7.1 Graphically determine the magnitude and type of member in the framework shown in FIGURE 7.1 below. Members: AE, BF, CF, DE and EF. SCALE: Space diagram: 1 : 100 Force diagram: 1 mm = 5 N 7.2 FIGURE 7.2 below shows a uniform beam. The beam is supported at points RL and RR. Three point loads of 4 N, 5 N and 3 N are exerted onto the beam. 7.2.1 Calculate the reactions at supports RL and RR. 7.2.2 Calculate the bending moments at point B, C and D. 7.2.3 Draw a bending-moment diagram of the beam. SCREENS: Space diagram: 10 mm = 1 m Bending-moment diagram: 5 mm = 1 N.m

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7.1 Graphically determine the magnitude and type of member in the framework shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2019 - Paper 1

Question 7

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7.1 Graphically determine the magnitude and type of member in the framework shown in FIGURE 7.1 below. Members: AE, BF, CF, DE and EF. SCALE: Space diagram: 1 : 100 ... show full transcript

Worked Solution & Example Answer:7.1 Graphically determine the magnitude and type of member in the framework shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2019 - Paper 1

Step 1

7.2.1 Calculate the reactions at supports RL and RR.

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Answer

To determine the reactions at the supports (RL and RR), we can apply the equilibrium equations for moments and forces. The sum of vertical forces should equal zero:

$ext{Sum of vertical forces: } R_L + R_R - (4 ext{ N} + 5 ext{ N} + 3 ext{ N}) = 0$

Calculating the total downward force: $4 ext{ N} + 5 ext{ N} + 3 ext{ N} = 12 ext{ N}$

Therefore, $R_L + R_R = 12 ext{ N}$

Next, taking moments around RL: $ext{Moment at } R_L: R_R imes 6 - 4 imes 3 - 5 imes 6 - 3 imes 9 = 0$

This gives: $R_R = 6.25 ext{ N} ext{ confirm this with the equilibrium of vertical forces}.$

Substituting back, we find: $R_L = 4 ext{ N} + 5 ext{ N} + 3 ext{ N} - R_R = 12 ext{ N} - 6.25 ext{ N} = 5.75 ext{ N}.$

Step 2

7.2.2 Calculate the bending moments at point B, C and D.

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For point B: Using the reaction at RL: $B_M = R_L imes 3 - (4 ext{ N} imes 3) = 6.25 imes 3 - 12 = 18.75 ext{ N.m}.$

For point C: The bending moment at C can be analyzed: $C_M = R_L imes 6 - (4 ext{ N} imes 3) - (5 ext{ N} imes 3) = 6.25 imes 6 - 12 - 15 = 25.5 ext{ N.m}.$

For point D: $D_M = R_L imes 9 - (4 ext{ N} imes 6) - (5 ext{ N} imes 3) = 6.25 imes 9 - 24 - 15 = 17.25 ext{ N.m}.$

Step 3

7.2.3 Draw a bending-moment diagram of the beam.

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To construct the bending-moment diagram, plot the calculated moments:

At point A (0 m), the moment is 0 N.m.
At point B (3 m), the moment is 18.75 N.m.
At point C (6 m), the moment drops to 25.5 N.m.
At point D (9 m), the moment goes down to 17.25 N.m.
At point E (12 m), the moment returns to 0 N.m.

This results in a smooth curve that represents the bending moments along the length of the beam.

7.1 Graphically determine the magnitude and type of member in the framework shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2019 - Paper 1

Worked Solution & Example Answer:7.1 Graphically determine the magnitude and type of member in the framework shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2019 - Paper 1

7.2.1 Calculate the reactions at supports RL and RR.

7.2.2 Calculate the bending moments at point B, C and D.

7.2.3 Draw a bending-moment diagram of the beam.

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