A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

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Asked on July 12, 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 · 2026-07-12T17:28:54.360Z

here are the solutions to your engineering science questions. QUESTION 1: STATICS 1.1 A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam. As an AI, I cannot physically draw the diagrams. However, I can describe them and provide the critical values. Loaded Beam Description: The beam is a 6m cantilever, fixed at the left end (let's call it A) and free at the right end (B). A concentrated downward load of 35 kN is applied at the free end (B). A uniformly distributed downward load (UDL) of 9 kN/m acts over the first 3 m from the fixed end (from A to a point C, 3 m from A). Shear Force Diagram (SFD) Description: The shear force starts at 0 kN at the free end (B). Just to the left of the free end (B), the shear force drops to -35 kN due to the concentrated load. From x=6 m (free end) to x=3 m (start of UDL), the shear force remains constant at -35 kN. From x=3 m to x=0 m (fixed end A), the shear force decreases linearly due to the UDL. At x=3 m, the shear force is -35 kN. At x=0 m (fixed end A), the shear force is -35 kN - (9 kN/m × 3 m) = -35 kN - 27 kN = -62 kN. The diagram will show a horizontal line at -35 kN from x=3 m to x=6 m, and a linearly decreasing line from -35 kN to -62 kN from x=0 m to x=3 m. Bending Moment Diagram (BMD) Description: The bending moment starts at 0 kNm at the free end (B). From x=6 m (free end) to x=3 m (start of UDL), the bending moment decreases linearly. At x=3 m, the bending moment is -35 kN × (6 m - 3 m) = -35 kN × 3 m = -105 kNm. From x=3 m to x=0 m (fixed end A), the bending moment decreases parabolically due to the UDL. At x=0 m (fixed end A), the bending moment is - (35 kN × 6 m) - (9 kN/m × 3 m × 1.5 m) = -210 kNm - 40.5 kNm = -250.5 kNm. The diagram will show a linearly decreasing line from 0 kNm at x=6 m to -105 kNm at x=3 m, and then a parabolically decreasing curve from -105 kNm to -250.5 kNm at x=0 m. The entire diagram will be in the negative region. 1.2 Calculate the position of the centroid of the lamina shown below. The lamina consists of a rectangle and a triangle. Let's set the origin at the bottom-left corner of the rectangle. Shape 1: Rectangle Dimensions: 60 mm × 50 mm Area A_1 = 60 mm × 50 mm = 3000 mm^2 Centroid x_1 = 60 mm2 = 30 mm Centroid y_1 = 50 mm2 = 25 mm Shape 2: Triangle Base (vertical) = 50 mm, Height (horizontal) = 50 mm Area A_2 = (1)/(2) × 50 mm × 50 mm = 1250 mm^2 Centroid x_2: From the origin, it's 60 mm (rectangle width) plus (1)/(3) of the triangle's height. x_2 = 60 mm + 50 mm3 = 60 mm + 16.667 mm = 76.667 mm Centroid y_2: From the origin, it's (1)/(3) of the triangle's base (since the base is along the y-axis from y=0 to y=50). y_2 = 50 mm3 = 16.667 mm Composite Centroid: Total Area A = A_1 + A_2 = 3000 mm^2 + 1250 mm^2 = 4250 mm^2 Step 1: Calculate the x-coordinate of the centroid. x = (A_1 x_1 + A_2 x_2)/(A_1 + A_2) x = (3000 mm^2 × 30 mm) + (1250 mm^2 × 76.667 mm)4250 mm^2 x = 90000 mm^3 + 95833.75 mm^34250 mm^2 = 185833.75 mm^34250 mm^2 x ≈ 43.726 mm Step 2: Calculate the y-coordinate of the centroid. y = (A_1 y_1 + A_2 y_2)/(A_1 + A_2) y = (3000 mm^2 × 25 mm) + (1250 mm^2 × 16.667 mm)4250 mm^2 y = 75000 mm^3 + 20833.75 mm^34250 mm^2 = 95833.75 mm^34250 mm^2 y ≈ 22.549 mm The position of the centroid is (x, y) = (43.73 mm, 22.55 mm) from the bottom-left corner. QUESTION 2: HYDRAULICS 2.1 A force of 400N on the plunger of a hydraulic press causes a force of 14.5kN to be exerted on the ram. The diameter of the ram is 400mm. Calculate: Given: F_p = 400 N, F_r = 14.5 kN = 14500 N, D_r = 400 mm = 0.4 m. 2.1.1 The liquid pressure Step 1: Calculate the area of the ram. A_r = ((D_r)/(2))^2 = (0.4 m2)^2 = (0.2 m)^2 = 0.04 m^2 ≈ 0.12566 m^2 Step 2: Calculate the liquid pressure using