Question

A column with a wide-flange section has a flange width b = 400 mm , height h = 400 mm , web thickness tw = 13 mm , and flange thickness tf = 21 mm (Figure 1). Calculate the stresses at a point 65 mm above the neutral axis if the section supports a tensile normal force N = 3 kN at the centroid, shear force V = 7.4 kN , and bending moment M = 4 kN⋅m as shown (Figure 2).
11 Review Learning Goal: To calculate the normal and shear stresses at a point on the cross section of a column. The state of stress at a point is a description of the normal and shear stresses at that point. The normal stresses are generally due to both internal normal force and internal bending moment. The net result can be obtained using the principle of superposition as long as the deflections remain small and the response is elastic. A column with a wide-flange section has a flange width b = 400 mm, height h = 400 mm, web thickness tv = 13 mm, and flange thickness to = 21 mm (Figure 1). Calculate the stresses at a point 65 mm above the neutral axis if the section supports a tensile normal force N = 3 kN at the centroid, shear force V = 7.4 kN, and bending moment M = 4 kNm as shown (Figure 2). Part A - Normal stress Calculate the normal stress at the point due to the internal normal force on the section. Express your answer with appropriate units to three significant figures. View Available Hint(s) normal = 0.140 MPa Submit Previous Answers Correct Part B - Shear stress Calculate the magnitude of the shear stress at the point due to the internal shear on the section. Express your answer with appropriate units to three significant figures. View Available Hint(s) Figure < 1 of 2 НА ? y I= 1.324 MPa H Submit Previous Answers X Incorrect; Try Again; 5 attempts remaining Activate Windows Go to Settings to activate Windows, Part C - Combined normal stress

Answer 1

Solution: 21 mm Akismam 65mm 400 mm N-A 358 mm M 400mm Given. N = 3KN, N=7.4 KN, M=4KN.m Step 1: Area of section At = 21454 mm 21 x 400 x 2 + 358% 13 Normal zle 3x103 N/mm² (MPa) 21454 Normal = 0.1398 mpa (Tensile) 4K Nim Step 2: Bending moment a At point os mm above Neutral anis, e compressive Bending stress will occur. = MY so, 3 I = 400x (400) - 2x 193.5x (3 583 12 12 I = 653.6168106 mmy

Here, y = 65 mm bending (4x108) 65 MPa 653.616 x 106 -0.3978 mpa So, combined Nomal stress 6 Normant bending 0.1398 to.3978 = 0.5376 MPa. So, combined normal stress in 0.5376 MPa. Steps shear stress TE VAT) At point "A" I waz4 kN] لا At point 65 mm abone Neutral aniq Ау 121 X 400)x (189.5) + 13x (200-21-65) * (200-21065 +65) A ý = 1772804 mm b=13 m 0 177 2604 653-6164 108 x 13 2= losuu mpa So, Shear stress at point 65 mm above Neutral anig is losty mpa. sor a = 704x10²