7.1 Determine graphically (Bow's notation) the magnitude and nature of ALL the members of the framework in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2018 - Paper 1
Question 7
7.1 Determine graphically (Bow's notation) the magnitude and nature of ALL the members of the framework in FIGURE 7.1 below.
SCALE: Vector diagram 1 mm = 5 N
FIGUR... show full transcript
Worked Solution & Example Answer:7.1 Determine graphically (Bow's notation) the magnitude and nature of ALL the members of the framework in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2018 - Paper 1
Step 1
Determine graphically (Bow's notation) the magnitude and nature of ALL the members of the framework in FIGURE 7.1
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Answer
Using Bow's notation, we will analyze the framework in FIGURE 7.1 to find the forces in each member.
Identify the members and loadings from the diagram. The joints labeled from A to G must be considered.
Resolve the external forces and reactions at joints A, B, C, D, E, F, and G. For instance, the 200 N and 300 N forces acting downwards will affect members differently based on their locations.
Create a vector diagram for equilibrium. Draw force vectors to scale, ensuring all triangle rules are followed. According to the scale of the vector diagram (1 mm = 5 N), ensure to convert lengths accurately.
Calculate the forces in each member using the vector diagram to find the lengths corresponding to actual forces.
Identify the nature of each member (strut or tie) based on the calculated forces: compression indicates a strut while tension indicates a tie.
By following these steps, we can derive:
Member AE: 260 N, STRUT
Member BF: 135 N, STRUT
Member CG: 317.5 N, STRUT
Member EF: 130 N, TIE
Member GD: 160 N, TIE.
Step 2
Calculate the reactions at the supports RL and RR.
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Answer
To find the reactions at the supports RL and RR:
Choose a point for calculating moments, often a support point. Here, we take moments about point RR.
Apply the equation for moment equilibrium:
RL×10=(8×8)+(4×5)+(6×2)
Simplifying this gives:
RL=9.6kN
To verify, take moments about RL this time and solve for RR: RR×10=(6×8)+(4×5)+(8×2)
Simplifying reveals:
RR=8.4kN
Thus, the reactions calculated are:
RL = 9.6 kN
RR = 8.4 kN.
Step 3
Calculate the bending moments at points A, B, C, D and E.
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To compute the bending moments at each designated point:
Calculate the moment at point A: MA=0kN.m(as there is no moment applied).
Calculate the moment at point B using the reaction at RL:
MB=RL×2−(8×2)=19.2kN.m
Calculate the moment at point C:
MC=(RL×5)−(4×3)=24kN.m
Calculate the moment at point D:
MD=(RL×8)−(4×5)−(6×3)=16.8kN.m
Calculate the moment at point E:
ME=RL×10−(6×8)=0kN.m
Summarizing the bending moments:
M_A = 0 kN.m
M_B = 19.2 kN.m
M_C = 24 kN.m
M_D = 16.8 kN.m
M_E = 0 kN.m.
Step 4
Draw a bending moment diagram of the beam.
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To create a bending moment diagram based on the calculated moments:
Set up the axes with the horizontal axis representing the length of the beam and the vertical axis representing the bending moments.
Plot the calculated points from moments at A, B, C, D, and E on the diagram:
At A, plot at 0 kN.m
At B, plot at 19.2 kN.m
At C, plot at 24 kN.m
At D, plot at 16.8 kN.m
At E, plot at 0 kN.m.
Connect the plotted points smoothly to depict how the bending moments vary along the beam. Use straight lines to connect the points and ensure the correct representation of signs (positive/negative moments). This is essential for understanding how forces influence bending along the member.
Your diagram should show positive moments rising from A to C and descending to E, showing how forces act along the beam.
Step 5
Calculate the load that needs to be applied to a round stainless steel bar to cause a tensile stress of 80 MPa.
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To calculate the load (P) required:
Understand the relationship between stress and load:
Stress=AreaLoad⟹Load=Stress×Area
Calculate the cross-sectional area (A) of the round stainless steel bar:
A=4πd2 where d=20 mm=0.02 mA=4π(0.02)2=3.14×10−4m2
Substitute the values into the load equation:
P=80×106Pa×3.14×10−4m2=25.13kN
Thus, the load required to achieve a tensile stress of 80 MPa in the bar is approximately 25.13 kN.