the force on the ram. P = (F_r)/(A_r) = 14500 N0.04 m^2 P ≈ 115380.2 Pa P ≈ 115.38 kPa The liquid pressure is 115.38 kPa. 2.1.2 The diameter of the plunger in mm Step 1: Calculate the area of the plunger using the plunger force and the calculated pressure. A_p = (F_p)/(P) = 400 N115380.2 Pa ≈ 0.0034668 m^2 Step 2: Calculate the diameter of the plunger. D_p = 2 sqrt((A_p)/()) = 2 sqrt(0.0034668 m)^2 ≈ 2 sqrt(0.0011035 m)^2 D_p ≈ 2 × 0.033219 m ≈ 0.066438 m Convert to millimeters: D_p ≈ 0.066438 m × 1000 mm/m ≈ 66.44 mm The diameter of the plunger is 66.44 mm. 2.2 The diameter of the ram of a hydraulic press is 380mm and it must lift a vehicle 500m up. The vehicle has a mass of 2000kg. Calculate the following: Given: D_r = 380 mm = 0.38 m, h = 500 m, m = 2000 kg, g = 9.81 m/s^2. 2.2.1 The pressure in the ram Step 1: Calculate the force (weight) of the vehicle. F = m × g = 2000 kg × 9.81 m/s^2 = 19620 N Step 2: Calculate the area of the ram. A_r = ((D_r)/(2))^2 = (0.38 m2)^2 = (0.19 m)^2 = 0.0361 m^2 ≈ 0.11341 m^2 Step 3: Calculate the pressure in the ram. P = (F)/(A_r) = 19620 N0.0361 m^2 P ≈ 173000 Pa P ≈ 173.0 kPa The pressure in the ram is 173.0 kPa. 2.2.2 The volume of liquid displaced to do this work Step 1: Calculate the volume of liquid displaced. V = A_r × h = (0.0361 m^2) × 500 m V ≈ 0.11341 m^2 × 500 m ≈ 56.705 m^3 The volume of liquid displaced is 56.71 m^3. 2.2.3 The work done if the press is 85% efficient Given efficiency = 85\% = 0.85. Step 1: Calculate the output work (work done on the vehicle). W_out = F × h = 19620 N × 500 m = 9810000 J W_out = 9.81 MJ Step 2: Calculate the input work (work done by the press) considering efficiency. W_in = W_out = 9810000 J0.85 W_in ≈ 11541176.47 J W_in ≈ 11.54 MJ The work done if the press is 85% efficient is 11.54 MJ. 2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m^3 and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m. Given: D = 100 mm = 0.1 m, L = 300 mm = 0.3 m, N = 80 rpm, = 1000 kg/m^3, = 68\% = 0.68, H = 100 m, g = 9.81 m/s^2. A double-acting pump displaces liquid on both the forward and return strokes. Step 1: Calculate the cross-sectional area of the cylinder. A = ((D)/(2))^2 = (0.1 m2)^2 = (0.05 m)^2 = 0.0025 m^2 ≈ 0.007854 m^2 Step 2: Calculate the volume displaced per revolution. V_rev = 2 × A × L = 2 × (0.0025 m^2) × 0.3 m = 0.0015 m^3 ≈ 0.004712 m^3 Step 3: Calculate the volume flow rate in m^3/s. Q = V_rev × N = (0.0015 m^3/rev) × (80 rev/min) × (1 min60 s) Q = (0.12)/(60) m^3/s = 0.002 m^3/s ≈ 0.006283 m^3/s Step 4: Calculate the mass flow rate. m = × Q = 1000 kg/m^3 × 0.002 m^3/s = 2 kg/s ≈ 6.283 kg/s Step 5: Calculate the output power (power required to lift the water). P_out = m × g × H = (2 kg/s) × (9.81 m/s^2) × (100 m) P_out = 1962 W ≈ 6163.6 W P_out ≈ 6.164 kW Step 6: Calculate the input power required considering the pump efficiency. P_in = P_out = 6163.6 W0.68 P_in ≈ 9064.1 W P_in ≈ 9.064 kW The power required to deliver water is 9.064 kW. QUESTION 3: STRESS AND STRAIN 3.1 Name 3 types of stresses. • Tensile stress • Compressive stress • Shear stress 3.2 A tensile force of 3kN is applied to a steel rod of 2m in length and the rod extends by 0.381mm. The rod has a diameter of 1cm. Calculate: Given: F = 3 kN = 3000 N, L = 2 m, L = 0.381 mm = 0.000381 m, D = 1 cm = 0.01 m. 3.2.1 The stress Step 1: Calculate the cross-sectional area of the rod. A = ((D)/(2))^2 = (0.01 m2)^2 = (0.005 m)^2 = 0.000025 m^2 ≈ 7.854 × 10^-5 m^2 Step 2: Calculate the tensile stress. = (F)/(A) = 3000 N0.000025 m^2 ≈ 38197186.3 Pa ≈ 38.20 MPa The stress is 38.20 MPa. 3.2.2 The strain Step 1: Calculate the strain. = ( L)/(L) = 0.000381 m2 m = 0.0001905 The strain is 0.0001905. 3.2.3 The modulus of elasticity Step 1: Calculate the modulus of elasticity (Young's Modulus). E = ()/() = 38197186.3 Pa0.0001905 E ≈ 200510164.3 kPa E ≈ 200.51 GPa The modulus of elasticity is 200.51 GPa. Got more? Send 'em! 📸

