A loaded truss is shown.
Which of the following is the redundant member?
- \(AB\)
- \(BD\)
- \(BE\)
- \(CD\)
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A truss is loaded as shown.
Showing working, complete the table. (6 marks)
\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & 73.2\ \text{kN} & \text{Compression} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & 71.138\ \text{kN} & \text{Tension} \\
\hline
\end{array}
\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & 73.2\ \text{kN} & \text{Compression} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & 71.138\ \text{kN} & \text{Tension} \\
\hline
\end{array}
\(\text{Consider member}\ EF:\)
→ \(\text{Force diagram closes with two collinear forces}\)
→ \(FG\ \text{is a zero force member}\)
→ \(EF = 73.2\ \text{kN (compression)} \)
\(\text{Consider member}\ CH:\)
\(+ \uparrow \Sigma F_{V}\) | \(=0\) | |
\(0\) | \(=-125 + 73.2 \times \sin\,60^{\circ} + CH \times \sin\,60^{\circ}\) | |
\(CH \times \sin\,60^{\circ}\) | \(=61.607\) | |
\(CH\) | \(= \dfrac{61.607}{\sin\,60^{\circ}} \) | |
\(=71.138\ \text{kN (tension)} \) |
The diagram shows a child with a mass of 45 kg hanging 2 metres from the left end of a structure, and an adult with a mass of 85 kg hanging 1 metre from the right end. \begin{array} {ll}
\text{R}_\text{L} = \text{............................... N} & \text{Direction ...............................} \\
& \\
\text{R}_\text{R} = \text{............................... N} & \text{Direction ...............................} \end{array}
i. \( \stackrel {\curvearrowright} {\sum{ \text{M}}{^{+}_\text{L}}}: \)
\(0\)
\(=(2 \times 450)+(4 \times 850)-(\text{R}_\text{R} \times 5) \)
\(5 \times \text{R}_\text{R}\)
\(=900 + 3400\)
\(\text{R}_\text{R}\)
\(=860\ \text{N} \uparrow \)
What are redundant truss members? (2 marks)
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Redundant truss members:
→ A truss that carries no load under the conditions present.
→ It is important to note that as load conditions change, the truss may not actually remain redundant.
Redundant truss members:
→ A truss that carries no load under the conditions present.
→ It is important to note that as load conditions change, the truss may not actually remain redundant.
A simplified diagram of a pin jointed truss of a hammerhead crane is shown. It is lifting a load of 54 kN.
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i. Reaction at `A` = 249 kN ↑
Reaction at `B` = 123 kN @ 12.7° acting down and to the right.
ii. Force in `M` = 121 kN (Compression)
i.
`text{R}_(BH)-27=0`
`text{R}_(BH)=27`
`text{From the diagram,}\ \ text{R}_(BH)=27\ text{kN →}`
+\({\circlearrowleft}\)`sumM_B` | `=0` | |
`0` | `=27xx2+54xx5-75xx1-R_Axx1` | |
`R_A` | `=54+270-75` | |
`=249\ text{kN ↑}` | ||
+ ↑`sumF_V` | `=0` | |
`0` | `=249-75-54+R_B` | |
`R_B` | `=129-249` | |
`=-120\ text{kN}` | ||
`=120\ text{kN ↓}` |
`R_A=249\ text{kN @ 0° ↑}`
`text{Reaction at}\ B:`
Reaction at `A` = 249 kN acting vertically, up
Reaction at `B` = 123 kN @ 12.7° acting down and to the right.
ii. `theta = tan^(-1)(20/40)=26.6°`
`F` | `=sqrt(108^2+54^2)` | |
`=sqrt(14\ 580)` | ||
`=120.747…` |
Force in `M` = 121 kN (Compression)
The force `F` is moved from joint `W` to joint `X`, as shown.
Which row of the table correctly describes the changes in the internal force in member `XY` and the reaction force at `W` as a result of force `F` moving from `W` to `X` ?
`B`
→ Initially there is not force in member XY and no reaction at W.
→ When F is moved, a turning moment is created and needs to be countered by the reaction at W.
→ Member XY is also placed in tension, therefore both the force and reaction increase.
`=>B`
A pin-jointed truss designed to support a roadside sign is shown.
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i. `R_B=37\ text{N} larr`
`R_A=44.7\ text{kN}, \ theta=26.6^(@)`
ii. `F_C=17\ text{kN (compression)}`
iii. Suitability of concrete
→ Easily poured and formed around the sign.
→ Relatively quick to cure.
→ Weather resistant.
→ Hardens and sets with high strength and hardness.
\[\textbf{i}. \ \ \ce{->[\ce{+}]} \sum M_A = 0 \]
`3xx1.5-20 xx3+R_(B)xx1.5=0`
`4.5-60+1.5R_(B)=0`
`1.5R_(B)` | `=55.5` | |
`R_(B)` | `=(55.5)/(1.5)` | |
`=37\ text{N} larr` |
`uarr sumF_(V)=0`
`R_(AV)=20\ text{kN} +uarr`
\[\ce{->[\ce{+}]} \sum F_H = 0 \]
`R_(AH)=37+3=40\ text{kN} rarr^(+)`
`R_A` | `=sqrt(20^(2)+40^(2))` | |
`=sqrt2000` | ||
`=44.7\ text{kN}` |
`tan\ theta` | `=(R_(AV))/(R_(AH))` | |
`=20/40` | ||
`=0.5` | ||
`:.theta` | `=26.6^(@)` |
iii. Suitability of concrete
→ Easily poured and formed around the sign.
