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ENGINEERING, CS 2025 HSC 26c

A truss is loaded as shown.
 

Determine the magnitude and nature of the force in member \(\text{M}\).   (6 marks)

Show Answers Only

\(\text{Magnitude} = 5869.7 \ \text{N}\)

\(\text{Nature}\ =\ \text{Tension}\)

Show Worked Solution

Find reactions at supports — take moments about \(\text{A}\):

\(\Sigma M_A\) \(= 0\)  
  \(0 = +(9 \times 1.4)+(15\sin60° \times 2.8)-(15\cos60° \times 0.7)-R_B \times 2.8\)  
  \(0 = 12.6+36.373-5.25-2.8R_B\)  
\(R_B\) \(= \dfrac{43.723}{2.8}\)  
\(R_B\) \(= 15.615 \ \text{kN} \uparrow\)  

 
Consider RHS of section plane — sum vertical forces:

\(\Sigma F_V\) \(= 0\)  
  \(0 = 15.615-15\sin60°-F_M\sin26.565°\)  
\(F_M\) \(= \dfrac{2.625}{\sin26.565°}\)  
\(F_M\) \(= 5.8697 \ \text{kN}\)  
\(F_M\) \(= 5869.7 \ \text{N (Tension)}\)  

♦♦ Mean mark 51%.

Find \(R_B\) as above \(= 15.615 \ \text{kN} \uparrow\)

     

Consider Joint C — sum vertical forces:

\(\Sigma F_V\) \(= 0\)  
  \(0 = 15.615-15\sin60°-F_M\sin26.565°\)  
\(F_M\) \(= \dfrac{2.625}{\sin26.565°}\)  
\(F_M\) \(= 5.8697 \ \text{kN}\)  
\(F_M\) \(= 5869.7 \ \text{N (Tension)}\)  

   

ENGINEERING, CS 2024 HSC 25c

A pin-jointed truss is shown. Member  \(AB\)  has been determined to be 250 N in compression.
 

Complete the table. You must use the method specified to solve each component.   (6 marks)

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \textit{}\\
\hline
\rule{0pt}{2.5ex} \text{External} & \text{Mathematical} & \\
\text{reaction at \(E\)} \rule[-1ex]{0pt}{0pt} & \text{} &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Direction: ...........................} \\
\hline
\end{array}

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \\
\hline
\rule{0pt}{2.5ex} \text{Internal} & \text{Methods of} & \\
\text{reaction of}  & \text{section} &\\
\text{member of \(CF\)} & &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Nature of force (T or C): ...........................} \\
\hline
\end{array}

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \\
\hline
\rule{0pt}{2.5ex} \text{Internal} & \text{Graphical} & \\
\text{reaction of}  &  &\\
\text{member \(BG\)} & &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Nature of force (T or C): ...........................} \\
\hline
\end{array}

Show Answers Only

External reaction at \(E\) (Mathematical method):

\( \Sigma \text{M}_\text{A}\Large{⤸} ^{\small{+}}\) \(=0=(200 \times 2)+(75 \times 10)-(129.9 \times 3.46)-\left(\text{R}_\text{E} \times 12\right) \)  
\(0\) \(=400+750-450-\text{R}_\text{E} \times 12 \)  
\(\text{R}_\text{E}\) \(=\dfrac{700}{12} = 58.33\ \text{N} \uparrow \)  

 

Internal reaction of \(CF\) (Method of sections):

\( \Sigma \text{F}_\text{V} \uparrow^{+}=0=\left(\sin \, 60^{\circ} \times \text{CF}\right)-75+58.33 \)

\(-\sin \, 60^{\circ} \times \text{CF} \) \(=-75+58.33 \)  
\(\text{CF}\) \(=\dfrac{16.67}{\sin \, 60^{\circ}} =19.25\ \text{N (T)} \)  
♦♦ Mean mark 46%.

