A steel tube, 25 mm outside diameter and 12mm inside diameter, carries an axial tensile load of 40 kN. What will be the stress in the bar? What further increase in load is possible if the stress in the bar is limited to 225 MN/m?

Okay, let's solve the mechanics questions from the image.

Question 3b) (Part 1 of 2)

a) A steel tube, 25 mm outside diameter and 12mm inside diameter, carries an axial tensile load of 40 kN. What will be the stress in the bar?

Step 1: Calculate the cross-sectional area of the tube.
The area of a hollow tube is given by $A = \frac{\pi}{4} (D_o^2 - D_i^2)$, where $D_o$ is the outside diameter and $D_i$ is the inside diameter.
Given: $D_o = 25 \text{ mm} = 0.025 \text{ m}$ and $D_i = 12 \text{ mm} = 0.012 \text{ m}$.

$$A = \frac{\pi}{4} ((0.025 \text{ m})^2 - (0.012 \text{ m})^2)$$ $$A = \frac{\pi}{4} (0.000625 \text{ m}^2 - 0.000144 \text{ m}^2)$$ $$A = \frac{\pi}{4} (0.000481 \text{ m}^2)$$ $$A \approx 3.776 \times 10^{-4} \text{ m}^2$$

Step 2: Calculate the stress in the bar.
Stress ($\sigma$) is defined as force ($P$) per unit area ($A$).
Given: $P = 40 \text{ kN} = 40 \times 10^3 \text{ N}$.

$$\sigma = \frac{P}{A}$$ $$\sigma = \frac{40 \times 10^3 \text{ N}}{3.776 \times 10^{-4} \text{ m}^2}$$ $$\sigma \approx 105.93 \times 10^6 \text{ N/m}^2$$ $$\sigma \approx \boxed{105.93 \text{ MN/m}^2}$$

b) What further increase in load is possible if the stress in the bar is limited to 225 MN/m²?

Step 3: Calculate the maximum possible load.
Given: Maximum allowable stress $\sigma_{max} = 225 \text{ MN/m}^2 = 225 \times 10^6 \text{ N/m}^2$.

$$P_{max} = \sigma_{max} \times A$$ $$P_{max} = (225 \times 10^6 \text{ N/m}^2) \times (3.776 \times 10^{-4} \text{ m}^2)$$ $$P_{max} \approx 84960 \text{ N}$$ $$P_{max} \approx 84.96 \text{ kN}$$

Step 4: Calculate the further increase in load.
Further increase in load ($\Delta P$) is the difference between the maximum possible load and the initial load.

$$\Delta P = P_{max} - P_{initial}$$ $$\Delta P = 84.96 \text{ kN} - 40 \text{ kN}$$ $$\Delta P = \boxed{44.96 \text{ kN}}$$

Question 4a)

A beam AB, 1.2m long, is simply-supported at its ends A and B and carries two concentrated loads, one of 10 kN at C, the other 15 kN at D. Point C is 0.4m from A, point D is 1 m from A. Draw the S.F. and B.M. diagrams for the beam inserting principal values.

Step 1: Calculate reactions at supports A and B.
Let $R_A$ and $R_B$ be the reactions at A and B.
Total length $L = 1.2 \text{ m}$.
Load $P_C = 10 \text{ kN}$ at $x_C = 0.4 \text{ m}$ from A.
Load $P_D = 15 \text{ kN}$ at $x_D = 1 \text{ m}$ from A.

Sum of vertical forces:

$$R_A + R_B = 10 \text{ kN} + 15 \text{ kN} = 25 \text{ kN} \quad (1)$$

Sum of moments about A (clockwise positive):

$$\sum M_A = 0$$ $$(10 \text{ kN} \times 0.4 \text{ m}) + (15 \text{ kN} \times 1 \text{ m}) - (R_B \times 1.2 \text{ m}) = 0$$ $$4 \text{ kN}\cdot\text{m} + 15 \text{ kN}\cdot\text{m} - 1.2 R_B = 0$$ $$19 \text{ kN}\cdot\text{m} = 1.2 R_B$$ $$R_B = \frac{19}{1.2} \text{ kN} \approx 15.833 \text{ kN}$$

Substitute $R_B$ into (1):

$$R_A = 25 \text{ kN} - 15.833 \text{ kN} \approx 9.167 \text{ kN}$$

The reactions are: $R_A = \boxed{9.167 \text{ kN}}$ and $R_B = \boxed{15.833 \text{ kN}}$.

Step 2: Calculate Shear Force (SF) values.

