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?

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Asked on July 7, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-07T11:15:06.737Z

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 = ()/(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 mm = 0.025 m and D_i = 12 mm = 0.012 m. A = ()/(4) ((0.025 m)^2 - (0.012 m)^2) A = ()/(4) (0.000625 m^2 - 0.000144 m^2) A = ()/(4) (0.000481 m^2) A ≈ 3.776 × 10^-4 m^2 Step 2: Calculate the stress in the bar. Stress () is defined as force (P) per unit area (A). Given: P = 40 kN = 40 × 10^3 N. = (P)/(A) = 40 × 10^3 N3.776 × 10^-4 m^2 ≈ 105.93 × 10^6 N/m^2 ≈ 105.93 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 _max = 225 MN/m^2 = 225 × 10^6 N/m^2. P_max = _max × A P_max = (225 × 10^6 N/m^2) × (3.776 × 10^-4 m^2) P_max ≈ 84960 N P_max ≈ 84.96 kN Step 4: Calculate the further increase in load. Further increase in load ( P) is the difference between the maximum possible load and the initial load. P = P_max - P_initial P = 84.96 kN - 40 kN P = 44.96 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 m. Load P_C = 10 kN at x_C = 0.4 m from A. Load P_D = 15 kN at x_D = 1 m from A. Sum of vertical forces: R_A + R_B = 10 kN + 15 kN = 25 kN (1) Sum of moments about A (clockwise positive): M_A = 0 (10 kN × 0.4 m) + (15 kN × 1 m) - (R_B × 1.2 m) = 0 4 kN·m + 15 kN·m - 1.2 R_B = 0 19 kN·m = 1.2 R_B R_B = (19)/(1.2) kN ≈ 15.833 kN Substitute R_B into (1): R_A = 25 kN - 15.833 kN ≈ 9.167 kN The reactions are: R_A = 9.167 kN and R_B = 15.833 kN. Step 2: Calculate Shear Force (SF) values. SF_A (just right of A) = R_A = 9.167 kN SF_AC (between A and C) = 9.167 kN SF_C (just right of C) = 9.167 kN - 10 kN = -0.833 kN SF_CD (between C and D) = -0.833 kN SF_D (just right of D) = -0.833 kN - 15 kN = -15.833 kN SF_DB (between D and B) = -15.833 kN SF_B (just left of B) = -15.833 kN (balanced by R_B) Step 3: Calculate Bending Moment (BM) values. BM_A = 0 kN·m (simply supported end) BM_C (at x=0.4 m) = R_A × 0.4 m = 9.167 kN × 0.4 m = 3.667 kN·m BM_D (at x=1 m) = (R_A × 1 m) - (10 kN × (1 m - 0.4 m)) = (9.167 kN × 1 m) - (10 kN × 0.6 m) = 9.167 kN·m - 6 kN·m = 3.167 kN·m BM_B = 0 kN·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 m from A. Support B is at x_B = 5 m from A. Load P_A = 5 kN at A (x=0). Span BC length = 5 m - 1 m = 4 m. Load P_D = 20 kN at D, center of span BC. So x_D = 1 m + 4 m2 = 3 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 kN + 20 kN = 25 kN (1) Sum of moments about C (clockwise positive): M_C = 0 (P_A × 1 m) + (P_D × (3 m - 1 m)) - (R_B × (5 m - 1 m)) = 0 (5 kN × 1 m) + (20 kN × 2 m) - (R_B × 4 m) = 0 5 kN·m + 40 kN·m - 4 R_B = 0 45 kN·m = 4 R_B R_B = (45)/(4) kN = 11.25 kN Substitute R_B into (1): R_C = 25 kN - 11.25 kN = 13.75 kN The reactions are: R_C = 13.75 kN and R_B = 11.25 kN. Step 2: Calculate Shear Force (SF) values. SF_A (just right of A) = -P_A = -5 kN (downward load) SF_AC (between A and C) = -5 kN SF_C (just right of C) = -5 kN + R_C = -5 kN + 13.75 kN = 8.75 kN SF_CD (between C and D) = 8.75 kN SF_D (just right of D) = 8.75 kN - P_D = 8.75 kN - 20 kN = -11.25 kN SF_DB (between D and B) = -11.25 kN SF_B (just left of B) = -11.25 kN (balanced by R_B) Step 3: Calculate Bending Moment (BM) values. BM_A = 0 kN·m (free end) BM_C (at x=1 m) = -P_A × 1 m = -5 kN × 1 m = -5 kN·m (hogging) BM_D (at x=3 m) = (-P_A × 3 m) + (R_C × (3 m - 1 m)) = (-5 kN × 3 m) + (13.75 kN × 2 m) = -15 kN·m + 27.5 kN·m = 12.5 kN·m (sagging) BM_B = 0 kN·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 kN/m over AB (from x=0 to x=3 m). Counterclockwise couple of 80 kN·m at B. Triangular load over BC (from x=3 m to x=9 m). The load intensity at B is 48 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 kN/m × 3 m = 30 kN. Acts at 1.5 m from A. Triangular load over BC: W_BC = (1)/(2) × base × height = (1)/(2) × 6 m × 48 kN/m = 144 kN. This load acts at (1)/(3) of the base from the higher end (B) or (2)/(3) of the base from the zero end (C). So, it acts at 3 m + (1)/(3) × 6 m = 3 m + 2 m = 5 m from A. Sum of vertical forces: F_y = 0 R_B + R_C = W_AB + W_BC R_B + R_C = 30 kN + 144 kN = 174 kN (1) Sum of moments about B (clockwise positive): M_B = 0 (W_AB × (3 m - 1.5 m)) + (80 kN·m) - (W_BC × (5 m - 3 m)) + (R_C × (9 m - 3 m)) = 0 (30 kN × 1.5 m) + (80 kN·m) - (144 kN × 2 m) + (R_C × 6 m) = 0 45 kN·m + 80 kN·m - 288 kN·m + 6 R_C = 0 125 kN·m - 288 kN·m + 6 R_C = 0 -163 kN·m + 6 R_C = 0 6 R_C = 163 kN·m R_C = (163)/(6) kN ≈ 27.167 kN Substitute R_C into (1): R_B ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._