Internal reaction of \(BG\) (Graphical):
 

\(BG \approx 19\ \text{N (T)} \)

Show Worked Solution

External reaction at \(E\) (Mathematical method):

\( \Sigma \text{M}_\text{A}\large{⤸} ^{\small{+}}\) \(=0=(200 \times 2)+(75 \times 10)-(129.9 \times 3.46)-\left(\text{R}_\text{E} \times 12\right) \)  
\(0\) \(=400+750-450-\text{R}_\text{E} \times 12 \)  
\(\text{R}_\text{E}\) \(=\dfrac{700}{12} = 58.33\ \text{N} \uparrow \)  

 

Internal reaction of \(CF\) (Method of sections):

\( \Sigma \text{F}_\text{V} \uparrow^{+}=0=\left(\sin \, 60^{\circ} \times \text{CF}\right)-75+58.33 \)

\(-\sin \, 60^{\circ} \times \text{CF} \) \(=-75+58.33 \)  
\(\text{CF}\) \(=\dfrac{16.67}{\sin \, 60^{\circ}} =19.25\ \text{N (T)} \)  
♦♦ Mean mark 46%.

Internal reaction of \(BG\) (Graphical):
 

\(BG \approx 19\ \text{N (T)} \)

ENGINEERING, CS 2023 HSC 26c

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} &  & \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} &  & \\
\hline
\end{array}
Show Answers Only

\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}

Show Worked Solution

\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}

\(\text{Consider member}\ EF:\)

               

  • \(\text{Force diagram closes with two collinear forces}\)
  • \(FG\ \text{is a zero force member}\)
  • \(EF = 73.2\ \text{kN (compression)} \)
♦♦♦ Mean mark 36%.

\(\text{Consider member}\ CH:\)
 

\(+ \uparrow \Sigma F_{V}\) \(=0\)  
\(0\) \(=-125 + 73.2 \times \sin\,60^{\circ} + CH \times \sin\,60^{\circ}\)  
\(CH \times \sin\,60^{\circ}\) \(=61.607\)  
\(CH\) \(= \dfrac{61.607}{\sin\,60^{\circ}} =71.138\ \text{kN (tension)} \)  

ENGINEERING, CS 2023 HSC 22b

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.

 

  1. Calculate the reactions at \(\text{R}_\text{L}\) and \(\text{R}_\text{R}\).   (3 marks)
      

\begin{array} {ll}
\text{R}_\text{L} = \text{............................... N} & \text{Direction ...............................} \\
 &  \\
\text{R}_\text{R} = \text{............................... N} & \text{Direction ...............................} \end{array}

 

  1. Complete the shear force and bending moment diagrams of the scenario described.   (3 marks)

 

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i.    \( \text{R}_{\text{L}}=860\ \text{N} \uparrow \)

\( \text{R}_{\text{L}}= 440\ \text{N} \uparrow \)

  
ii.    

Show Worked Solution

i.    \(  \stackrel {\curvearrowright} {\sum{ \text{M}}{^{+}_\text{L}}}: \)

\(0\) \(=(2 \times 450)+(4 \times 850)-(\text{R}_\text{R} \times 5) \)  
\(5 \times \text{R}_\text{R}\) \(=900 + 3400\)  
\(\text{R}_\text{R}\) \(=860\ \text{N} \uparrow \)  

 

\( \sum \text{F}_\text{V} \uparrow:\)

\(0\) \(= -450-850+860+\text{R}_\text{L} \)  
\( \text{R}_{\text{L}}\) \(= 440\ \text{N} \uparrow \)  

 
ii.    

ENGINEERING, CS 2016 HSC 26b

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. 
Show Worked Solution

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. 

♦ Mean mark 46%.

ENGINEERING, CS 2016 HSC 26a

A simplified diagram of a pin jointed truss of a hammerhead crane is shown. It is lifting a load of 54 kN.
 

  1. Calculate the reactions at supports \(A\) and \(B\).   (4 marks)
  1. Determine the magnitude and nature of the force in member \(M\).   (2 marks)
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i.     \(\text{Reaction at} \ \ A=249 \ \text{kN}\uparrow\)

\(\text{Reaction at} \ \ B=123 \ \text{kN @ 12.7}^{\circ} \ \text{acting down and to the right.}\)

ii.    \(\text{Force in} \ \ M=121 \ \text{kN (Compression)}\)

Show Worked Solution

i.

