Statics Chapter Four Moment By Laith Batarseh Home Next Previous End.

Statics Chapter Four Moment By Laith Batarseh Home Next Previous End

Moment Scalar The acting force Moment arm Statics Definition
Moment is defined as the tendency of a body lies under force to rotate about a point not on the line of the action of that force (i.e. there is a distance between the force and the rotation point ) Moment is a vector quantity Description Moment depends on two variables: The acting force Home Next Previous End Moment arm

Moment Scalar Force Arm Tendency to rotate Statics Description Home
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Moment Scalar Statics Tendency for rotation Home Next Previous End

Moment Scalar F D Statics Magnitude Moment magnitude (M) = F.D Home
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Moment Scalar Statics Direction Home Next Previous End

Moment Scalar Statics Solving procedures
1. Define the magnitudes of force (F) and arm (D) 2. Assume the positive direction (eg. Counter clock wise) 3. Find the magnitude of moment (M) as F.D 4. Give the moment the correct sign according to the tendency for rotation Home Next Previous End

Find the moment caused by the following forces about point O
Moment Scalar Statics Example [1] Find the moment caused by the following forces about point O 100 N 0.5m 2m (a) O 100 N 0.5m 2m (b) O Home Next Previous End

Assume the CCW direction is the positive direction.
Moment Scalar Statics Example [1] Assume the CCW direction is the positive direction. 100 N 0.5m 2m (a) O 100 N 0.5m 2m (b) O Branch (a) Mo = F.d = -(100N)(0.5m) Mo=-50 N.m=50N.m CW + Home Next Previous End Branch (b) Mo=F.d = (100N)(2m) Mo=200 N.m CCW +

Principle Of Moments Statics Principle of Moments
some times called Vrigonon’s theorem (Vrigonon is French mathematician ). State that the moment of a force about a point equals the summation of the moments created by the force components In two dimensional problems: the magnitude is found as M = F.d and the direction is found by the right hand rule In three dimensional problems: the moment vector is found by M =rxf and the direction is determined by the vector notation (ie. i,j and k directions) Home Next Previous End

Principle Of Moments Statics Example [1] Find the moment caused by the following forces about point O Home Next Previous End

Principle Of Moments + +
Statics Example [1] + Mo,1 = 100 sin(30) (10) = 500 N.m + Mo,2 =- 100 cos(30) (5) =- 433N.m M = Mo,1+Mo,2= =67N CCW Home Next Previous End

Principle Of Moments 100 cos(40) 120 sin(60) 100 sin(40) 120 cos(60) +
Statics Example [2] 1. Force analysis 100 cos(40) 1.2 m 0.3 m 100 sin(40) 120 cos(60) 120 sin(60) 2. Moment calculations + ∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW Home Next Previous End

Moment Scalar F1 d1 d2 d3 F3 F2 + Statics Moment resultant
Mo = ∑Mo = M1 + M2 – M3 = F1d1+F2d2 – F3d3 + Home Next Previous End

Moment Scalar Statics Example [2] Find the moment caused by the following forces about point O 2m 3m 5m 1m 30o 100 N 50 N 60 N 75 N O Home Next Previous End

Moment Scalar O + Statics Example [2] 2m 3m 5m 1m 30o 100 N 50 N 60 N
Home Next Previous End

Moment Scalar Statics Exercise Find the moment caused by the following forces about point O 100 N 300 N 5m 2m 45o 30o O 0.3m Home Next Previous End

Moment Scalar O + Statics Exercise 100 N 300 N 5m 2m 0.3m
300 sin (45)N 300 cos (45)N 100sin (30)N 100cos (30)N + Home Next Previous End

Moment vector Statics Cross product
Cross product is a mathematical operation can be done on vectors Cross product for one time is done for two vectors The cross product of two vectors is a vector perpendicular to the plane of A and B The notation of vector A cross vector B is: C = AxB where C is the resultant vector from the cross product the vector C can be represented as : C =CUc where Uc is a unit vector in a direction perpendicular to the plane that contains both A and B. The value of the scalar quantity C is given as : C=A.B.sin(ϕ) where ϕ is the angle between A and B. The cross product is controlled by the right-hand rule Home Next Previous End

