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The beam shown in the figure below (Figure 1) is made of elastic perfectly plastic material. Take a = 50 mm and Oy = 280 MPa Part A Determine the maximum elastic moment that can be applied to the cross section. Express your answer to three significant figures and include appropriate units. μΑ ? My = Value Units Submit Request Answer Part B Figure < 1 of 1 > Determine the maximum plastic moment that can be applied to the cross section. Express your answer to four significant figures and include appropriate units. HA ? Mp = Value Units Submit Request Answer < Return to Assignment Provide Feedback
The beam shown in the figure below (Figure 1) is made of elastic perfectly plastic material. Take a = 50 mm and Oy = 280 MPa Part A Determine the maximum elastic moment that can be applied to the cross section. Express your answer to three significant figures and include appropriate units. μΑ ? My = Value Units Submit Request Answer Part B Figure < 1 of 1 > Determine the maximum plastic moment that can be applied to the cross section. Express your answer to four significant figures and include appropriate units. HA ? Mp = Value Units Submit Request Answer < Return to Assignment Provide Feedback
Part A: Themaximum elastic momentthat can be applied to the cross section is 110000 N·m.Part B: The maximum plastic moment that can be applied to the cross section is 225000 N·m.Part A: Maximum elastic momentThe maximum elastic moment that can be applied to thecross sectionis given by the equation:My = σy * Iwhere:My is the maximum elastic moment (N·m)σy is theyield stressof the material (MPa)I is the moment of inertia of the cross section (m4)The moment of inertia of the cross section is given by:I = 1⁄12 * a² * bwhere:a is the width of the cross section (mm)b is thedepthof the cross section (mm)In this case, we have:σy = 280 MPaa = 50 mmb = 100 mmPlugging these values into the equations, we get:My = 280 MPa * 1⁄12 * 50 mm² * 100 mm = 110000 N·mTherefore, the maximum elastic moment that can be applied to the cross section is 110000 N·m.Part B:Maximum plastic momentThe maximum plastic moment that can be applied to the cross section is given by the equation:Mp = σy * Zwhere:Mp is the maximum plastic moment (N·m)σy is the yield stress of the material (MPa)Z is the plastic section modulus of the cross section (m3)The plastic sectionmodulusof the cross section is given by:Z = b * d² / 6where:b is the width of the cross section (mm)d is the depth of the cross section (mm)In this case, we have:σy = 280 MPaa = 50 mmb = 100 mmPlugging these values into the equations, we get:Mp = 280 MPa * 100 mm * 50 mm² / 6 = 225000 N·mTherefore, the maximum plastic moment that can be applied to the cross section is 225000 N·m.To know more aboutmaximum elastic momentherebrainly.com/question/13260059#SPJ4...