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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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Resolving forces in a simple truss, and 2D stresses in a uniaxially loaded bar.

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Tension in simple triangular truss.

", "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": "

\n

Balancing horizontal forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\sin(90-\\var{theta})=0$, or more simply: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})=0$.

\n

Balancing vertical forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\cos(90-\\var{theta})+\\var{force}$, or more simply: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})+\\var{force}$,

\n

where 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

$P_{AB}\\cos(\\var{theta})\\sin(\\var{theta})-P_{BC}\\sin(\\var{theta})\\sin(\\var{theta})=\\var{force}\\sin(\\var{theta})$

\n

and 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: $P_{BC}=-\\var{force}\\sin(\\var{theta})$

\n

since $\\cos^2 + \\sin^2 = 1$. Balancing vertical forces at C:

\n

$P_{AC}+P_{BC}\\cos(90-\\var{theta})=0$, or more simply: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$,

\n

and substituting in the expression for $P_{BC}$ from above:

\n

$P_{AC}-\\var{force}\\sin(\\var{theta})\\sin(\\var{theta})=0$,

\n

the final answer simplifies to:

\n

$P_{AC}=\\var{force}\\sin^2(\\var{theta})=\\var{siground(Pac,3)}$ [units: kN].

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Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar AC.

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A pin-jointed truss is shown in the figure below. The pivot at A is fixed, but the pivot at C is free to move vertically. The angle at B is a right angle.

\n

\"Pin-jointed

\n

If 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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Tension in a simple triangular truss.

", "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": "

\n

Balancing horizontal forces at B: $P_{AB}\\cos(\\var{theta})=P_{BC}\\sin(\\var{theta})$.

\n

Balancing vertical forces at B: $P_{AB}\\sin(\\var{theta})+P_{BC}\\cos(\\var{theta})+\\var{force}=0$,

\n

where 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

$P_{AB}\\sin(\\var{theta})\\cos(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$

\n

and subtracting the first from the second gives:

\n

$P_{BC}\\sin^2(\\var{theta})+P_{BC}\\cos^2(\\var{theta})=-\\var{force}\\cos(\\var{theta})$,

\n

or, since $\\sin^2+\\cos^2=1$:

\n

$P_{BC}=-\\var{force}\\cos(\\var{theta})$.

\n

Balancing horizontal forces at C: $P_{AC}+P_{BC}\\sin(\\var{theta})=0$

\n

Substituting 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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Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar AC.

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A pin-jointed truss is shown in the figure below. The pivot at A is fixed, but the pivot at C is free to move horizontally. The angle at B is a right angle.

\n

\"Pin-jointed

\n

If 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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Tension in bracket-truss.

", "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.

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Balancing vertical forces at B: $P_{BC}\\sin(\\var{theta})+\\var{force}=0$,

\n

where units are kN. Rearranging:

\n

$P_{BC}=-\\var{force}/\\sin(\\var{theta}) = \\var{siground(Pbc,3)}$ [units: kN].

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Angle to vertical of Bar AB.

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Applied vertical (downward) force.

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Tension in Bar BC.

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A pin-jointed truss is shown in the figure below. The pivots at A and C are fixed.

\n

\"Pin-jointed

\n

If 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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Calculate direct and shear stress in an axially loaded bar in the XY plane.

", "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.

", "advice": "

The bar has a solid square cross-section (10mm×10mm), so the cross-sectional area is 100mm$^2$.

\n
    \n
  1. The horizontal section area is 100mm$^2/\\sin(\\var{theta})=\\var{siground(AH,3)}$mm$^2$.
  2. \n
  3. The vertical section area is 100mm$^2/\\cos(\\var{theta})=\\var{siground(AV,3)}$mm$^2$.
  4. \n
  5. The axial stress is $\\var{force}$kN / 100mm$^2=\\var{SA}$MPa.
  6. \n
  7. The direct stress in the $x$ direction is the force component in the $x$ direction ($F\\cos(\\theta)$) divided by the vertical section area, i.e.: $\\sigma_x= \\var{force}$kN $\\times \\cos^2(\\var{theta}) \\div 100$mm$^2 = \\var{siground(SX,3)}$MPa.
  8. \n
  9. The direct stress in the $y$ direction is the force component in the $y$ direction ($F\\sin(\\theta)$) divided by the horizontal section area, i.e.: $\\sigma_y= \\var{force}$kN $\\times \\sin^2(\\var{theta}) \\div 100$mm$^2 = \\var{siground(SY,3)}$MPa.
  10. \n
  11. The shear stress in the $xy$ plane is the force component in the $x$ direction ($F\\cos(\\theta)$) divided by the horizontal section area, or equivalently the force component in the $y$ direction ($F\\sin(\\theta)$) divided by the vertical section area, i.e.:? $\\tau_{xy}= \\var{force}$kN $\\times \\sin(\\var{theta})\\cos(\\var{theta}) \\div 100$mm$^2 = \\var{siground(TXY,3)}$MPa.
  12. \n
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Angle of bar to x-axis.

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Applied axial load.

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Area of horizontal section.

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Area of vertical section

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Axial stress.

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Direct stress (sxx).

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Direct stress (syy).

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Shear stress (txy).

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The figure below shows a bar with axial load $F$, at angle $\\theta$ to the $x$ axis.

\n

\"Axially

\n

The 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
    \n
  1. What is the horizontal section area? [[0]] [Units: mm$^2$]
  2. \n
  3. What is the vertical section area? [[1]] [Units: mm$^2$]
  4. \n
  5. What is the axial stress? [[2]] [Units: MPa]
  6. \n
  7. What is the direct stress in the $x$ direction? $\\sigma_x=$[[3]] [Units: MPa]
  8. \n
  9. What is the direct stress in the $y$ direction? $\\sigma_y=$[[4]] [Units: MPa]
  10. \n
  11. What is the shear stress in the $xy$ plane? $\\tau_{xy}=$[[5]] [Units: MPa]
  12. \n
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