Moment vector Uc ϕ Statics Graphical representation Home Previous Next
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z K=ixj j=kxi y i=jxk x Moment vector Statics
Cartesian vector formulation i=jxk j=kxi K=ixj y x z Home Next Previous End

Moment vector A = Ax i + Ay j + Az k B = Bx i + By j + Bz k
Statics Cartesian vector formulation A = Ax i + Ay j + Az k B = Bx i + By j + Bz k AxB=(Ay Bz -Az By )i-(Ax Bz - Az Bx )j + (Ax By - Ay Bx )k Home Next Previous End

Best for three dimensional problems
Moment vector Statics Moment – vector formulation F r θ d M O x y Mo = rxF Magnitude: Mo = rFsin(θ) = Fd Direction: perpendicular to x-y plane (z-direction) Matrix notation: Resultant moment:MRo = ∑(rxF) Best for three dimensional problems Home Next Previous End

Moment vector Statics Example [1] Find the moment caused by the following forces about point O F = [5i + 10j + 6k]N z x y 3m 2m 1m O Home Next Previous End

Moment vector Statics Example [1]
1. Formulate the position vector (r) : r = 3i+2j+1k 2. Find the moment vector (Mo) by matrix notation r = 3i+2j+1k F = [5i + 10j + 6k]N Home Next Previous End

Moment vector Statics Example [2] Find the moment caused by the following forces about point O x y z O F = [-5i + 5j -5k]N Home Next Previous End

Moment vector Statics Example [2]
1. Formulate the position vector (r) : r= 15i + 10j +6k 2. Find the moment vector (Mo) by matrix notation Home Next Previous End

Moment is a vector can be found by cross product and matrix notation
Moment vector Statics Summary Moment is a vector can be found by cross product and matrix notation Matrix notation: Home Next Previous End

Principle Of Moments Statics Principle of Moments
some times called Vrigonon’s theorem (Vrigonon is French mathematician ). State that the moment of a force about a point equals the summation of the moments created by the force components In two dimensional problems: the magnitude is found as M = F.d and the direction is found by the right hand rule In three dimensional problems: the moment vector is found by M =rxf and the direction is determined by the vector notation (ie. i,j and k directions) Home Next Previous End

Principle Of Moments + +

Principle Of Moments 100N 120N
Statics Example [2] The following is a gate. In which direction this gate will rotate? 100N 120N 1.2 m 0.3 m 60o 40o Home Next Previous End

Principle Of Moments 100 cos(40) 120 sin(60) 100 sin(40) 120 cos(60) +
Statics Example [2] 1. Force analysis 100 cos(40) 1.2 m 0.3 m 100 sin(40) 120 cos(60) 120 sin(60) 2. Moment calculations + ∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW Home Next Previous End

Principle Of Moments Statics Example [2] r1= 15i + 10j +6k F1 = [5j]N
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Moment of force about axis
Statics In many real cases, the force tendency to rotate is about a specified axis. Example : Moment components: Mo,1 = (100)(10) (about y-axis) Mo,2 = (100)(15) (about x-axis) Mo,3 = (about z-axis) Home Next Previous End

Statics Magnitude Scalar analysis: M = F.d Vector analysis: Where: ua is a unit vector defining the direction of a-axis and given as ua = ua,x i + ua,y j + ua,z k To find the moment vector (Ma): Ma. ua Home Next Previous End

Statics Example [2] Determine the magnitude of the moments of the force F about the x, y, and z axes. Solve the problem using a Cartesian vector approach using a scalar approach. Home Next Previous End

Statics Example [2] using a Cartesian vector approach rAB = {5i + 4j -3k} m For the axes: x, y and z the unit vectors are i, j and k respectively. Mx = i . (rAB x F) My = j . (rAB x F) Mz = k . (rAB x F) Home Next Previous End

Statics Example [2] using a scalar approach Mx = ∑Mx = 10(3) – 4(4) = 14 N.m My = ∑My = -5(3) + 4(5) = 5 N.m Mz = ∑Mz = -5(4) + 10(5) = 30N.m Home Next Previous End

