotates, centrifugal forces act outward, creating tensile stresses that increase geometric stiffness. However, the term 'spin softening' refers to the decrease in stiffness when the rotation speed is high enough that the centrifugal forces cause a reduction in effective stiffness for certain modes. Actually, for a rotating beam, the natural frequencies generally increase with rotation speed due to centrifugal stiffening. Spin softening is a misnomer; it is the opposite effect where rotation reduces stiffness in some cases (e.g., for a rotating ring). In most rotating beams, centrifugal stiffening dominates.
8. A water jet hits a stationary flat plate and splits into two streams. Use the Reynolds transport theorem to explain how to find the force on the plate.
To find the force, we apply the Reynolds transport theorem to the momentum equation. We choose a control volume that encloses the plate and the jet. The theorem says the net force on the control volume equals the rate of change of momentum inside plus the net momentum flow out. For steady flow, the rate of change inside is zero. So the force on the plate equals the difference between the incoming momentum flow (mass flow rate times velocity) and the outgoing momentum flows. Since the plate is symmetric, the outgoing streams have equal velocities but opposite directions, so their horizontal momentum cancels. The force is simply the incoming momentum flow, which is the mass flow rate of the jet times its velocity.
9. A two-span continuous beam has a uniform load on the left span only. Describe the steps to find the final moments at the interior support using moment distribution.
First, lock all joints and calculate fixed-end moments: for the left span, FEM = wL^2/12 at both ends; for the right span, FEM = 0. At the interior support, the unbalanced moment is the sum of the left and right FEMs. Then, determine distribution factors: if the beam has constant EI, each span has stiffness 4EI/L, so distribution factors are equal if spans are equal. Distribute the unbalanced moment to each span (with opposite sign). Then carry over half of the distributed moment to the far ends. Repeat the process: at the interior support, the new unbalanced moment is the carry-over from the left and right spans. Continue until the carry-over moments are very small. Sum all contributions to get final moments.
10. A reaction has a negative ΔH (exothermic) and a positive ΔS. Using the Gibbs-Helmholtz equation, explain how ΔG changes with increasing temperature.
For a reaction with negative ΔH (exothermic) and positive ΔS, the Gibbs-Helmholtz equation (∂(ΔG/T)/∂T)_P = -ΔH/T^2 tells us that since ΔH is negative, -ΔH/T^2 is positive. So, as temperature increases, ΔG/T increases. But ΔG itself? Since ΔG = ΔH - TΔS, at low temperatures the negative ΔH dominates, making ΔG negative. As temperature rises, the -TΔS term becomes more negative (since ΔS positive), so ΔG becomes more negative. Actually, check: ΔG = ΔH - TΔS. If ΔH is negative and ΔS positive, then ΔG becomes more negative with increasing T. The Gibbs-Helmholtz equation gives the same: d(ΔG)/dT = -ΔS, so since ΔS positive, ΔG decreases with T. So ΔG becomes more negative, making the reaction more spontaneous.
11. In coupled thermoelasticity, why does temperature change cause stress, and stress cause temperature change?
In coupled thermoelasticity, temperature change causes thermal expansion, which leads to stress if the material is constrained. Conversely, stress can cause temperature change through the thermoelastic effect: when a material is compressed, it heats up, and when stretched, it cools. This coupling is important in high-speed processes like impact or rapid heating, where the interaction between mechanical and thermal fields cannot be ignored. The coupled equations include both the heat equation with a term for mechanical work and the elasticity equations with thermal strain. For most engineering problems, the coupling is weak and can be neglected, but in precision applications like laser heating, it matters.
12. Use the Gibbs-Duhem equation to explain why the vapor pressure of a solvent in an ideal solution decreases when you add a non-volatile solute.
In an ideal solution, the Gibbs-Duhem equation relates the chemical potentials of solvent and solute. When you add a non-volatile solute, the mole fraction of the solvent decreases. At constant temperature, the chemical potential of the solvent is related to its vapor pressure. The Gibbs-Duhem equation shows that if the chemical potential of the solute increases (because its concentration increases), the chemical potential of the solvent must decrease. A lower chemical potential for the solvent means its vapor pressure is lower. So, adding a non-volatile solute reduces the solvent's vapor pressure, which is Raoult's law. The equation ensures that the changes in chemical potentials are consistent.