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Several exams in the past have started with a truss question followed by stresses in the truss members.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore obtained through trigonometry.
", "advice": "\nBalancing horizontal forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\sin(90-\\var{theta})=0$, or more simply: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})=0$.
\nBalancing vertical forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\cos(90-\\var{theta})+\\var{force}$, or more simply: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})+\\var{force}$,
\nwhere the units are kN. Multiplying the first by $\\cos(\\var{theta})$ and the second by $\\sin(\\var{theta})$, we get:
\n$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos(\\var{theta})\\cos(\\var{theta})=0$
\n$P_{AB}\\cos(\\var{theta})\\sin(\\var{theta})-P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=\\var{force}\\sin(\\var{theta})$
\nand subtracting the second from the first:
\n$P_{BC}\\cos(\\var{theta})\\cos(\\var{theta})+P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=-\\var{force}\\sin(\\var{theta})$, or more simply: $P_{BC}=-\\var{force}\\sin(\\var{theta})$
\nsince $\\cos^2 + \\sin^2 = 1$. Balancing vertical forces at C:
\n$P_{AC}+P_{BC}\\cos(90-\\var{theta})=0$, or more simply: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$,
\nand substituting in the expression for $P_{BC}$ from above:
\n$P_{AC}-\\var{force}\\sin(\\var{theta})\\sin(\\var{theta})=0$,
\nthe final answer simplifies to:
\n$P_{AC}=\\var{force}\\sin^2(\\var{theta})=\\var{siground(Pac,3)}$ [units: kN].
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\n\nIf the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle of Bar AB to the vertical, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar AC?
\n[[0]] [Units: kN]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore obtained through trigonometry.
", "advice": "\nBalancing horizontal forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})$.
\nBalancing vertical forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})+\\var{force}=0$,
\nwhere units are kN. Multiplying the first by $\\sin(\\var{theta})$ and the second by $\\cos(\\var{theta})$ gives:
\n$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})-P_{BC}\\sin^2(\\var{theta})=0$
\n$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$
\nand subtracting the first from the second gives:
\n$P_{BC}\\sin^2(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$,
\nor, since $\\sin^2+\\cos^2=1$:
\n$P_{BC}=-\\var{force}\\cos(\\var{theta})$.
\nBalancing horizontal forces at C: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$
\nSubstituting in the expression for $P_{BC}$ from above, and rearranging:
\n$P_{AC}=\\var{force}\\sin(\\var{theta})\\cos(\\var{theta}) = \\var{siground(Pac,3)}$ [units: kN].
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\n\nIf the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle of Bar AB to the horizontal, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar AC?
\n[[0]] [Units: kN]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore obtained through trigonometry.
", "advice": "Balancing vertical forces at B: $P_{BC}\\sin(\\var{theta})+\\var{force}=0$,
\nwhere units are kN. Rearranging:
\n$P_{BC}=-\\var{force}/\\sin(\\var{theta}) = \\var{siground(Pbc,3)}$ [units: kN].
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\n\nIf the applied force, $F$, is $\\var{force}$ kN vertically down, and the angle between Bar AB and Bar BC, $\\theta$, is $\\var{theta}^\\circ$, what is the tension in Bar BC?
\n[[0]] [Units: kN]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Axial stress in a bar is a straightforward calculation, but $\\sigma_x$, $\\sigma_y$ and $\\tau_{xy}$ depend on the choice of $x$ and $y$ axes and the orientation of the bar.
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\nAngle of bar to x-axis.
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", "templateType": "anything", "can_override": false}, "SA": {"name": "SA", "group": "Ungrouped variables", "definition": "1000*force/100", "description": "Axial stress.
", "templateType": "anything", "can_override": false}, "SX": {"name": "SX", "group": "Ungrouped variables", "definition": "1000*(force/100)*(cos(radians(theta)))^2", "description": "Direct stress (sxx).
", "templateType": "anything", "can_override": false}, "SY": {"name": "SY", "group": "Ungrouped variables", "definition": "1000*(force/100)*(sin(radians(theta)))^2", "description": "Direct stress (syy).
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", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["theta", "force", "AH", "AV", "SA", "SX", "SY", "TXY"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "The figure below shows a bar with axial load $F$, at angle $\\theta$ to the $x$ axis.
\n\nThe bar has a solid square cross-section (10mm×10mm). If the axial load is $F=\\var{force}$ kN, and the angle is $\\theta=\\var{theta}^\\circ$:
\n