Moment of couple F d -F Statics Definition
Couple is the moment generated by two forces has the same magnitude and opposite direction. F -F d Home Next Previous End

Moment of couple M F d -F Statics Scalar analysis
This analysis is considered for 2-D problems The magnitude of moment is found by: M=F.d where F is the force magnitude. the direction of couple moment is perpendicular to the plain that contain the F and d and it is found by the right hand rule. -F F d M Home Next Previous End

M1 = F1 . d1 M2 = F2 . d2 Moment of couple F1 d1 F2 d2 -F2 -F1 Statics
Example on scalar analysis F1 d1 F2 M1 = F1 . d1 M2 = F2 . d2 d2 -F2 -F1 Home Next Previous End

Moment of couple F -F O r rA rB A B
Statics Vector analysis This analysis is considered for 3-D problems The moment vector is found by: M=F x r where r is the position vector directed between the forces F and –F Note that the moment vector is dependent on the position vector directed between the forces F and –F (r). Derivation: M = rB x F + rA x –F = (rB - rA) x F But: (rB - rA) = r Then: M = rx F F -F O r rA rB A B Home Next Previous End

Moment of couple Statics Example [1]
Find the resultant moment couple produced by the following forces. All dimensions in m Home Next Previous End

Moment of couple + + Statics Example [1] Solution:
1. By traditional moment analysis ∑Mo = -(150)(7)-(150)(7) – (200)(9)-(200)(9) = N.m + 2.By cpouple moment analysis ∑Mc = -(150)(14) – (200)(18) = N.m + Home Next Previous End

Find the moment couple produced by the following force. All dimensions in m Home Next Previous End

Moment of couple Statics Example [2] 1. F = 150 j 2. r = 3k
3. M = r x F = 150 j x 3k = {450N.m} i Home Next Previous End M

= Moment of couple F1 -F1 d1 F1.d1 = F2.d2 d2 F2 -F2 Statics
Equivalent couples in many of life applications, an equivalent couple is required to solve some technical problems such as space and size. Equivalent couples are the couples that have the same magnitude and same direction As you can see, the relation between the forces and the arm distances in equivalent coupels is reverse (for example, as we reduce the moment arm, the required force for equivalent couple increases) F1 -F1 d1 F1.d1 = F2.d2 d2 F2 -F2 = Home Next Previous End

Moment of couple M2 M1 M1 MR M2
Statics Resultant couple Resultant couple is the vectorial summation of the couples act on the body as you can see. In simple situation as shown in the figure, the parallelogram is used to sum the moments and in more complicated cases or three dimensional problems, the Cartesian notation is used. M2 M1 M1 MR Home Next Previous End M2

Moment of couple 200N 150 N 100N Statics Example [1]
The member shown in the figure is subjected to three coupling forces : 150, 200 and 100 N. the moment arms are shown. If All dimensions in m, find the resultant couple moment produced by the following forces. 150 N 100N 200N Home Next Previous End

Moment of couple + Statics Example [1] Solution:
First, the Moment direction: Second, the calculate the coupled moments M1 = -(200)(13) = N.m M2 = -(150)(7) = N.m M3 = - (100)(8) = -800 N.m Finally, calculate the moment sum (resultant) MT = =-4450 N.m MT = 4450 N.m (clockwise) + Home Next Previous End

The member shown in the figure is subjected to two coupling forces : 200 and 100 N. the moment arms are shown. Find the resultant couple moment produced by the following forces. 3m 6m 0.6m 45o 100 N 200 N Home Next Previous End

Moment of couple + Statics Example [2] Solution:
First, the Moment direction: Second, the calculate the coupled moments M1 = -(100)(0.6) = -60 N.m M2,x = (200 cos(45))(0.6) = 85 N.m M2,y = -(200 sin(45))(3) = -424 N.m Finally, calculate the moment sum (resultant) MT = =-399 N.m MT =399 N.m (clockwise) + Home Next Previous End

