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**Forces and Moments MET 2214 Ok. Lets get started.**

My name is Dr Simin Nasseri, a faculty member here at MET and I am going to teach this course MET ????, Course name. Statics (MET 2214) Prof. S. Nasseri

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Moments and Forces Part 2 Statics (MET 2214) Prof. S. Nasseri

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**Calculating the moment using rectangular components**

The moment of a force F about the axis passing through point O and perpendicular to the plane containing O and F can be expressed using the cross product: Mo = r x F The magnitude of the moment is the area shown below: Statics (MET 2214) Prof. S. Nasseri

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**Calculating the moment using rectangular components**

Mo = | r x F | = r F sinθ r = x i + y j + z k F = Fx i + Fy j + Fz k Statics (MET 2214) Prof. S. Nasseri

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Resultant Moment Statics (MET 2214) Prof. S. Nasseri

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Moments and Forces Part 3 Statics (MET 2214) Prof. S. Nasseri

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Moment about an axis Sometimes the moment about a point is known and you are supposed to calculate its component about an axis. To find the moment, consider the dot product of Mo and unit vector along axis a: O: any point on a-a Statics (MET 2214) Prof. S. Nasseri

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Moment about an axis You can also find the tangent force Fθ and then r x Fθ is the moment about aa: Statics (MET 2214) Prof. S. Nasseri

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Example 1 Force F causes a moment MO about point O. What is the component of MO along axis oy (My)? Statics (MET 2214) Prof. S. Nasseri

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**Scalar Analysis There are 2 methods to find My Scalar Analysis**

Vector Analysis (1) Scalar Analysis (first way): MO = (20)(0.5) = 10 N.m Imagine that we have found the direction of this moment (shown in figure) MO tends to turn the pipe around axis ob. The component of MO along the y-axis, My, tends to unscrew the pipe from the flange at O. Thus it is important to know its value. My = (3/5)(10) = 6 N.m Statics (MET 2214) Prof. S. Nasseri

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**Scalar Analysis Scalar Analysis (second way):**

To find My directly (not form MO) it is necessary to determine the moment-arm, knowing that the distance from F to the y-axis is 0.3m: My = (20)(0.3) = 6 N.m In general, If the line of action of a force F is perpendicular to any specific axis aa thus: Ma = F .da Statics (MET 2214) Prof. S. Nasseri

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Vector Analysis First, use the cross product formula to calculate the moment about O: MO = rA × F MO = (0.3i + 0.4j) × (-20k) MO = {-8i + 6j} N.m Then use the dot product of MO and the unit vector along y-axis to get My: My = MO . ua My = (-8i + 6j). (j) My = 6 N.m Statics (MET 2214) Prof. S. Nasseri

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Vector Analysis We can always combine the two previous equations that you saw into one: MO = rA × F yielding a scalar My = MO . Ua My = (rA × F) . ua Or My = ua . (rA × F) Which is called triple scalar product Statics (MET 2214) Prof. S. Nasseri

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**Remember the triple scalar product from Math section?!!**

Magnitude of moment M about axis aa’ Vector of moment M about axis aa’ Statics (MET 2214) Prof. S. Nasseri

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Example 2 Statics (MET 2214) Prof. S. Nasseri

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Example 2 Statics (MET 2214) Prof. S. Nasseri

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Example 2 Statics (MET 2214) Prof. S. Nasseri

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CE 201 - Statics Lecture 12. MOMENT OF A FORCE ABOUT A SPECIFIED AXIS Scalar Analysis Mo = (20 N) (0.5 m) = 10 N.m (F tends to turn about the Ob axis)

CE 201 - Statics Lecture 12. MOMENT OF A FORCE ABOUT A SPECIFIED AXIS Scalar Analysis Mo = (20 N) (0.5 m) = 10 N.m (F tends to turn about the Ob axis)

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