J_-_Moment_about_an_Axis

J_-_Moment_about_an_Axis - FORCE SYSTEM RESULTANTS MOMENT...

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Unformatted text preview: FORCE SYSTEM RESULTANTS MOMENT ABOUT AN AXIS MOMENT LEARNING OBJECTIVES LEARNING Be able to determine the moment of a Be force about a specified axis using scalar and vector methods and PRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurements Trigonometry concepts Vector concepts Rectangular components concepts Moment concept Cross product and dot product MOMENT OF A FORCE ABOUT A SPECIFIED AXIS MOMENT (Scalar Method) M ob = F ⋅ db Where Mob is the moment of the force F about the Ob axis and db is the perpendicular or the shortest distance from the force to the axis EXAMPLE The Moment arm about the y-axis is 0.3 m, My = 6 N-m The Moment arm about the Ob-axis is 0.5 m, Mob = 10 N-m The Moment arm about the x-axis, is 0.4 m, Mx = -8 N-m MOMENT OF A FORCE ABOUT A SPECIFIED AXIS MOMENT (Vector Method) M OB = rOA × F ( M Ob = u Ob • rOA × F ( ) ) Where uOb is the unit vector in the direction of the Ob-axis M Ob i = ( u bx i + u by j + u bz k ) • rx Fx j ry Fy k u bx rz = r x Fz Fx u by ry Fy u bz rz Fz EXAMPLE 1 rOb =− i +3 j 4 u Ob u Ob rOb − 4i + 3 j = = 2 2 rOb ( 4) + 3 −4 3 = i+ j 5 5 M Ob = u Ob • ( r × F ) M Ob uobx = rX FX uoby rY FY uobz rZ FZ 4 5 = 0.3 0 3 5 0.4 0 0 0 - 20 EXAMPLE 1 M Ob 4 5 = 0.3 0 3 5 0.4 0 0 0 - 20 M Ob 4 3 = − ( 0.4 × −20 − 0 × 0 ) − ( 0.3 × −20 − 0 × 0 ) = 10 N − m 5 5 EXAMPLE 2 EXAMPLE Determine the moment of the force F about the Determine Oa axis using vector method Oa SOLUTION 1 SOLUTION m ar Determine the directional and Determine the unit vectors in the direction of the Oa axis : direction rOa = 4 j + 3k u Oa rOa = rOa Determine the moment arm vector u Oa 4 3 = = j+ k 2 2 5 5 4 +3 r 4 j + 3k rarm = i − 2 j + 6k SOLUTION 1 – continued continued Determine the moment Determine about the Oa axis about M Oa = u Oa • (rarm × F ) M Oa ux = rx Fx 0 =1 50 uy ry Fy 4 5 −2 − 20 uz rz Fz 3 5 6 20 arm r 4 3 = 0[(−2)(20) − (−20)(6)] − [(1)(20) − (6)(50)] + [(1)(−20) − (−2)(50)] 5 5 = 0 + 224 + 48 = 272 N - m SOLUTION 1 – continued continued arm In vector form Moa can be obtained as Moa(uoa) r M Oa M Oa 4 3 = u Oa ( M oa ) = 272 0i + j + k 5 5 = 217.6 j + 163.2k EXAMPLE 2 EXAMPLE Determine the moment of the given 20 lb Determine force about the hinge using vector method force SOLUTION 2 SOLUTION Determine the unit directional Determine vector in the direction of hinge : vector uhing e u hinge = i Determine the force vector using the Determine following 3 steps: following 1. Find the position vector rBA Find 2. Find the unit vector uBA Find 3. Find the force vector F = FuBA rBA = 3i − 3 cos 20° j + ( 4 − 3 sin 20°) k = 3i − 2.819 j + 2.974k u BA = 3i − 2.819 j + 2.974k 3 + ( − 2.819 ) + 2.974 2 2 2 = 0.5907i − 0.5551j + 0.5856k F = F ⋅ u BA = ( 20)( 0.5907i − 0.5551j + 0.5856k ) F =11.81i −11.10 j +11.71k lb SOLUTION 2 – continued continued SOLUTION Determine the moment arm or the position vector rOB or uhing e OB rOB = 3 cos 20° j + 3 sin 20°k Determine the moment at the hinge Mhinge r M hinge = u hinge • (rarm × F) = ux rx Fx uy ry Fy uz rz = Fz 1 0 11.81 0 0 3 cos 20° 3 sin 20° = 44.4 lb - ft - 11.10 11.71 ALTERNATIVE SOLUTION 2 ALTERNATIVE Determine the unit vector Determine in the direction of hinge : in uhing e u hinge =i As before, the force vector F As is is F = {11.81i −11.10 j +11.71k} lb ALTERNATIVE SOLUTION 2 – continued continued Determine the position vector rarm from O to A vector arm rarm = rOA = 3i + 4k Determine Mhinge r M hinge = u hinge • (rarm × F ) = ux rx Fx uy ry Fy uz 1 0 0 rz = 3 0 4 = 44.4 lb - ft Fz 11.81 −11.10 11.71 ...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.

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