• $SF_A$ (just right of A) $= R_A = 9.167 \text{ kN}$
• $SF_{AC}$ (between A and C) $= 9.167 \text{ kN}$
• $SF_C$ (just right of C) $= 9.167 \text{ kN} - 10 \text{ kN} = -0.833 \text{ kN}$
• $SF_{CD}$ (between C and D) $= -0.833 \text{ kN}$
• $SF_D$ (just right of D) $= -0.833 \text{ kN} - 15 \text{ kN} = -15.833 \text{ kN}$
• $SF_{DB}$ (between D and B) $= -15.833 \text{ kN}$
• $SF_B$ (just left of B) $= -15.833 \text{ kN}$ (balanced by $R_B$)

Step 3: Calculate Bending Moment (BM) values.

• $BM_A = 0 \text{ kN}\cdot\text{m}$ (simply supported end)
• $BM_C$ (at $x=0.4 \text{ m}$) $= R_A \times 0.4 \text{ m} = 9.167 \text{ kN} \times 0.4 \text{ m} = \boxed{3.667 \text{ kN}\cdot\text{m}}$
• $BM_D$ (at $x=1 \text{ m}$) $= (R_A \times 1 \text{ m}) - (10 \text{ kN} \times (1 \text{ m} - 0.4 \text{ m}))$ $= (9.167 \text{ kN} \times 1 \text{ m}) - (10 \text{ kN} \times 0.6 \text{ m})$ $= 9.167 \text{ kN}\cdot\text{m} - 6 \text{ kN}\cdot\text{m} = \boxed{3.167 \text{ kN}\cdot\text{m}}$
• $BM_B = 0 \text{ kN}\cdot\text{m}$ (simply supported end)

Question 4b)

A beam AB, 5 m long, is simply-supported at the end B and at a point C, 1 m from A. It carries vertical loads of 5 KN at A and 20kN at D, the center of the span BC. Draw S.F. and B.M. diagrams for the beam inserting principal values.

Step 1: Calculate reactions at supports C and B.
Beam length AB = 5 m.
Support C is at $x_C = 1 \text{ m}$ from A.
Support B is at $x_B = 5 \text{ m}$ from A.
Load $P_A = 5 \text{ kN}$ at A ($x=0$).
Span BC length = $5 \text{ m} - 1 \text{ m} = 4 \text{ m}$.
Load $P_D = 20 \text{ kN}$ at D, center of span BC. So $x_D = 1 \text{ m} + \frac{4 \text{ m}}{2} = 3 \text{ m}$ from A.

Let $R_C$ and $R_B$ be the reactions at C and B.

Sum of vertical forces:

$$R_C + R_B = P_A + P_D$$ $$R_C + R_B = 5 \text{ kN} + 20 \text{ kN} = 25 \text{ kN} \quad (1)$$

Sum of moments about C (clockwise positive):

$$\sum M_C = 0$$ $$(P_A \times 1 \text{ m}) + (P_D \times (3 \text{ m} - 1 \text{ m})) - (R_B \times (5 \text{ m} - 1 \text{ m})) = 0$$ $$(5 \text{ kN} \times 1 \text{ m}) + (20 \text{ kN} \times 2 \text{ m}) - (R_B \times 4 \text{ m}) = 0$$ $$5 \text{ kN}\cdot\text{m} + 40 \text{ kN}\cdot\text{m} - 4 R_B = 0$$ $$45 \text{ kN}\cdot\text{m} = 4 R_B$$ $$R_B = \frac{45}{4} \text{ kN} = 11.25 \text{ kN}$$

Substitute $R_B$ into (1):

$$R_C = 25 \text{ kN} - 11.25 \text{ kN} = 13.75 \text{ kN}$$

The reactions are: $R_C = \boxed{13.75 \text{ kN}}$ and $R_B = \boxed{11.25 \text{ kN}}$.

Step 2: Calculate Shear Force (SF) values.

• $SF_A$ (just right of A) $= -P_A = -5 \text{ kN}$ (downward load)
• $SF_{AC}$ (between A and C) $= -5 \text{ kN}$
• $SF_C$ (just right of C) $= -5 \text{ kN} + R_C = -5 \text{ kN} + 13.75 \text{ kN} = 8.75 \text{ kN}$
• $SF_{CD}$ (between C and D) $= 8.75 \text{ kN}$
• $SF_D$ (just right of D) $= 8.75 \text{ kN} - P_D = 8.75 \text{ kN} - 20 \text{ kN} = -11.25 \text{ kN}$
• $SF_{DB}$ (between D and B) $= -11.25 \text{ kN}$
• $SF_B$ (just left of B) $= -11.25 \text{ kN}$ (balanced by $R_B$)

Step 3: Calculate Bending Moment (BM) values.

