Unformatted text preview: MOMENT OF A FORCE
SCALAR AND VECTOR FORMULATION LEARNING OBJECTIVES Be able to understand and define moment Be able to determine the moment of a force in 2D and 3D cases PREREQUISITE KNOWLEDGE Units of measurements Trigonometry concepts Vector concepts Rectangular components concepts MOMENT OF A FORCE A moment of a force around a point or an axis is a measure of the tendency of the force to cause rotation of the point or axis. The moment of a force around a point or an axis can be calculated as the magnitude of the force times the distance between the force and the point or the axis. The sign of a moment follows the righthand rule, clockwise is negative and counterclockwise is positive. MOMENT OF A FORCE AROUND A POINT
F The force F located at the perpendicular distance d1 from point A, causes the point to rotate clockwise, its moment is M = Fd d2 A d1 Q The force Q located at the perpendicular distance d2 from point A causes the point to rotate counterclockwise, its magnitude is M = Qd MOMENT OF A FORCE AROUND AN AXIS Direction of a moment follows the righthand rule Magnitude of moment Mo = F d Where d is the perpendicular distance projected from a reference point or axis to the force EXAMPLE 1
Determine Mx, My and Mz
Z Mx = My = 0 Nm Mz = (20 N)(2 m) = 40 Nm
Y 2m 20 N X EXAMPLE 2
Determine Mx, My and Mz :
Z 8m 10 N Mx = My = 0 Nm Mz = 80 Nm Y X EXAMPLE 3
Determine Mx, My and Mz :
Z Mx = My = Mz = 0 Nm 20 00 N X Y EXAMPLE 4
Determine Mx, My and Mz :
Z Mx = 40 Nm My = 0 Nm
2m X 20 N
5m Y Mz = 100 Nm EXAMPLE 5
Determine Mx, My and Mz :
Z Mx = 80 Nm My = 100 Nm
Y 20 N 4m 5m Mz = 0 Nm X Determine Mx, My and Mz : 3 d Z = 4 = 2.4 m 5
3m EXAMPLE 6
Z 4m 3 m 4m 20 N X 5m Y Mx = 0 Nm My = 0 Nm Mz = 48 Nm DESIGN EXAMPLES USING MOMENT CALCULATION EXAMPLE 7  BALCONY
During a spring break in Florida, several students decided to stand on the balcony of their hotel as shown below to ....
3.6 ft W1 = 3 tons W2 = 1 tons 2 ft 2 ft 1 ton = 2000 lb The dead weights of the concrete balcony and concrete railing are 3 and 1 tons, respectively. The balcony was designed to withstand a maximum moment of 44000 lbft. For a factor of safety of 2, how many students can safely stand on the balcony? Assume the average weight of a student is 170 lbs. EXAMPLE 7 BALCONY (continued)
3.6 ft W1 = 3 tons W2 = 1 tons The total moment that can be applied is 44000/2 = 22000 lbft; where 2 is the factor of safety A 2 ft
2 ft 1 ton = 2000 lb If the maximum number of people that can stand on the balcony is (P), then the total applied moment at A due to the people weights and the dead weight of the balcony is : (6000)(2) + (2000)(4) + (170)(P)(3.6) = 22000 lbft P = 3.3 3 people EXAMPLE 8  SIMPLY SUPPORTED BEAM
200 lb
5.2 ft 5.2 ft 5 ft R = 100 lb 5 ft Shear = 100 lb MMax 5 ft R = 100 lb R = 100 lb A 200 pound force is applied at mid span of a simply supported concrete beam. Assume that the beam is weightless and the reaction R is the same at both supports. Calculate the maximum moment exerted on the beam. 1. Based on equilibrium, R = 100 lb. 2. Draw FBD of half the beam 3. Calculate the maximum moment (MMax) M Max = (100)(5) = 500 lbft EXAMPLE 8  SIMPLY SUPPORTED BEAM