Accepted Answer · 2026-07-12T17:28:54.360Z

here are the solutions to your engineering science questions. QUESTION 1: STATICS 1.1 A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam. As an AI, I cannot physically draw the diagrams. However, I can describe them and provide the critical values. Loaded Beam Description: The beam is a 6m cantilever, fixed at the left end (let's call it A) and free at the right end (B). A concentrated downward load of 35 kN is applied at the free end (B). A uniformly distributed downward load (UDL) of 9 kN/m acts over the first 3 m from the fixed end (from A to a point C, 3 m from A). Shear Force Diagram (SFD) Description: The shear force starts at 0 kN at the free end (B). Just to the left of the free end (B), the shear force drops to -35 kN due to the concentrated load. From x=6 m (free end) to x=3 m (start of UDL), the shear force remains constant at -35 kN. From x=3 m to x=0 m (fixed end A), the shear force decreases linearly due to the UDL. At x=3 m, the shear force is -35 kN. At x=0 m (fixed end A), the shear force is -35 kN - (9 kN/m × 3 m) = -35 kN - 27 kN = -62 kN. The diagram will show a horizontal line at -35 kN from x=3 m to x=6 m, and a linearly decreasing line from -35 kN to -62 kN from x=0 m to x=3 m. Bending Moment Diagram (BMD) Description: The bending moment starts at 0 kNm at the free end (B). From x=6 m (free end) to x=3 m (start of UDL), the bending moment decreases linearly. At x=3 m, the bending moment is -35 kN × (6 m - 3 m) = -35 kN × 3 m = -105 kNm. From x=3 m to x=0 m (fixed end A), the bending moment decreases parabolically due to the UDL. At x=0 m (fixed end A), the bending moment is - (35 kN × 6 m) - (9 kN/m × 3 m × 1.5 m) = -210 kNm - 40.5 kNm = -250.5 kNm. The diagram will show a linearly decreasing line from 0 kNm at x=6 m to -105 kNm at x=3 m, and then a parabolically decreasing curve from -105 kNm to -250.5 kNm at x=0 m. The entire diagram will be in the negative region. 1.2 Calculate the position of the centroid of the lamina shown below. The lamina consists of a rectangle and a triangle. Let's set the origin at the bottom-left corner of the rectangle. Shape 1: Rectangle Dimensions: 60 mm × 50 mm Area A_1 = 60 mm × 50 mm = 3000 mm^2 Centroid x_1 = 60 mm2 = 30 mm Centroid y_1 = 50 mm2 = 25 mm Shape 2: Triangle Base (vertical) = 50 mm, Height (horizontal) = 50 mm Area A_2 = (1)/(2) × 50 mm × 50 mm = 1250 mm^2 Centroid x_2: From the origin, it's 60 mm (rectangle width) plus (1)/(3) of the triangle's height. x_2 = 60 mm + 50 mm3 = 60 mm + 16.667 mm = 76.667 mm Centroid y_2: From the origin, it's (1)/(3) of the triangle's base (since the base is along the y-axis from y=0 to y=50). y_2 = 50 mm3 = 16.667 mm Composite Centroid: Total Area A = A_1 + A_2 = 3000 mm^2 + 1250 mm^2 = 4250 mm^2 Step 1: Calculate the x-coordinate of the centroid. x = (A_1 x_1 + A_2 x_2)/(A_1 + A_2) x = (3000 mm^2 × 30 mm) + (1250 mm^2 × 76.667 mm)4250 mm^2 x = 90000 mm^3 + 95833.75 mm^34250 mm^2 = 185833.75 mm^34250 mm^2 x ≈ 43.726 mm Step 2: Calculate the y-coordinate of the centroid. y = (A_1 y_1 + A_2 y_2)/(A_1 + A_2) y = (3000 mm^2 × 25 mm) + (1250 mm^2 × 16.667 mm)4250 mm^2 y = 75000 mm^3 + 20833.75 mm^34250 mm^2 = 95833.75 mm^34250 mm^2 y ≈ 22.549 mm The position of the centroid is (x, y) = (43.73 mm, 22.55 mm) from the bottom-left corner. QUESTION 2: HYDRAULICS 2.1 A force of 400N on the plunger of a hydraulic press causes a force of 14.5kN to be exerted on the ram. The diameter of the ram is 400mm. Calculate: Given: F_p = 400 N, F_r = 14.5 kN = 14500 N, D_r = 400 mm = 0.4 m. 2.1.1 The liquid pressure Step 1: Calculate the area of the ram. A_r = ((D_r)/(2))^2 = (0.4 m2)^2 = (0.2 m)^2 = 0.04 m^2 ≈ 0.12566 m^2 Step 2: Calculate the liquid pressure using the force on the ram. P = (F_r)/(A_r) = 14500 N0.04 m^2 P ≈ 115380.2 Pa P ≈ 115.38 