\(R _{B H}-27=0\)

\(R _{B H}=27\)

\(\text{From the diagram,} \ \ R_{B H}=27 \ \text{kN} \rightarrow\)

\(+\circlearrowleft \sum M_B\) \(=0\)  
\(0\) \(=27 \times 2+54 \times 5-75 \times 1-R_A \times 1\)  
\(R_A\) \(=54+270-75=249 \ \text{kN} \uparrow\)  
     
\(+\uparrow \sum F_V\) \(=0\)  
\(0\) \(=249-75-54+R_B\)  
\(R_B\) \(=129-249=-120 \ \text{kN}=120 \ \text{kN} \downarrow\)  

 

\(R_A=249 \ \text{kN @} \ 0^{\circ} \uparrow\)

\(\text{Reaction at} \ B:\)
 

\(\text{Reaction at}\ A = 249\ \text{kN acting vertically, up}\)

\(\text{Reaction at}\ B = 123\ \text{kN @ 12.7° acting down and to the right.}\)
  


Mean mark (i) 52%.

ii.  \(\theta=\tan ^{-1}\left(\dfrac{20}{40}\right)=26.6^{\circ}\)

   

 
\(F=\sqrt{108^2+54^2}=\sqrt{14580}=120.747 \ldots\)

\(\text{Force in}\ M=121\ \text{kN (Compression)}\)


♦ Mean mark (ii) 49%.

ENGINEERING, CS 2018 HSC 18 MC

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` ?
 

Show Answers Only

`B`

Show Worked Solution
  • 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`

ENGINEERING, CS 2017 HSC 25a

A pin-jointed truss designed to support a roadside sign is shown.
 

  1. Determine the magnitude and direction of the reactions at `A` and `B`.   (4 marks)
  1. Determine the magnitude and nature of the force in member `C`.   (2 marks)
  1. Explain why concrete would be a suitable material to support the roadside sign around point `D`.   (2 marks)
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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.
Show Worked Solution

\[\textbf{i}. \ \ \ce{->[\ce{+}]} \sum M_A = 0 \]

`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`  

  

Mean mark 56%.

`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^(@)`  

 

\[\textbf{ii}. \]

`text{Taking moments about X:}`

\({+ \circlearrowleft} \Sigma M_X = 0\)

`0` `=-1.5 xxF_(C)+1.5 xx20-1.5 xx3`  
`0` `=-F_(C)+20-3`  
`F_(C)` `=17\ text{kN (compression)}`  

♦♦ Mean mark (ii) 38%.

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.

♦ Mean mark (iii) 52%.

ENGINEERING, CS 2019 HSC 22c

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)

Show Answers Only
\(M\) \(=0.324 \ \text{kN (tension)}\)
\(M\) \(=14.33 \ \text{kN (compression)}\)
Show Worked Solution

\(\text{Forces at Joint} \  A\)

\(\text{Horizontal forces}=0\)

\(\therefore \ \text{To calculate vertical force at} \ A \ \rightarrow \ \text{use moments.}\)


♦♦ Mean mark 36%.
\({\circlearrowright}+\Sigma M_C\) \(=0\)  
\(0\) \(=-(12 \times 4)+(R_A \times 12)-(10 \times 7)-(3 \times 18)\)  
\(12R_A\) \(=48+70+54\)  
\(R_A\) \(=\dfrac{172}{12}=14.33 \text{kN}\uparrow\)  

    
\(\text{Forces in Member} \  N \rightarrow \ \text{method of joints at} \ A\)

  • \(\text{No horizontal forces}\)
  • \(\text{Member}\ AC \ \text{redundant and carrying no load}\)
  • \(F_{\text{up}} = F_{\text{down}}\)

\(\therefore \ \text{Member} \ AB \ \text{in compression (the force acting down on joint} \ A \ \text{from member} \ AB \  \text{is 14.33 kN)}\)

\(\therefore \ \text{Force in} \ N = 14.33 \ \text{kN (compression)}\)

  
\(\text{Using Method of Sections} \rightarrow \ \text{take moments about Joint} \ H:\)

\(\text{Find the perpendicular distance} \ d:\)

\(BH^2\) \(=18^2+6^2\)  
\(BH\) \(=\sqrt{360}\)  
\(\sin 40.6^{\circ}\) \(=\dfrac{d}{\sqrt{360}}\)  
\(d\) \(=\sqrt{360} \times \sin 40.6^{\circ}=12.348 \ \text{m}\)  