Accepted Answer · 2026-07-07T11:15:06.737Z

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 = ()/(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 mm = 0.025 m and D_i = 12 mm = 0.012 m. A = ()/(4) ((0.025 m)^2 - (0.012 m)^2) A = ()/(4) (0.000625 m^2 - 0.000144 m^2) A = ()/(4) (0.000481 m^2) A ≈ 3.776 × 10^-4 m^2 Step 2: Calculate the stress in the bar. Stress () is defined as force (P) per unit area (A). Given: P = 40 kN = 40 × 10^3 N. = (P)/(A) = 40 × 10^3 N3.776 × 10^-4 m^2 ≈ 105.93 × 10^6 N/m^2 ≈ 105.93 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 _max = 225 MN/m^2 = 225 × 10^6 N/m^2. P_max = _max × A P_max = (225 × 10^6 N/m^2) × (3.776 × 10^-4 m^2) P_max ≈ 84960 N P_max ≈ 84.96 kN Step 4: Calculate the further increase in load. Further increase in load ( P) is the difference between the maximum possible load and the initial load. P = P_max - P_initial P = 84.96 kN - 40 kN P = 44.96 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 m. Load P_C = 10 kN at x_C = 0.4 m from A. Load P_D = 15 kN at x_D = 1 m from A. Sum of vertical forces: R_A + R_B = 10 kN + 15 kN = 25 kN (1) Sum of moments about A (clockwise positive): M_A = 0 (10 kN × 0.4 m) + (15 kN × 1 m) - (R_B × 1.2 m) = 0 4 kN·m + 15 kN·m - 1.2 R_B = 0 19 kN·m = 1.2 R_B R_B = (19)/(1.2) kN ≈ 15.833 kN Substitute R_B into (1): R_A = 25 kN - 15.833 kN ≈ 9.167 kN The reactions are: R_A = 9.167 kN and R_B = 15.833 kN. Step 2: Calculate Shear Force (SF) values. SF_A (just right of A) = R_A = 9.167 kN SF_AC (between A and C) = 9.167 kN SF_C (just right of C) = 9.167 kN - 10 kN = -0.833 kN SF_CD (between C and D) = -0.833 kN SF_D (just right of D) = -0.833 kN - 15 kN = -15.833 kN SF_DB (between D and B) = -15.833 kN SF_B (just left of B) = -15.833 kN (balanced by R_B) Step 3: Calculate Bending Moment (BM) values. BM_A = 0 kN·m (simply supported end) BM_C (at x=0.4 m) = R_A × 0.4 m = 9.167 kN × 0.4 m = 3.667 kN·m BM_D (at x=1 m) = (R_A × 1 m) - (10 kN × (1 m - 0.4 m)) = (9.167 kN × 1 m) - (10 kN × 0.6 m) = 9.167 kN·m - 6 kN·m = 3.167 kN·m BM_B = 0 kN·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 m from A. Support B is at x_B = 5 m from A. Load P_A = 5 kN at A (x=0). Span BC length = 5 m - 1 m = 4 m. Load P_D = 20 kN at D, center of span BC. So x_D = 1 m + 4 m2 = 3 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 