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Calculate the reactions at RL and RR - NSC Mechanical Technology Welding and Metalwork - Question 6 - 2021 - Paper 1

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Calculate the reactions at RL and RR. Calculate the bending moments at points B, C and D. Construct the bending moment diagram for points B, C and D by using scale... show full transcript

Worked Solution & Example Answer:Calculate the reactions at RL and RR - NSC Mechanical Technology Welding and Metalwork - Question 6 - 2021 - Paper 1

Step 1

Calculate the reactions at RL and RR.

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Answer

To calculate the reactions at supports RL and RR, we can use the equilibrium equations for the beam. Let R1 be the reaction at RL and R2 at RR.

Using the summation of vertical forces:

R1+R2=80N+60N+100NR_1 + R_2 = 80 N + 60 N + 100 N

$$Using moments about point R: \ $$M_R = 0 = 80 N \times 3 m + 60 N \times 5 m + 100 N \times 7 m - R_2 \times 2 m$$ Solving these equations will give us the values for R1 and R2.

Step 2

Calculate the bending moments at points B, C and D.

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Answer

The bending moment at any point can be calculated using the internal moment equation.

  • At point B: The moment due to the load before point B is:

MB=R1×3mM_B = R_1 \times 3 m

  • At point C: The moment will sum the contributions from both R1 and the 80 N load:

MC=R1×5m80N×2mM_C = R_1 \times 5 m - 80 N \times 2 m

  • At point D: The contributions will include R1, the 80 N, and 60 N load:

MD=R1×7m80N×4m60N×2mM_D = R_1 \times 7 m - 80 N \times 4 m - 60 N \times 2 m

Step 3

Construct the bending moment diagram for points B, C and D by using scale 1 m = 1 cm and 10 N m = 1 cm.

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Answer

To construct the bending moment diagram:

  1. Calculate the points B, C, and D moments using the values derived above.
  2. Translate these calculated moments into the diagram using the provided scaling factor (1 m = 1 cm and 10 N m = 1 cm).
  3. Plot the bending moments accordingly on the X-axis and connect these points to visualize the diagram.

Step 4

7.2.1 The diameter of the bar.

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Answer

Using the tensile stress formula:

σ=FA\sigma = \frac{F}{A}

Where

  • σ=20MPa\,\sigma = 20 MPa
  • F=40kNF = 40 kN

The area A can be expressed as:

A=FσA = \frac{F}{\sigma}

Solving for A gives us the cross-sectional area. The diameter can be found using:

A=πd24A = \frac{\pi d^2}{4}.

Step 5

7.2.2 The strain.

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Answer

Strain (ϵ\epsilon) is defined as the change in length (ΔL\Delta L) over the original length (L):

ϵ=ΔLL\epsilon = \frac{\Delta L}{L}

We can express the change in length with Young's modulus:

ϵ=σE\epsilon = \frac{\sigma}{E}

Substituting in the values for stress and Young's modulus:

ϵ=20MPa90GPa\epsilon = \frac{20 MPa}{90 GPa}.

Step 6

7.2.3 The change in length in mm.

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Answer

To find the change in length (ΔL\Delta L), we use the formula derived for strain:

ΔL=ϵ×L\Delta L = \epsilon \times L

Substituting in our strain calculation and the original length of 2 m to find the change in length in mm.

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