• $BM_A = 0 \text{ kN}\cdot\text{m}$ (free end)
• $BM_C$ (at $x=1 \text{ m}$) $= -P_A \times 1 \text{ m} = -5 \text{ kN} \times 1 \text{ m} = \boxed{-5 \text{ kN}\cdot\text{m}}$ (hogging)
• $BM_D$ (at $x=3 \text{ m}$) $= (-P_A \times 3 \text{ m}) + (R_C \times (3 \text{ m} - 1 \text{ m}))$ $= (-5 \text{ kN} \times 3 \text{ m}) + (13.75 \text{ kN} \times 2 \text{ m})$ $= -15 \text{ kN}\cdot\text{m} + 27.5 \text{ kN}\cdot\text{m} = \boxed{12.5 \text{ kN}\cdot\text{m}}$ (sagging)
• $BM_B = 0 \text{ kN}\cdot\text{m}$ (simply supported end)

Question 5)

A beam ABC is 9 m long and supported at B and C, 6 m apart as shown in Fig. 3.18. The beam carries a triangular distribution of load over the portion BC together with an applied counterclockwise couple of moment 80 KN m at Band a u.d.l. Of 10 KN/m over AB, as shown. Draw the S.F. and B.M. diagrams for the beam.

Note: The image provided only shows the diagram for Question 5, but the text for Question 5 is cut off. I will proceed with the information visible in the diagram and the partial text.

From the diagram:

• Beam length AC = 9 m.
• Support B is at 3 m from A.
• Support C is at 9 m from A (6 m from B).
• UDL of $10 \text{ kN/m}$ over AB (from $x=0$ to $x=3 \text{ m}$).
• Counterclockwise couple of $80 \text{ kN}\cdot\text{m}$ at B.
• Triangular load over BC (from $x=3 \text{ m}$ to $x=9 \text{ m}$). The load intensity at B is $48 \text{ kN/m}$ and decreases linearly to 0 at C.

Step 1: Calculate reactions at supports B and C.
Let $R_B$ and $R_C$ be the reactions at B and C.

Convert distributed loads to equivalent point loads:

• UDL over AB: $W_{AB} = 10 \text{ kN/m} \times 3 \text{ m} = 30 \text{ kN}$. Acts at $1.5 \text{ m}$ from A.
• Triangular load over BC: $W_{BC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \text{ m} \times 48 \text{ kN/m} = 144 \text{ kN}$. This load acts at $\frac{1}{3}$ of the base from the higher end (B) or $\frac{2}{3}$ of the base from the zero end (C). So, it acts at $3 \text{ m} + \frac{1}{3} \times 6 \text{ m} = 3 \text{ m} + 2 \text{ m} = 5 \text{ m}$ from A.

Sum of vertical forces:

$$\sum F_y = 0$$ $$R_B + R_C = W_{AB} + W_{BC}$$ $$R_B + R_C = 30 \text{ kN} + 144 \text{ kN} = 174 \text{ kN} \quad (1)$$

Sum of moments about B (clockwise positive):

$$\sum M_B = 0$$ $$(W_{AB} \times (3 \text{ m} - 1.5 \text{ m})) + (80 \text{ kN}\cdot\text{m}) - (W_{BC} \times (5 \text{ m} - 3 \text{ m})) + (R_C \times (9 \text{ m} - 3 \text{ m})) = 0$$ $$(30 \text{ kN} \times 1.5 \text{ m}) + (80 \text{ kN}\cdot\text{m}) - (144 \text{ kN} \times 2 \text{ m}) + (R_C \times 6 \text{ m}) = 0$$ $$45 \text{ kN}\cdot\text{m} + 80 \text{ kN}\cdot\text{m} - 288 \text{ kN}\cdot\text{m} + 6 R_C = 0$$ $$125 \text{ kN}\cdot\text{m} - 288 \text{ kN}\cdot\text{m} + 6 R_C = 0$$ $$-163 \text{ kN}\cdot\text{m} + 6 R_C = 0$$ $$6 R_C = 163 \text{ kN}\cdot\text{m}$$ $$R_C = \frac{163}{6} \text{ kN} \approx 27.167 \text{ kN}$$

Substitute $R_C$ into (1):

$$$$