kPa The liquid pressure is 115.38 kPa. 2.1.2 The diameter of the plunger in mm Step 1: Calculate the area of the plunger using the plunger force and the calculated pressure. A_p = (F_p)/(P) = 400 N115380.2 Pa ≈ 0.0034668 m^2 Step 2: Calculate the diameter of the plunger. D_p = 2 sqrt((A_p)/()) = 2 sqrt(0.0034668 m)^2 ≈ 2 sqrt(0.0011035 m)^2 D_p ≈ 2 × 0.033219 m ≈ 0.066438 m Convert to millimeters: D_p ≈ 0.066438 m × 1000 mm/m ≈ 66.44 mm The diameter of the plunger is 66.44 mm. 2.2 The diameter of the ram of a hydraulic press is 380mm and it must lift a vehicle 500m up. The vehicle has a mass of 2000kg. Calculate the following: Given: D_r = 380 mm = 0.38 m, h = 500 m, m = 2000 kg, g = 9.81 m/s^2. 2.2.1 The pressure in the ram Step 1: Calculate the force (weight) of the vehicle. F = m × g = 2000 kg × 9.81 m/s^2 = 19620 N Step 2: Calculate the area of the ram. A_r = ((D_r)/(2))^2 = (0.38 m2)^2 = (0.19 m)^2 = 0.0361 m^2 ≈ 0.11341 m^2 Step 3: Calculate the pressure in the ram. P = (F)/(A_r) = 19620 N0.0361 m^2 P ≈ 173000 Pa P ≈ 173.0 kPa The pressure in the ram is 173.0 kPa. 2.2.2 The volume of liquid displaced to do this work Step 1: Calculate the volume of liquid displaced. V = A_r × h = (0.0361 m^2) × 500 m V ≈ 0.11341 m^2 × 500 m ≈ 56.705 m^3 The volume of liquid displaced is 56.71 m^3. 2.2.3 The work done if the press is 85% efficient Given efficiency = 85\% = 0.85. Step 1: Calculate the output work (work done on the vehicle). W_out = F × h = 19620 N × 500 m = 9810000 J W_out = 9.81 MJ Step 2: Calculate the input work (work done by the press) considering efficiency. W_in = W_out = 9810000 J0.85 W_in ≈ 11541176.47 J W_in ≈ 11.54 MJ The work done if the press is 85% efficient is 11.54 MJ. 2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m^3 and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m. Given: D = 100 mm = 0.1 m, L = 300 mm = 0.3 m, N = 80 rpm, = 1000 kg/m^3, = 68\% = 0.68, H = 100 m, g = 9.81 m/s^2. A double-acting pump displaces liquid on both the forward and return strokes. Step 1: Calculate the cross-sectional area of the cylinder. A = ((D)/(2))^2 = (0.1 m2)^2 = (0.05 m)^2 = 0.0025 m^2 ≈ 0.007854 m^2 Step 2: Calculate the volume displaced per revolution. V_rev = 2 × A × L = 2 × (0.0025 m^2) × 0.3 m = 0.0015 m^3 ≈ 0.004712 m^3 Step 3: Calculate the volume flow rate in m^3/s. Q = V_rev × N = (0.0015 m^3/rev) × (80 rev/min) × (1 min60 s) Q = (0.12)/(60) m^3/s = 0.002 m^3/s ≈ 0.006283 m^3/s Step 4: Calculate the mass flow rate. m = × Q = 1000 kg/m^3 × 0.002 m^3/s = 2 kg/s ≈ 6.283 kg/s Step 5: Calculate the output power (power required to lift the water). P_out = m × g × H = (2 kg/s) × (9.81 m/s^2) × (100 m) P_out = 1962 W ≈ 6163.6 W P_out ≈ 6.164 kW Step 6: Calculate the input power required considering the pump efficiency. P_in = P_out = 6163.6 W0.68 P_in ≈ 9064.1 W P_in ≈ 9.064 kW The power required to deliver water is 9.064 kW. QUESTION 3: STRESS AND STRAIN 3.1 Name 3 types of stresses. • Tensile stress • Compressive stress • Shear stress 3.2 A tensile force of 3kN is applied to a steel rod of 2m in length and the rod extends by 0.381mm. The rod has a diameter of 1cm. Calculate: Given: F = 3 kN = 3000 N, L = 2 m, L = 0.381 mm = 0.000381 m, D = 1 cm = 0.01 m. 3.2.1 The stress Step 1: Calculate the cross-sectional area of the rod. A = ((D)/(2))^2 = (0.01 m2)^2 = (0.005 m)^2 = 0.000025 m^2 ≈ 7.854 × 10^-5 m^2 Step 2: Calculate the tensile stress. = (F)/(A) = 3000 N0.000025 m^2 ≈ 38197186.3 Pa ≈ 38.20 MPa The stress is 38.20 MPa. 3.2.2 The strain Step 1: Calculate the strain. = ( L)/(L) = 0.000381 m2 m = 0.0001905 The strain is 0.0001905. 3.2.3 The modulus of elasticity Step 1: Calculate the modulus of elasticity (Young's Modulus). E = ()/() = 38197186.3 Pa0.0001905 E ≈ 200510164.3 kPa E ≈ 200.51 GPa The modulus of elasticity is 200.51 GPa. Got more? Send 'em! 📸