 

\({\circlearrowright}+\Sigma M_H\) \(=0\)
\(0\) \(=+(12 \times 2)+(M \times 12.348)-(7 \times 4)\)
\(12.348M\) \(=-24+28\)
\(M\) \(=\dfrac{4}{12.348}=0.324 \ \text{kN (tension)}\)

 

\(\therefore\) \(M\) \(=0.324 \text{kN (tension)}\)
  \(N\) \(=14.33 \text{kN (compression)}\)

ENGINEERING, CS 2022 HSC 26a

  1.  The diagram shows a tower crane being used in the construction of a building.
     

  1. Determine the number of 100 kg concrete blocks required to place the boom arm in equilibrium.   (2 marks)
  1. Under a different set of conditions, a wind force is applied, as shown in the diagram.
     

  1. Determine the magnitude and nature of the internal reaction in member A.   (6 marks)
Show Answers Only

i.   `12`

ii.  `A=25\ text{kN in compression}`

Show Worked Solution

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.


♦ Mean mark (i) 46%.

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}`


♦ Mean mark (ii) 43%.

ENGINEERING, CS 2020 HSC 25c

 

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)

Show Answers Only

\(C=6.24 \ \text{kN in tension}\)

Show Worked Solution

\(\text{Let the length of a member be 2 units}\)

\(\sin 60^{\circ}\) \(=\dfrac{d}{2}\)  
\(d\) \(=2 \times \sin 60^{\circ}=1.732 \ \text{units}\)  

 

\(+\circlearrowleft \Sigma M_B\) \(=0\)  
\(0\) \(=(3000 \times 1.732)+(20\,000 \times 4)+\left(-R_A \times 6\right)+(1200 \times 8)\)  
\(0\) \(=5196+80\,000-6 R_A+9600\)  
\(6R_A\) \(=94\,796\)  
\(R_A\) \(=15\,799.33=15.80 \ \text{kN} \ \uparrow\)  

\(\sin 60^{\circ}\) \(=\dfrac{y}{C}\)  
\(y\) \(=C \sin 60^{\circ}\)  
\(+\uparrow \Sigma F_Y\) \(=-1200+15799.33-20000-C \sin 60^{\circ}\)  
\(0\) \(=-5400.667-C \sin 60^{\circ}\)  
\(C \sin 60^{\circ}\) \(=-5401\)  
\(C\) \(=-\dfrac{5401}{\sin 60^{\circ}}=-6236.54\)  

 
\(\Rightarrow \ \text{Assumed direction was incorrect and reaction in} \ C \ \text {is away from the joint.}\)

\(\therefore C=6.24 kN \ \text{in tension}\)


♦ Mean mark 48%.

ENGINEERING, CS 2021 HSC 25b

A truss is fixed to a wall at `A` and `B` as shown. Ignore the mass of the truss.
 


 

  1. Determine the horizontal reaction at `A`.   (2 marks)
  1. Determine, using method of joints, the internal reaction in member `AC`. Indicate the nature of the force in the member.   (3 marks)
  1. Determine, using method of sections, the internal reaction in member `CE`. Indicate the nature of the force.   (3 marks)
Show Answers Only

i.    `3.75\ text{kN ←}`

ii.    `AC=4802\ text{N}`

iii.   `CE = 3\ text{kN (tension)}`

Show Worked Solution

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 ←}`  

 

ii.   Method of joints at A
 
     

→Σ`F_H` `=0`  
`0` `= -3750+AC xx sin(51.34°)`  
`AC` `=3750/sin(51.34°)=4802\ text{N}`  

♦♦ Mean mark (ii) 41%.

iii.  Method of Sections

\(\circlearrowright \Sigma \text{M}_\text{B}\) \(=0\)  
`0` `= (1500 xx 10)-(CE xx 5)`  
`CE` `=(15\ 000)/5=3000\ text{N}=3\ text{kN (tension)}`  

♦♦♦ Mean mark (iii) 30%.

ENGINEERING, CS 2021 HSC 7 MC

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?
 

Show Answers Only

`A`

Show Worked Solution
  • The sum of horizontal forces and the sum of vertical forces must both be equal to 0.

`=>A`