Moment of couple Statics 3-D problems
For three dimensional problems, it is better to use the Cartesian notation (I, j and k) to represent the monuments. The moment in a direction not one of the principle axes (x, y or z) can be represented as: M = Mu. Where u is a unit vector in the direction of the moment M. The resultant moment can be found finally by vector addition for the moments vectors Home Next Previous End

Moment of couple M1 M2 x z y θ
Statics 3-D problems M1 M2 x z y θ Example: The figure show an object subjected to two moments: M1 and M2. as you can see M2 has an angle θ from the y-axis. The moment M1 can be represented as: M1= M1i. While the moment M2 can be represented as: M2 = M2 (0i + cos(θ) j + sin(θ) k) Home Next Previous End

Simplification of a force and couple system
Statics Equivalent couples In many of situation where there a group of forces and moments acting on an object, it is seem more convenient to reduce the large number of forces and moments to one force and one moment. Physical meaning: replacing a system of forces and moments by a system of one force and one moment. Condition: the external effects produced by the forces and moments on the body for the original system are the same of the single force and moment in the new simplified system Home Next Previous End

Statics Simplification conditions F -F F F B A B Note: the acting force can be transport from one position to another on its line of action (i.e. force vector) F F A B A B M = F.d Note: the force acts on a member can be transport from one position to another on a line perpendicular force vector by adding the moment generated by the original force (i.e. M=F.d) Home Next Previous End

Statics Simplification conditions Assume an object as shown in Fig.a is subjected to two forces ( F1 and F2) and one moment M. The forces F1 and F2 has a position vectors r1 and r2 respectively from the rotation point O to the line of action for each force. F1 O F2 r1 r2 M Home Next Previous End Fig.a

Statics Simplification conditions To convert the previous system into one force-moment system we must: first move each force to the point of rotation O. this step include adding the moments produced by both forces (M1 and M2 respectively )at the rotation point as shown in the Fig.b. O F2 M M1 = r1 x F1 M2 = r2 x F2 F1 Home Next Previous End Fig.b

Statics Simplification conditions Then all forces and moments are summed using the following formulas: FR = ∑F MR,O = ∑MO + ∑M O FR MR,o Home Next Previous End Fig.c

Statics Example [1] Figure below shows a plate a group of forces (100, 150, 200 and 300 N) If All dimensions in m, Simplify the following force system to single force and moment system about point O Home Next Previous End

Statics Example [1] Solution: First, calculate the resultant force FR ∑Fx = 300 – 100 = 200 N ∑Fy = = 350 N Second, calculate the resultant moment MR MR =-(200)(13) – (300)(4) –(150)(5) – (100)(4) = 4950 N.m Finally, represent the new force and moment on the original system as shown. Home Next Previous End

Statics Example [2]: couple resultant The system shown in the figure is subjected to two coupling forces : 200 and 100 N. the moment arms are as shown. Simplify the following force system to single force and moment system about point O 3m 6m 0.6m 45o 100 N 200 N 399 N.m Home Next Previous End

Statics Example [2] Solution: First, the Moment direction: Second, calculate the coupled moments M1 = -(100)(0.6) = -60 N.m M2,x = (200 cos(45))(0.6) = 85 N.m M2,y = -(200 sin(45))(3) = -424 N.m Finally, calculate the moment sum (resultant) MT = =-399 N.m MT =399 N.m (clockwise) Note that the resultant force (FR ) = 0 which is true for all the coupled forces systems + Home Next Previous End

Statics Example [3] The three forces act on the pipe assembly. If F1 = 50 N and F2 = 80 N, replace this force system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form.. Home Next Previous End

Statics Example [3] Solution: First, calculate the resultant force FR Finally, calculate the resultant moment MR using cross product Home Next Previous End

Statics Special cases: Concurrent forces: forces that’s lines of action intersect at a common point Concurrent forces are simply summed to find FR and as seen the moment is zero due to the passing of forces lines of action through the rotation point F1 F2 F4 F3 O FR O = Home Next Previous End

