L7 - including notes

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1045

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Mechanical Engineering

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Oct 30, 2023

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ECOR1045 — STATICS LECTURE 7 - MOMENT ABOUT AN AXIS, COUPLES AND TRIPLE SCALAR PRODUCT e Mixed Triple Product ( Scalar —ffiP[e, Praol,u(f ) To find the moment of force F about y-axis using a vector analysis, first, we determine the moment of force about any point O on the y-axis applying M = # x F. Then, the component My along the y axis is the projection of Mo onto the y axis. Hence, it is calculated as; Myzj'(’f"xfi) Z \ Moment Axis The generalized expression can be written as - d - = - _LJ Ma=[ A2 + YT +AT] [T A A A A )\% —> (0 MpaA<A+$ a_( U Veeto— Y + hon —> corponent of oSN vector F), = (,ompa/\,mL ofi ~Horee VeeAo—

ECOR1045 — STATICS Example: Determine the magnitude of the moment that the force F exerts about the y-axis of the shaft. Solve the problem using a Cartesian vector approach and using a scalar approach. = A 1Y P e N o 7z K S loc. o R %60530: '/5«% N [6sin0= & IN @ZMJ = 13.86 (02 $in49 — g (0.0 Cosks) = .84 N ® \ Md e = x?) e = (02coss)T ~ 0.05T D25 e =(0 44T ~0.057- OALLE) 1 o) Fo(3ut+9T ) N My=g-(5xF) = | 2 L 0 014 =005 —044] 1384 O & > —4 [ 04uaxg ~ 014 x13.8 ) '—%0.526 Nm 3

ECOR1045 — STATICS Example: Determine the magnitude of the moment of the force F about the OA axis if: F =300N i— 200N j + 150N k }I 00 -0 45p > 0.4 ( 0.4xI50 — (zooxo.Z)) — 0.8 0.5x150 + @‘ZXSOa) 0 5 -32 Nw My =72 N

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ECOR1045 - STATICS COUPLES A couple is a special case of moments. A couple is defined as two parallel forces that have the same magnitude but opposite directions and are separated by a perpendicular distance d. T&l + : Fhs e 354 d : % wF It does not produce any translation, but it does create rotation. The resultant force of a couple is zero, the only effect of a couple is to produce a rotation or tendency of rotation. The moment produced by a couple is called a couple moment. Scalar Formulation: The moment of a couple M is defined as having a magnitude of M =Fd where, F is the magnitude of one of the forces and d is the perpendicular distance or moment arm between the forces. The direction of the couple moment is determined by the right-hand rule. In all cases, M will act perpendicular to the plane containing these forces. Vector Formulation: The moment of a couple M can also be defined by the vector cross product applying cross product. X 1 =y X o Example: The forces that two hands apply to turn a steering wheel are often (or should be) a couple. Each hand grips the wheel at points on opposite sides of the shaft. When they apply a force that is equal in magnitude yet opposite in direction the wheel rotates. If both hands applied a force in the same direction, the sum of the moments would equal to zero and to cause a translation with a ? magnitude of 2F.

ECOR1045 — STATICS The magnitude of the moment of a couple is the same for all points in the plane of the couple. ARM =d { d/2 i df { | 1 A i 2 F F MefFd &Moo Fld) + Fly) = Fd i /45 OOM sec/ '('La mnjnfil«a& 01[ —FLL cau{)/Q 1% Yflalbpm-/' of Hha rz#ej‘aog lcation (fa?#) [+ 1 M; aTM v £ | Important things to remember about COUPLES: G2Mo= —Fx + F(dax) = FA e The resultant of a number of couples is their algebraic sum. (A couple is also a vector whether it is presented as two forces or a single pure moment!) e A couple CANNOT be put in equilibrium by a single force! If a single force is added to the system that balances the sum of the moments, one of the other two equations of equilibrium will not be satisfied.

ECOR1045 — STATICS Resolution of a force into a force and a couple: If a force is moved off its line of action, a couple must be added to the force system so that the new system generates the same moments as the old system. The resultant of a force and couple system: \/ Since couple moments are vectors, their resultant can be determined by vector addition. MR=M1+M2 If we have more than two couples act on the body, we may generalize this concept and write the vector resultant. My = Z(? e 7’3

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ECOR1045 — STATICS Example: Two couples act on the frame. If d = 4 ft, determine the resultant couple moment. Compute the resultant by resolving each force into x and y components by: a) Finding the moment of each couple; and, b) Summing the moment of all force components about point B. | 31ft 41t 401b ..... (0 4 =4 Lt] 305 2 (5 6 /\ 0 Yo’ 60 Ib ,,,,, (B (tzMy = 430) = 43(3) +40e0s30(5) — ocor30(Y) =7 »%.9 b ¢

ECOR1045 — STATICS Simplification of a Force and Couple System! Sometimes it is convenient to reduce a system of forces and couple moments acting on a body to a simpler form by replacing it with an equivalent system consisting of a single resultant force acting at a specific point and a resultant couple moment. In other words, a system of several forces and couple moments acting on a body can be reduced to an equivalent single resultant force acting at a point O and a resultant couple moment. Fp = F Equation 1 R Z Equation 2 (Mg)o =XM,+3¥M Equation -1 states that the resultant force of the system is equivalent to the sum of all the forces. Equation -2 states that the resultant couple moment of the system is equivalent to the sum of all the couple moments ), M plus the moments of all the forces ), M, about the point O.

ECOR1045 — STATICS Example: Three hikers are shown crossing a footbridge. Knowing that the weights of the hikers at points C, D and E are 800 N, 700 N and 540 N respectively. Determine: a) The horizontal distance from A to the line of action of the resultant of the three weights when a=1.1m b) The value of a so that the loads on the bridge supports A and B are equal. - Gm a) 1S 1A \Lf\L wiidh.s o = 540 800+ 00 ": X %: . Fo= 2090 N | . /fiz"i = 800((5)-1‘ 7100(2.6)-7L '31{0(42@”) Lok X =7, éO‘?? n sooN FOON F4oN .({FZM,,C_—_ ) b/ ' . _800)(13"_?00({\?4'4),-. f 1 —swpsiesa)+ i), A=A M 10

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ECOR1045 — STATICS Simplification of 2D and 3D Concurrent Force Systems * Concurrent force systems have their lines of action intersecting at a point, in this case O. » These forces can be replaced by a resultant force with a line of action through point O. Fy F; F Simplification of 2D Coplanar Force Systems » Coplanar force systems consist of forces with their lines of action in the same plane, but not concurrent. » The coplanar force systems can be replaced by a resultant force with a line of action in the same plane, and the sum of moments of the forces. ¥, F, Simplification of Parallel Force Systems * Consider a parallel force system with all forces parallel to the z-axis: * The resultant force at point O must also be parallel to z-axis. 11

ECOR1045 — STATICS * The resultant moment (MR), lies in a perpendicular plane to z-axis. * The resultant moment and resultant force can be replaced by the resultant force acting at a point d from O. &N Lot 3 ] * 12