University Physics I · topic UPI-12.07
Structural equilibrium applications
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Structural Equilibrium Applications: predict a finite change
A truss is a structure of straight members joined by pins and loaded only at those pins, so that each member carries a single axial force, tension or compression, found from the equilibrium of the joints. Use the supplied topic model Y_out = p/(2*sind(th)), where Y_out is RAFTER PUSH P/(2 sin θ) (kN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Rise angle θ, measured in °. Keep p (Apex load P) = 12 kN; cap (Rafter buckling capacity) = 30 kN fixed. For this exercise use only 19.7813 ° ≤ th ≤ 55.2188 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at th = 27.5775 ° and 32.5388 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δth across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RAFTER PUSH P/(2 sin θ) against th
Predictions from the supplied Structural Equilibrium Applications model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RAFTER PUSH P/(2 sin θ) / kN
Horizontal axis (x): th: Rise angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Rise angle θ / ° | RAFTER PUSH P/(2 sin θ) / kN |
|---|---|
| 19.78125 | 17.72892476161191 |
| 28.640625 | 12.51788329277872 |
| 37.5 | 9.85607779022749 |
| 46.359375 | 8.290926263472311 |
| 55.21875 | 7.305171703128504 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of th separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At th = 27.5775 °, substitute into Y_out = p/(2*sind(th)): Y₀ = 12.9604 kN.
- At th = 32.5388 °, the same fixed-parameter model gives Y₁ = 11.1551 kN.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.8053 kN; the input change is Δth = 4.96125 °.
- Average sensitivity is ΔY_out/Δth = -0.363881 kN per °. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 12.9604 kN, Y₁ = 11.1551 kN; ΔY_out = -1.8053 kN; average sensitivity = -0.363881 kN per °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Structural Equilibrium Applications: infer an input with uncertainty
A truss is a structure of straight members joined by pins and loaded only at those pins, so that each member carries a single axial force, tension or compression, found from the equilibrium of the joints. Use the supplied topic model Y_out = p/(2*sind(th)), where Y_out is RAFTER PUSH P/(2 sin θ) (kN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Rise angle θ, measured in °. Keep p (Apex load P) = 12 kN; cap (Rafter buckling capacity) = 30 kN fixed. For this exercise use only 19.7813 ° ≤ th ≤ 55.2188 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 9.073212 kN, with a bounded reading uncertainty ±0.26521 kN. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00354375 °, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RAFTER PUSH P/(2 sin θ) against th
Predictions from the supplied Structural Equilibrium Applications model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RAFTER PUSH P/(2 sin θ) / kN
Horizontal axis (x): th: Rise angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Rise angle θ / ° | RAFTER PUSH P/(2 sin θ) / kN |
|---|---|
| 19.78125 | 17.72892476161191 |
| 28.640625 | 12.51788329277872 |
| 37.5 | 9.85607779022749 |
| 46.359375 | 8.290926263472311 |
| 55.21875 | 7.305171703128504 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(th) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(19.7813) = 17.7289 kN and Y_out(55.2188) = 7.30517 kN. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 19.7813 + (9.07321 − 17.7289)(55.2188 − 19.7813)/(7.30517 − 17.7289) = 49.208 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives th = 41.3981 °; substituting back gives Y_out = 9.073212 kN.
- The interpolation error in the input is |th_line − th| = 7.80984 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.808002 kN and 9.338422 kN. Sort the roots even if the response decreases: 39.979 ° ≤ th ≤ 42.9374 °.
- The lower and upper input deviations are 1.41912 ° and 1.53926 °. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer th ≈ 41.3981 °; compatible interval [39.979, 42.9374] °. Endpoint interpolation gives 49.208 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.