Chapter Objectives Concept of moment of a force in two and three dimensions Method for finding the moment of a force about a specified axis. Define the moment of a couple Determine the resultants of non-concurrent force systems Reduce a simple distributed loading to a resultant force having a specified location

Chapter Outline Moment of a Force – Scalar Formation Cross Product Moment of Force – Vector Formulation Principle of Moments Moment of a Force about a Specified Axis Moment of a Couple Simplification of a Force and Couple System Further Simplification of a Force and Couple System Reduction of a Simple Distributed Loading Moment is a combination of a physical quantity and a distance. = the product of quantity (as a force) and the distance to a particular axis or point.

4.1 Moment of a Force – Scalar Formation Moment of a force about a specified point or axis is a quantity indicating an attempt to rotate an object about the point or axis.

4.1 Moment of a Force – Scalar Formation Magnitude Magnitude of MO (Moment about O from F) is d = Perpendicular distance measured from O to a line passing the force Direction Find the direction by right hand rule

4.1 Moment of a Force – Scalar Formation Resultant Moment MRo = The sum of moments from all forces

Example 4.1 For each case, determine the moment of the force about point O. Line of action is extended as a dashed line to establish moment arm d. Tendency to rotate is indicated and the orbit is shown as a curl.

Solution

Example 4.2 Determine the resultant moment of the forces about point O. Assume positive moment as counter-clockwise;

4.2 Cross Product The cross product of the vectors A and B become the vector C C = A X B Magnitude 0≤ θ ≤ 180°, θ is the angle between A and B C = AB sinθ

4.2 Cross Product Direction Vector C is perpendicular to the plane containting Vector A and B according to the right hand rule (sweep from A to B and the direction is indicated by the thumb) Vector C can be writing in a vector form as C = A X B = (AB sin θ)uC uC is a unit vector of vector C

4.2 Cross Product B X A = -C Laws of Operations 1. Commutative law not applied A X B ≠ B X A But A X B = - (B X A) Cross product B X A give a vector that is opposite direction to vector C B X A = -C

4.2 Cross Product Laws of Operations 2. Multiplication by a Scalar a( A X B ) = (aA) X B = A X (aB) = ( A X B )a 3. Distributive Law A X (B + D) = (A X B) + (A X D) And (B + D) X A = (B X A) + (D X A)

4.2 Cross Product Cartesian Vector Formulation i j k

4.2 Cross Product Cartesian Vector Formulation A more compact determinant in the form as

4.3 Moment of Force - Vector Formulation Moment of a force F about O can be found by a cross product MO = r X F Magnitude cross product, MO = rF sin θ Consider if F or r can be moved along a line of action (sliding vector) และ d = r sinθ, MO = rF sinθ = F (rsinθ) = Fd Direction Direction of MO can be found by the right hand rule *Note: - Sweep r to F and the thumb indiction the direction of the moment MO

4.3 Moment of Force - Vector Formulation Principle of Transmissibility For a a force F acting at A, the moment about O is MO = rA x F F can be moved along a line of action MO = r1 X F = r2 X F = r3 X F

4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation For force expressed in Cartesian form, หาค่า determinant, MO = (ryFz – rzFy)i + (rzFx - rxFz)j + (rxFy – ryFx)k

4.3 Moment of Force - Vector Formulation The resultant moment of forces The resultant moment about O is the sum of all moments about O MRo = ∑(r x F)

Example 4.4 Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector. A (0, 5, 0) B (4, 5, -2)

Solution Position vectors are directed from point O to each force as shown. These vectors are The resultant moment about O is

4.4 Principles of Moments Varignon’s Theorem “The moment of a resultant force is equivalent to the sum of moment from the compenent forces From F = F1 + F2

Example 4.5 Determine the moment of the force about point O.

Solution The moment arm d can be found from trigonometry, Since the force tends to rotate or orbit clockwise about point O, the moment is directed into the page. or

4.5 Moment of a Force about a Specified Axis Moment of a force about an axis is perpendicular to the plane containing the position vector and the force The component moment about a specified axis can be found by Scalar notation or vector Cartesian vector e.g. My from Mo

4.5 Moment of a Force about a Specified Axis Using scalar notation From the right hand rule My is pointed in y direction For any axis a The force will not generate a moment if the force is parallel to the specified axis

4.5 Moment of a Force about a Specified Axis Using vector Cartesian notation Find the moment about O The component in y can be found from a projection on the axis y For any axis a

4.5 Moment of a Force about a Specified Axis For a vector Cartesian notation is the component of a unit vector is the component of a position vector r (โมเมนต์รอบแกนที่สนใจ ไม่ใช่รอบจุดที่สนใจ)

4.5 Moment of a Force about a Specified Axis In determinant form, If the result is positive, the direction of a moment is in the same direction as the specified axis Then

Example 4.8 Determine the moment produced by the force F which tends to rotate the rod about the AB axis. A (0, 0, 0) B (0.4, 0.2 0)

Solution Unit vector defines the direction of the AB axis of the rod, where rC or rD can be used For simplicity, choose rD The force is

Solution

Check for different position vector r Using FD F (-0.8, -0.4 ,0) Using ED D (0.6,0,0) E (0.8, 0.4 ,0)