Statics Special cases: Coplanar forces: forces share the same plane Coplanar forces produce moments about the point of rotation and are summed to find FR . All the moments produced by the acting forces are summed to find the equivalent moment M. F1 F2 F4 F3 O FR O M = Home Next Previous End

Statics Parallel forces system: FR = ∑F O z MR,O b a FR = ∑F O z b a d F2 F1 F3 F4 O z A reverse process can be done to transform the single force – moment system into a single force with moment arm from the rotation point Home Next Previous End

Statics Analysis procedures Establish the coordinate system (x, y and z axes). It is preferred to put the origin of this system at the rotation point. Force summation find the resultant force by summing the acting forces. You may resolve the forces to their rectangular components. Moment summation The resultant moment is the summation of the moments acting on the body and the moments produced by the acting forces. Home Next Previous End

Statics Special cases: In three dimensional systems, we can find an equivalent force and moment. However, in general cases the moments and force are not perpendicular to each other. Because of that, it become impossible to reduce the system to single force with moment arm from the rotation point. Home Next Previous End

Statics Example [1] Replace the following forces-moment system to a single force system. 5m 10 kN 7 kN 30o 2m 8m Home Next Previous End

Statics Example [1] Solution: First, calculate the resultant force F ∑Fx = cos(30) = 16.06kN ∑Fy = - 7 sin(30) = 3.5 kN Second, calculate the resultant moment MR MR =-(7sin30)(5) -(7cos30)(8)-(10)(8-2) = 126 kN.m Finally, you can represent the new force and moment on the original system. Xo= MR/FR = 126/16.43 = 7.67 m from the base point Home Next Previous End

Statics Example [2] Replace the following forces-moment system to a single force system. 3m x z y 200N 300 N 600 N 4m Home Next Previous End

Statics Example [2] Solution: First, calculate the resultant force F FR = – 200 = (700k) N Second, calculate the resultant moment Mx and My Mx =(300)(0) – (200)(3) + (600)(6) = 3000 N.m = FRy → y = 4.29m My =-(300)(3) + (200)(0) - (600)(4) = 3300N.m = FRx→ x = 4.71m Home Next Previous End

Statics Example [2] Solution: Finally, you can represent the new force on the original system. x z y 700 N 4.29m 4.72 m Home Next Previous End

Distributed Loads x w L w = w(x) x w L dF = dA dx w = w(x) Statics
Uniform loading along a single axis x w L w = w(x) Distributed force is a force acting on a line or surface of the rigid body. The value of this force (w) is represented by a function in terms of dimensions. For example: w(x). The force function could be linear or none linear x w L dF = dA dx w = w(x) We can represent the distributed force by a single. To do that we first take an infinitesimal segment of the distributed force (dF) which equal the infinitesimal segment of the area under the force function as you can see on the screen

Distributed Loads w FR w = w(x) x Statics Magnitude of resultant force
Let us first assume a distributed force w(x) acting on the member as shown in the fig. w w = w(x) FR Now, assume there is an equivalent force called FR for the distrusted force w(x) and it is located at a distance equal x’ x The magnitude of FR can obtained by integrating the function w(x) over the distance x: As you can see from the above equation that the magnitude of FR equal the area under the curve w(x)

Distributed Loads w FR w = w(x) x Statics Location of resultant force
The location of the resultant force (d) can be found using the principle of centroid (will be discussed later) as: x FR w w = w(x) For this stage, the location of the centroid for the given shape will be given

Distributed Loads Statics Analysis procedures To analyze the distributed forces you have to follow the following procedures: distributed load is defined as function w = w(x) with unit of N/m or lbf/ft. The effect of distributed load is simplified by single concentrated force acts at certain point in the body The resultant force equals the area under the loading diagram and acts on the centroid of this area

Distributed Loads x w 200 N/m 3m (b) x w 200 N/m 2m (a) Statics
Example 1: Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figures. x w 200 N/m 3m (b) x w 200 N/m 2m (a)

Distributed Loads w 200 N/m x 2m w 400 N x 1m Statics Example 1:
Solution: part a x w 200 N/m 2m FR = area under the loading diagram FR = (200 N/m) (2m) = 400 N (x’) at the center of load rectangle x‘ = 1m x w 1m 400 N You can note that the centroid of a rectangular area is its geometric center

Distributed Loads 200 N/m w x 3m w 300 N x 2m Statics Example 1:
Solution: part b Similar to part a FR = area under the loading diagram FR = (1/2)(200 N/m) (3m) = 300 N (x’) at the centroid of triangle load x‘ = (2/3)(3) = 2m x w 2m 300 N You can note that the centroid of a triangular area is located at a distance equal 1/3 of its height fro its base.

