"CESE 656 – Advanced Mechanicsfor Structural EngineersCivil Structural EngineeringUniversity of Alabama at BirminghamShear Center Shear Center for Thin-walledOpen Nonsymmetric Sections2 Shear Center CESE 656 Example – Nonsymmetric Section? Example 8.2 in your textbook ? Shear Center is located by applying a shear force, V, in eachprincipal direction? When the plane of loads is along a principal axis, the NA isknown to be perpendicular to the plane of loads? Solving the problem this way, while conceptuallystraightforward tends to be geometrically complex? Let’s look at another way to solve this problem using a moreconvenient set of axes for the cross-section geometry andwithout needing to calculate I , I , or I x y xy3 Shear Center CESE 656 Example – Unequal Leg Channel UsingGeometric Axes? Locate the shear center for the unequal leg angle shown? As shown in the figure, the centroid is located 25 mm right of D and 40 mm above D2 2 400? ? 50 + 20 0? ? ( 25 ) ? ? = = 25 ??? 2 100 0 ? ? 2 2 400? ? 50 + 2 00? ? ( 1 00 ) ? ? = = 40 ??? 2 100 0 ? ? ? Note that although they are not needed, and? noting that t is small4 ? ? = 1,733,333 ? ?? 4 ? ? = 875,000 ? ?? 4 ? ? = -500,000 ? ??? 4 Shear Center CESE 656?? ?? ?? ??Example – Unequal Leg Channel UsingGeometric Axes? First assume that the x-axis is the neutral axis? Determine the first moment of the area at each critical location (A, B, C, D)? Since? = , the shear flow is proportional to the first moment of the area, Q? ? ? = ? = 0 (Free surfaces)? ? 3 ? ? = 50 ?? 4 ?? 60 ?? = 12,000 ? ?? 3 ? ? = ? + 4 ?? 60 ?? 30 ?? = 19,200 ? ?? ? 3 ? ? = 100 ?? 4 ?? 40 ?? = 16,000 ? ?? 3 ? ? = ? + 4 ?? 40 ?? 20 ?? = 19,200 ? ? ?? ? ? Shear flow is linear in elements parallel to the N.A.BA? Shear flow is parabolic in elements perpendicular to the N.A.BCAECED5 Shear Center CESE 656??Example – Unequal Leg Channel UsingGeometric Axes? Determine the shear force in each section? The total shear force in each section is proportional to the area under the distribution of the firstmoment of the area1 3 4 ? ? = 50 ?? 12,000 ? ? = 300,000 ? ?? ? 2 2 3 3 3 ? ? = 60 ?? 12,000 ? ? + 60 ?? 19,200 ? ? - 12,000 ? ?? ? 3 2 3 3 3 4 + 40 ?? 16,000 ? ? + 40 ?? 19,200 ? ? - 16,000 ? ? = 1,733,333 ? ?3 1 3 4 ? ? = 100 ?? 16,000 ? ? = 800,000 ? ? ? ? 2 ? ? = 1,733,333 = ?? ? B? ? = -800,000 + 300,000 = -500,000 = ?A? ? ? BCAECED6 Shear Center CESE 656 Example – Unequal Leg Channel UsingGeometric Axes? Determine the line of action of the V 1? ? = 1,733,333 ?? ? ? = 500,000 ?? ? ? = 300,000 ?? ? ? M caused by the internal shear flow = M caused by VD D ? ? = 300,000 100 = 1,733,333 ? ? ? = 17.31 ?? ( ?)? ? ? = 300,000 100 = 500,000 ? ? ? = 60.00 ?? ( ?)? ? Equation for the line of action of V is1 ? ? ? + = 1 ? ? = -3.46 7? + 601 7. 3 1 6 0. 0 0 BABVACCEED7 Shear Center CESE 656?? ??Example – Unequal Leg Channel UsingGeometric Axes? Next assume that the y-axis is the neutral axis? Determine the first moment of the area at each critical location (A, B, C, D)? Since? = , the shear flow is proportional to the first moment of the area, Q? ? ? = ? = 0 (Free surfaces)? ? ? ? = 0 (Since AB is symmetric about y)? 3 ? ? = 25 ?? 4 ?? 12.5 ?? = 1,250 ? ?? 3 ? ? = 100 ?? 4 ?? 25 ?? = 10,000 ? ?? 3 ? ? = 75 ?? 4 ?? 37.5 ?? = 11,250 ? ?? ? Shear flow is linear in elements parallel to the N.A.BA? Shear flow is parabolic in elements perpendicular to the N.A.BCF AECEGD8 Shear Center CESE 656??Example – Unequal Leg Channel UsingGeometric Axes? Determine the shear force in each section? The total shear force in each section is proportional to the area under the distribution of the firstmoment of the area2 3 4 ? ? = 50 ?? 1,250 ? ? = 41,667 ? ?? ? 3 1 3 4 ? ? = 100 ?? 10,000 ? ? = 500,000 ? ?? ? 2 2 3 3 ? ? = 75 ?? 11,250 ? ? + 25 ?? 10,000 ? ?? ? 3 2 3 3 4+ 25 ?? (11,250 ? ? - 10,000 ? ? ) = 833,333 ? ? 3 ? ? = -500,000 = ?? ? ? B? ? = 41,667 + 833,333 = 875,000 = ?? ? ABCF AECEGD9 Shear Center CESE 656 Example – Unequal Leg Channel UsingGeometric Axes? Determine the line of action of the V 2? ? = 500,000 ?? ? ? = 875,000 ?? ? ? = 41,667 ?? ? ? M caused by the internal shear flow = M caused by VD D ? ? = 41,667 100 ?? = -500,0 00? ? ? = -8.33 ?? ( ?)? ? ? = 41,667 100 ?? = -875,000 ? ? ? = -4.76 ?? ( ?)? ? Equation for the line of action of V is2 ? ? ? + = 1 ? ? = -0.57 1? - 4.76- 8. 3 3 - 4. 7 6 BABCF AECVEGD10 Shear Center CESE 656 "