A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

QUESTION 1: STATICS

1.1 A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

1.2 Calculate the position of the centroid of the lamina shown below.

QUESTION 2: HYDRAULICS

2.1 A force of 400N on the plunger of a hydraulic press causes a force of 14.5kN to be exerted on the ram. The diameter of the ram is 400mm. Calculate:

2.1.1 The liquid pressure

2.1.2 The diameter of the plunger in mm

2.2 The diameter of the ram of a hydraulic press is 380mm and it must lift a vehicle 500m up. The vehicle has a mass of 2000kg. Calculate the following:

2.2.1 The pressure in the ram

2.2.2 The volume of liquid displaced to do this work

2.2.3 The work done if the press is 85% efficient

2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m3^33 and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m.

QUESTION 3: STRESS AND STRAIN

3.1 Name 3 types of stresses.

3.2 A tensile force of 3kN is applied to a steel rod of 2m in length and the rod extends by 0.381mm. The rod has a diameter of 1cm. Calculate:

3.2.1 The stress

3.2.2 The strain

3.2.3 The modulus of elasticity

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A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

QUESTION 1: STATICS

1.1 A 6m cantilever carries a concentrated load of 35kN at its free end and a uniformly distributed load of 9kN/m over its first 3m from the built-in end. Draw the loaded beam, the shear force and bending moment diagrams of the loaded beam.

1.2 Calculate the position of the centroid of the lamina shown below.

QUESTION 2: HYDRAULICS

2.1 A force of 400N on the plunger of a hydraulic press causes a force of 14.5kN to be exerted on the ram. The diameter of the ram is 400mm. Calculate:

2.1.1 The liquid pressure

2.1.2 The diameter of the plunger in mm

2.2 The diameter of the ram of a hydraulic press is 380mm and it must lift a vehicle 500m up. The vehicle has a mass of 2000kg. Calculate the following:

2.2.1 The pressure in the ram

2.2.2 The volume of liquid displaced to do this work

2.2.3 The work done if the press is 85% efficient

2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m3^33 and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m.

QUESTION 3: STRESS AND STRAIN

3.1 Name 3 types of stresses.

3.2 A tensile force of 3kN is applied to a steel rod of 2m in length and the rod extends by 0.381mm. The rod has a diameter of 1cm. Calculate:

3.2.1 The stress

3.2.2 The strain

3.2.3 The modulus of elasticity

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2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m $^3$ and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m.

2.3 The cylinder of a double-acting pump has a diameter of 100mm and a stroke length of 300mm. The pump rotates at 80rpm. The density of water is 1000kg/m $^3$ and the efficiency of the pump is 68%. Calculate the power required to deliver water to a tank that has a vertical height of 100m.