kN + 20 kN = 25 kN (1) Sum of moments about C (clockwise positive): M_C = 0 (P_A × 1 m) + (P_D × (3 m - 1 m)) - (R_B × (5 m - 1 m)) = 0 (5 kN × 1 m) + (20 kN × 2 m) - (R_B × 4 m) = 0 5 kN·m + 40 kN·m - 4 R_B = 0 45 kN·m = 4 R_B R_B = (45)/(4) kN = 11.25 kN Substitute R_B into (1): R_C = 25 kN - 11.25 kN = 13.75 kN The reactions are: R_C = 13.75 kN and R_B = 11.25 kN. Step 2: Calculate Shear Force (SF) values. SF_A (just right of A) = -P_A = -5 kN (downward load) SF_AC (between A and C) = -5 kN SF_C (just right of C) = -5 kN + R_C = -5 kN + 13.75 kN = 8.75 kN SF_CD (between C and D) = 8.75 kN SF_D (just right of D) = 8.75 kN - P_D = 8.75 kN - 20 kN = -11.25 kN SF_DB (between D and B) = -11.25 kN SF_B (just left of B) = -11.25 kN (balanced by R_B) Step 3: Calculate Bending Moment (BM) values. BM_A = 0 kN·m (free end) BM_C (at x=1 m) = -P_A × 1 m = -5 kN × 1 m = -5 kN·m (hogging) BM_D (at x=3 m) = (-P_A × 3 m) + (R_C × (3 m - 1 m)) = (-5 kN × 3 m) + (13.75 kN × 2 m) = -15 kN·m + 27.5 kN·m = 12.5 kN·m (sagging) BM_B = 0 kN·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 kN/m over AB (from x=0 to x=3 m). Counterclockwise couple of 80 kN·m at B. Triangular load over BC (from x=3 m to x=9 m). The load intensity at B is 48 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 kN/m × 3 m = 30 kN. Acts at 1.5 m from A. Triangular load over BC: W_BC = (1)/(2) × base × height = (1)/(2) × 6 m × 48 kN/m = 144 kN. This load acts at (1)/(3) of the base from the higher end (B) or (2)/(3) of the base from the zero end (C). So, it acts at 3 m + (1)/(3) × 6 m = 3 m + 2 m = 5 m from A. Sum of vertical forces: F_y = 0 R_B + R_C = W_AB + W_BC R_B + R_C = 30 kN + 144 kN = 174 kN (1) Sum of moments about B (clockwise positive): M_B = 0 (W_AB × (3 m - 1.5 m)) + (80 kN·m) - (W_BC × (5 m - 3 m)) + (R_C × (9 m - 3 m)) = 0 (30 kN × 1.5 m) + (80 kN·m) - (144 kN × 2 m) + (R_C × 6 m) = 0 45 kN·m + 80 kN·m - 288 kN·m + 6 R_C = 0 125 kN·m - 288 kN·m + 6 R_C = 0 -163 kN·m + 6 R_C = 0 6 R_C = 163 kN·m R_C = (163)/(6) kN ≈ 27.167 kN Substitute R_C into (1): R_B ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._

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?

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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?

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