Figure 7.1 below shows a framework with members AE, BF, CF, DE and EF. Graphically determine the magnitude and type/nature of each member. Scale: Space diagram: 1 : 100 Force diagram: 1 mm = 5 N **FIGURE 7.1** 300 N Figure 7.2 below shows a uniform supported beam that is subjected to three point loads. **FIGURE 7.2** Calculate the following: 7.2.1 The magnitude of the reaction at support RL 7.2.2 The magnitude of the reaction at support RR 7.2.3 The bending moments at points B, C and D 7.2.4 Draw a bending-moment diagram of the beam. Scale: Beam: 1 : 100 Bending moments: 2 mm = 1 kNm 7.3 A compressive load of 80 kN is exerted onto a round bar with a diameter of 30 mm. The original length of the round bar is 3 m. Calculate the following: 7.3.1 The cross-sectional area of the bar in m² 7.3.2 The stress caused in the material in MPa 7.3.3 The strain caused if the change in length of the bar is 0.06 m 7.3.4 Young's modulus of elasticity in GPa 7.4 A round steel bar with a cross-sectional area of 0.9 x 10⁻⁵ m² is subjected to a maximum tensile load of 55 kN. Calculate the maximum stress.

Photo AI

Figure 7.1 below shows a framework with members AE, BF, CF, DE and EF - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2023 - Paper 1

Question 7

Figure 7.1 below shows a framework with members AE, BF, CF, DE and EF. Graphically determine the magnitude and type/nature of each member. Scale: Space diagram: 1 :... show full transcript

Worked Solution & Example Answer:Figure 7.1 below shows a framework with members AE, BF, CF, DE and EF - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2023 - Paper 1

Step 1

The magnitude of the reaction at support RL

96%

114 rated

Answer

To find the reaction at support RL, we use the moment about RR.

The equation is given by: $RL \times 10 = (8 \times 8) - (4 \times 5) - (6 \times 2)$

Calculating: $RL \times 10 = 96 - 48 - 12 = 36 \rightarrow RL = \frac{36}{10} = 3.6 \text{ kN}$

Now taking into account the upward forces must equal the downward forces: $RL + RR + 8 kN + 4 kN + 6 kN = 0$

Thus, substituting for RR we get: $3.6 + RR + 18 = 0 \rightarrow RR = 10.4 kN$ . Hence, the magnitude of the reaction at support RL is 9.6 kN.

Step 2

The magnitude of the reaction at support RR

99%

104 rated

Answer

For support RR, exploiting the same moment equation:

$RR \times 10 = (6 \times 8) + (4 \times 5) + (8 \times 2)$

Performing the calculations: $RR = \frac{(48 + 20 + 16)}{10} = \frac{84}{10} = 8.4 kN$

Thus, the magnitude of the reaction at support RR is 8.4 kN.

Step 3

The bending moments at points B, C and D

96%

101 rated

Answer

The bending moment at points can be calculated as:

At Point B: $M_B = (9.6 \text{ kN}) \times 2 = 19.2 kNm$
At Point C: $M_C = 9.6 \times 3 - 8 \times 3 = 24 kNm$
At Point D: $M_D = 9.6 \times 8 - 4.8 \times 6 = 16.8 kNm$

Hence the bending moments are:

At B: 19.2 kNm
At C: 24 kNm
At D: 16.8 kNm.

Step 4

Draw a bending-moment diagram of the beam

98%

120 rated

Answer

The bending-moment diagram is established by plotting the calculated values at points A, B, C, D, and E.

Start at zero at point A, with a moment of 19.2 kNm at B and 24 kNm at C, then downward to 16.8 kNm at D, returning to zero at E. The area under the diagram signifies various segments of moment influenced by loads.

Step 5

The cross-sectional area of the bar in m²

97%

117 rated

Answer

The cross-sectional area A of a round bar is given by: $A = \frac{\pi D^2}{4}$ Substituting the diameter D = 30 mm = 0.03 m: $A = \frac{\pi (0.03)^2}{4} \approx 0.0007069 m^2$

Thus the cross-sectional area is 0.0007069 m².

Step 6

The stress caused in the material in MPa

97%

121 rated

Answer

Stress can be calculated using: $\text{Stress} = \frac{\text{Load}}{\text{Area}}$ Substituting Load = 80 kN = 80,000 N, and the area previously calculated: $\text{Stress} = \frac{80000}{0.0007069} \approx 112.68 MPa$

Thus, the stress caused is 112.68 MPa.

Step 7

The strain caused if the change in length of the bar is 0.06 m

96%

114 rated

Answer

Strain is defined as: $\text{Strain} = \frac{\Delta L}{L_0}$ Where $\Delta L$ is the change in length (0.06 m), and $L_0$ is the original length (3 m): $\text{Strain} = \frac{0.06}{3} = 0.02$

Therefore, the strain caused is 0.02.

Step 8

Young's modulus of elasticity in GPa

99%

104 rated

Answer

Young's modulus can be calculated by: $E = \frac{\text{Stress}}{\text{Strain}}$ Substituting the stress and strain values: $E = \frac{112.68 \times 10^6}{0.02} \approx 5634.80 GPa$

Thus, Young's modulus of elasticity is 5634.80 GPa.

Step 9

Calculate the maximum stress

96%

101 rated

Answer

The maximum stress caused in the round steel bar due to a tensile load can be calculated as: $\text{Stress} = \frac{\text{Load}}{\text{Area}}$ Substituting Load = 55 kN = 55,000 N, and Area = 0.9 x 10⁻⁵ m²: $\text{Stress} = \frac{55000}{0.9 \times 10^{-5}} \approx 6111.11 MPa$

Therefore, the maximum stress is 6111.11 MPa.

Figure 7.1 below shows a framework with members AE, BF, CF, DE and EF - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2023 - Paper 1

Worked Solution & Example Answer:Figure 7.1 below shows a framework with members AE, BF, CF, DE and EF - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2023 - Paper 1

The magnitude of the reaction at support RL

The magnitude of the reaction at support RR

The bending moments at points B, C and D

Draw a bending-moment diagram of the beam

The cross-sectional area of the bar in m²

The stress caused in the material in MPa

The strain caused if the change in length of the bar is 0.06 m

Young's modulus of elasticity in GPa

Calculate the maximum stress

Join the NSC students using SimpleStudy...