→ Relatively quick to cure.
→ Weather resistant.
→ Hardens and sets with high strength and hardness.
The diagram shows some dimensions and forces associated with a telecommunications tower.
By considering any necessary reaction, calculate the magnitude of the forces in members `M` and `N`. State the nature of each force. Ignore the weight of the tower. (6 marks)
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“
Forces at Joint `A`
Horizontal forces `=0`
`:.` To calculate vertical force at `A ` → use moments.
\({\circlearrowright}\)`+SigmaM_C` | `=0` | |
`0` | `=-(12xx4)+(R_Axx12)-(10xx7)-(3xx18)` | |
`12R_A` | `=48+70+54` | |
`R_A` | `=172/12` | |
`=14.33\ text{kN}↑` |
Forces in Member `N` → method of joints at `A`
→ No horizontal forces
→ Member `AC` redundant and carrying no load
→ `F_(up) = F_(down)`
`:.` Member `AB` in compression (the force acting down on joint `A` from member `AB` is 14.33 kN)
`:.` Force in N = 14.33 kN (compression)
Using Method of Sections → take moments about Joint `H`
Find the perpendicular distance `d`
`BH^2` | `=18^2+6^2` | |
`BH` | `=sqrt{360}` | |
`sin\ 40.6º` | `=d/(sqrt{360})` | |
`d` | `=sqrt{360}xx sin\ 40.6º` | |
`d` | `=12.348\ text{m}` |
\({\circlearrowright}\)`+SigmaM_H` | `=0` |
`0` | `=+(12xx2)+(Mxx12.348)-(7xx4)` |
`12.348M` | `=-24+28` |
`M` | `=4/12.348` |
`M` | `=0.324\ text{kN (tension)}` |
`:.\ ` | `M` | `=0.324\ text{kN (tension)}` |
`N` | `=14.33\ text{kN (compression)}` |
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i. `12`
ii. `A=25\ text{kN in compression}`
i. Let the total weight of the concrete blocks `= x`
`50x` | `=9.23xx65` | |
`50x` | `=600` | |
`x` | `=600/50` | |
`=12\ text{kN ↓}` | ||
`=12\ 000\ text{N ↓}` | ||
`m` | `=1200\ text{kg}` |
∴ 12 × 100 kg concrete blocks are needed for the counterweight.
ii. Magnitude and nature of internal reaction
\( \circlearrowright+\Sigma M_R \) | `=0` | |
`0` | `=-(6xx5.5)+(R_Lxx1)+(10xx6)-(10.4xx7)` | |
`R_L` | `=45.8\ text{kN}↑` |
Taking the horizontal section shown:
\( \circlearrowright + \Sigma M_P \) | \(= 0\) | |
`0` | `=(Axx1)+(45.8xx1)-(10.4xx2)` | |
`A` | `=-25\ text{kN}` |
∴ `A=25\ text{kN in compression}`
A cycle bridge has been constructed using a Warren girder truss loaded as shown. The diagram is drawn to scale.
By considering necessary loads and reactions, calculate the magnitude and nature of the force in member `C`. (6 marks)
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`C` = 6.24 kN in tension
Let the length of a member be 2 units
`sin60°` | `=d/2` | |
`d` | `=2 xx sin60°` | |
`=1.732\ text{units}` |
\({+ \circlearrowleft \Sigma M_B}\) | `=0` | |
`0` | `=(3000xx1.732)+(20\ 000xx4)+(-R_Axx6)+(1200xx8)` | |
`0` | `=5196+80\ 000-6R_A+9600` | |
`6R_A` | `=94\ 796` | |
`R_A` | `=15\ 799.33` | |
`=15.80\ text{kN} ↑` |
`sin60°` | `= y/C` | |
`y` | `= C\ sin60°` | |
`+↑ SigmaF_Y` | `= – \ 1200 + 15\ 799.33 – 20\ 000 – C\ sin60°` | |
`0` | `= – \ 5400.667 – C\ sin60°` | |
`C\ sin60°` | `= – \ 5401` | |
`C` | `= – \ 5401/(sin60°)` | |
`= – \ 6236.54` |
∴ Assumed direction was incorrect and reaction in `C` is away from the joint.
∴ `C` = 6.24 kN in tension
A truss is fixed to a wall at `A` and `B` as shown. Ignore the mass of the truss.
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i. `3.75\ text{kN ←}`
ii. `AC=4802\ text{N}`
iii. `CE = 3\ text{kN (tension)}`
i. Horizontal reaction at `A`
\(\circlearrowright \Sigma \text{M}_\text{B}\) | \(= 0\) | |
`0` | `= -(A_H xx 4\ text{m}) + (1500\ text{N} xx 10\ text{m})` | |
`A_H` | `= 3750\ text{N}` | |
`= 3.75\ text{kN ←}` |
The diagram shows a section of a pin jointed truss in equilibrium. The force in Member 1 is 200 kN in compression.
Which row of the table identifies the magnitude and nature of the forces in Member 2 and Member 3?
`A`
→ The sum of horizontal forces and the sum of vertical forces must both be equal to 0.
`=>A`