Distributed Loads w 250 N/m x 2.5m Statics Example 2:
Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figure. x w 250 N/m w = (30)x2 N/m 2.5m

Distributed Loads w 250 N/m x 156.25 N w x Statics Example 2:
w = (30)x2 N/m 2.5m Solution: magnitude x w N Try to solve it by your self and verify the solution

Distributed Loads w 250 N/m x 156.25 N w x Statics Example 2:
w = (30)x2 N/m 2.5m Solution: location x w N 1.875 m Try to solve it by your self and verify the solution

Distributed Loads Statics Combined distributed loads In this lecture we will learn how find the resultant force from a combined distributed forces If you have a several disturbed loads , you can find the resultant force for the whole combination by finding the resultant force from each distributed force and then sum the resultant forces to obtain one equivalent force

Distributed Loads 400N/m w 300N/m x 2m 4m 2m Statics Example 1:
Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figure. 400N/m w 300N/m x 2m 4m 2m

Distributed Loads w 400N x Statics Example 1: Solution: first load
FR = area under the loading diagram FR = (0.5)(400 N/m) (2m) = 400 N (x’) at the centroid of triangle load x‘ = (1/3)(2) = 2/3m x w 2/3 m 400N

Distributed Loads w 1600N x Statics Example 1: Solution: second load
FR = area under the loading diagram FR = (400 N/m) (4m) = 1600 N (x’) at the centroid of load area x‘ = 2+(0.5)(4)= 4m x w 4m 1600N

Distributed Loads w 600N x Statics Example 1: Solution: third load
FR = area under the loading diagram FR = (300 N/m) (2m) = 600 N (x’) at the centroid of load area x‘ = 2+4+(0.5)(2)= 7m x w 7m 600N

Distributed Loads w 4m 600N 1600N 400N Statics Example 1:
Solution: representing all forces w 2/3m 4m 7m 600N 1600N 400N

Distributed Loads w 2600N Statics Example 1:
Solution: representing all forces FR =∑F = = 2600 N w x’=4.2m 2600N To find the location of this force we must use the technique of simplifying of a force and couple moment you learn previously MR = FR * x’ → x’ = MR/FR MR =(400)(2/3) + (1600)(4) + (600)(7) = N.m x‘ = 10867/2600 = 4.2 m Try to solve the same problem if the second load (400N/m) acts on the lower surface of the body. Be aware to the sum of forces and the direction of resultant moment

Distributed Loads Statics Example 2:
Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figure. 250 N/m 200 N/m 400 N/m 60o 45o 5m 10m

Distributed Loads Statics Example 2: 1250 N 1000 N 4000 N 60o 45o 5m
∑Fx = -(4000)(cos45) + (1000)(cos30) = N ∑Fy = – (4000)(sin45) – (1000)(sin30) = N FR = {(-2198)2 + (-4578)2}1/2 = 5078 N Ө = tan-1(-4578/-2198) = 64o

O Distributed Loads Statics Example 2:
to find the location of the resultant force, we can take the moment about a certain point in the body. Let us take point O. ∑Mo = -(1000)(cos30)(0.5)(5sin60) – (1000)(sin30)(0.5)(5cos60) – (1250)(2.5+5cos60) – (4000)(sin45)(5cos60+5+(0.5)10cos45) + (4000)(cos45)(0.5)(10sin45) = N.m x‘ = Mo/FR = /5078 = 5.9 m from point O O 1250 N 1000 N 4000 N 60o 45o 5m 10m (4000)(cos45) (4000)(sin45) (1000)(cos30) (1000)(sin30)

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