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48 4G III. 3:52 تیلیجرام Cancel e12b4d57-fff5-4ab0-98d6-d29... Your Question H.W 8 What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m². Puncher 20mm 아 5 25 mm thick The resisting area is the shaded area along the perimeter and the shear force Vis equal to the punching force P. Step 1 of 2:) Given. Punched out 20 Diameter of the hole, de 20 mm 20 x 10 m. Thickness, t = 25 mm = 25 x 10°³ m 25 mm Your Answer Shear stress = 350 MN/m² 350x 10° N/m² 3.5 x 10° N/m². Stress is the ratio of force and area. Stress Area A is. Force Stress Area 1× A. A = dt = 3.14 x (20 x 10³ (25x10 ) = 0.0015708 m². Thus, punching force is, P=TA (3.5 x 10") x 0.0015708 549780 N = 549.8 kN. Punching force, P 549.8 kN Step 2 of 2:) Punching force, P-549.8 kN The step-by-step solution is attached below Refer to the explanation section. Good luck

48 4G III. 3:52 تیلیجرام Cancel e12b4d57-fff5-4ab0-98d6-d29... Your Question H.W 8 What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m². Puncher 20mm 아 5 25 mm thick The resisting area is the shaded area along the perimeter and the shear force Vis equal to the punching force P. Step 1 of 2:) Given. Punched out 20 Diameter of the hole, de 20 mm 20 x 10 m. Thickness, t = 25 mm = 25 x 10°³ m 25 mm Your Answer Shear stress = 350 MN/m² 350x 10° N/m² 3.5 x 10° N/m². Stress is the ratio of force and area. Stress Area A is. Force Stress Area 1× A. A = dt = 3.14 x (20 x 10³ (25x10 ) = 0.0015708 m². Thus, punching force is, P=TA (3.5 x 10") x 0.0015708 549780 N = 549.8 kN. Punching force, P 549.8 kN Step 2 of 2:) Punching force, P-549.8 kN The step-by-step solution is attached below Refer to the explanation section. Good luck

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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48 4G III. 3:52 تیلیجرام Cancel e12b4d57-fff5-4ab0-98d6-d29... Your Question H.W 8 What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m². Puncher 20mm 아 5 25 mm thick The resisting area is the shaded area along the perimeter and the shear force Vis equal to the punching force P. Step 1 of 2:) Given. Punched out 20 Diameter of the hole, de 20 mm 20 x 10 m. Thickness, t = 25 mm = 25 x 10°³ m 25 mm Your Answer Shear stress = 350 MN/m² 350x 10° N/m² 3.5 x 10° N/m². Stress is the ratio of force and area. Stress Area A is. Force Stress Area 1× A. A = dt = 3.14 x (20 x 10³ (25x10 ) = 0.0015708 m². Thus, punching force is, P=TA (3.5 x 10") x 0.0015708 549780 N = 549.8 kN. Punching force, P 549.8 kN Step 2 of 2:) Punching force, P-549.8 kN The step-by-step solution is attached below Refer to the explanation section. Good luck
Transcribed Image Text:48 4G III. 3:52 تیلیجرام Cancel e12b4d57-fff5-4ab0-98d6-d29... Your Question H.W 8 What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m². Puncher 20mm 아 5 25 mm thick The resisting area is the shaded area along the perimeter and the shear force Vis equal to the punching force P. Step 1 of 2:) Given. Punched out 20 Diameter of the hole, de 20 mm 20 x 10 m. Thickness, t = 25 mm = 25 x 10°³ m 25 mm Your Answer Shear stress = 350 MN/m² 350x 10° N/m² 3.5 x 10° N/m². Stress is the ratio of force and area. Stress Area A is. Force Stress Area 1× A. A = dt = 3.14 x (20 x 10³ (25x10 ) = 0.0015708 m². Thus, punching force is, P=TA (3.5 x 10") x 0.0015708 549780 N = 549.8 kN. Punching force, P 549.8 kN Step 2 of 2:) Punching force, P-549.8 kN The step-by-step solution is attached below Refer to the explanation section. Good luck
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