4.6 Moment of a Couple Couple The forces are parallel Identical magnitude but opposite direction The distance between the forces is d The resultant force = 0 Attempt to rotate in a direction The moment of a couple is a sum of moments about a specified axis

4.6 Moment of a Couple Moment of a couple is a Free vector since it only depends on r, hence it can be applied anywhere and the result will be the same (This is different to a moment of a force as the result will be different for difference specified points or axes)

4.6 Moment of a Couple Magnitude of couple moment M = Fd Direction and sense are determined by right hand rule M acts perpendicular to plane containing the forces

4.6 Moment of a Couple In a vector notation For couple moment, M = r X F If consider the moment about A, moment from –F will be zero

4.6 Moment of a Couple Equivalent Couples Two couple moments are equivalent if they are equal Both couple moments are in the same planes

4.6 Moment of a Couple The resultant moment of couples Couple moments are free vector and can then be added using vector addition MR = M1 + M2 In general, MR = ∑(r X F)

Example 4.12 Determine the couple moment acting on the pipe. Segment AB is directed 30° below the x–y plane.

SOLUTION I (VECTOR ANALYSIS) Take moment about point O, Take moment about point A

SOLUTION II (SCALAR ANALYSIS) Take moment about point A or B, Apply right hand rule, M acts in the –j direction

4.7 Simplification of a Force and Couple System Equivalent force and moment will give the same effect as when applying the original force and moment Original system + Opposite actions Equivalent system Principle of Transmissibility

4.7 Simplification of a Force and Couple System Consider F1 F2 and couple moment M Can move the forces to O and then generate additionl couples about MO as M1=r1xF1 and M2=r2xF2 The equivalent force about O and the result couple moments can be written as For 2D in x–y plane and the couple moment perpendicular to this plane,

4.7 Simplification of a Force and Couple System Procedure for Analysis Define origin at O that point to the interested direction Sum forces Sum moments

Example 4.16 A structural member is subjected to a couple moment M and forces F1 and F2. Replace this system with an equivalent resultant force and couple moment acting at its base, point O.

Solution Express the forces and couple moments as Cartesian vectors.

Solution Force Summation.

4.8 Further Simplification of a Force and Couple System For equivalent force and moment, about point O, we can find one equivalent force and one moment O (FR) and (MR)O Can further simplified to have only one force if FR and (MR)O are perpendicular to each other This include Forces with a common point Coplanar forces Parallel forces

4.8 Further Simplification of a Force and Couple System Concurrent Force System All forces have a commom point O There is no moment, hence we can find an equivalent force

4.8 Further Simplification of a Force and Couple System Coplanar Force System All forces are in the same plane Moments of each force are perpendicular to the plan The resultant moment can be found by moving the resultant forces by d

4.8 Further Simplification of a Force and Couple System Parallel Force System ตัวอย่าง แรงทั้งหมดขนานกับแกน z แรงลัพท์ก็ขนานกับแกนนี้ โมเมนต์ของแรงเหล่านั้น และโมเมนต์ลัพท์วางอยู่บนระนาบ โมเมนต์ลัพท์สามารถแทนด้วยแรงที่เลื่อนตั้งฉากออกไปด้วยระยะ d แนวของ FR และ (MR)O ตั้งฉากกัน

4.8 Further Simplification of a Force and Couple System Reduction to a Wrench โดยทั่วไปแนวของ FR และ (MR)O ไม่ตั้งฉากกัน (ลดรูปต่อไม่ได้) (a) แตก (MR)O ออกเป็นแนวตั้งฉากและขนานกับ FR ได้ (M┴ , M ║ ) (b) แทน M┴ ได้ด้วยการย้าย FR ไปจุด P ด้วยระยะ d (c) ย้าย M ║ ไปจุด P ได้เนื่องจากเป็น Free vector FR และ M ║ พยายามที่จะเลื่อนและหมุนวัตถุรอบแกน เรียกว่า Wrench หรือ Screw

Example 4.18 The jib crane is subjected to three coplanar forces. Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC. 1.75 kN

Solution Force Summation

Solution For magnitude of resultant force, For direction of resultant force, Assume line of FR (assume y or x)

Solution Moment Summation  Summation of moments about point A,

Solution Moment Summation  Principle of Transmissibility

4.9 Reduction of a Simple Distributed Loading For large areas, they are modelled as distributed loadings It is the force per unit area, which is equivalent to pressure Unit Pascal (Pa): 1 Pa = 1N/m2 Uniform Loading Along a Single Axis This is the most common problem

4.9 Reduction of a Simple Distributed Loading Magnitude of Resultant Force For an infinitestimal dF acting on an infinitestimal dA For an entire length L, the resultant force is the integral over the length Essentially, the resultant force is the area under the graph

4.9 Reduction of a Simple Distributed Loading Location of Resultant Force (for Centroid ) MR = ∑MO dF causes the moment xdF = x w(x) dx about O The the sum of moment, Solving for

Ex. Determine the magnitude and location of the equivalent resultant force acting on the shaft. Solution For the small differential area element, For resultant force

Solution For location of line of action, Checking,