200 lb
5.2 ft 5.2 ft 5 ft R = 100 lb 5 ft Plot the moment and shear diagram along the beam. Step 1 draw a free body diagram of x feet of the beam where x < 5 ft R = 100 lb Shear = 100 lb M X < 5 ft R = 100 lb At x ft from the support, the moment and shear can be calculated as follows: M = 100x lbft (positive) S = 100 lb downward (positive) EXAMPLE 8  SIMPLY SUPPORTED BEAM
200 lb
5.2 ft 5.2 ft 5 ft R = 100 lb 5 ft R = 100 lb Step 2 draw a free body diagram of x feet of the beam where x > 5 ft 200 lb 5 ft X > 5 ft R = 100 lb At x ft from the support, the moment Shear = and shear can be calculated as follows:
100 lb M M = {100x 200(x5)} lbft S = 100 200 = 100 lb (upward) EXAMPLE 8  SIMPLY SUPPORTED BEAM
200 lb
5.2 ft 5.2 ft 5 ft R = 100 lb
100 50 Shear Force (lb) 0 50 100 0 2 4 6 8 10 Distance from point A (ft) 200 lb
5.2 ft 5.2 ft 5 ft R = 100 lb 5 ft 5 ft R = 100 lb
500 400 Moment (lbft) 300 200 100 0 0 2 4 6 8 R = 100 lb 10 Distance from point A (ft) Shear diagram Moment diagram CALCULATION OF MOMENT So far, the moment has been calculated using scalar approach. For a complex system, it is more convenient to determine the moment using a different approach, "Vector Formulation". This requires crossproduct of vectors or forces. CROSS PRODUCT OF VECTORS Cross product of two vectors yields a third vector normal to the original vectors with the direction that follows the righthand rule CROSS PRODUCT OF VECTORS
A = A i + A j+ A k
x y z Given B = B i + B j+ B k
x y z i A =A B B
i A B j A B k A B i A B j A B k A B x x
i A B j A B
j A B y y
k A B k A B z z Recall x x y y z z x x y y z z x x y y z z CROSS PRODUCT OF VECTORS
i A B = Ax Bx j Ay By k Az Bz = ( Ay Bz  B y Az ) i  ( Ax Bz  Bx Az ) j + ( Ax B y  Bx Ay )k Note: 1) A x B B x A but A x B =  B x A 2) Magnitude of A x B = AB sin EXAMPLE 7
A = 3i + 2 j + 1k and B = 4i  5k
Determine A x B: i j k A B = 3 2 1 = (  10  0 ) i  (  15  4 ) j + ( 0  8) k 4 0 5 = 10i + 19j 8k Determine B x A :
Given B A =  A B = 10i  19j + 8k MOMENT OF A FORCE
(VECTOR FORMULATION) The Direction of moments still follow the righthand rule Vector of moment Mo = r F
Where r is a position vector extending from the reference point or axis to the force (r does not have to be normal to the force) EXAMPLE 8
Determine the moment of the 20 N force about the Z axis. Z F = 20 N
O
3X m
4 mY F =F +F
X
Y F F=20N 5m Y X 3 4 F = 20 i + 20 j + 0k 5 5 F = 12i + 16 j + 0k F EXAMPLE 8  continued
Z F = 12i + 16 j + 0k
O
3X m
4 mY
F F=20N 5m F X Y i M Z = r1 F = 3 j 0 k 0 = 48 N  m k  12 16 0 EXAMPLE 8  continued
Z F = 12i + 16 j + 0k
O
3X m
4 mY
F F=20N 5m F X Y i M Z = r2 F = 0 j 4 k 0 = 48 N  m k  12 16 0
The results are the same for both position vectors (also the same as the results obtained from scalar formulation). EXAMPLE 8  continued
Z Take r1 = 3 i
i j k M Z = r1 F = 3 0 0 = 48 N  m k  12 16 0
Now, take r2 = 4 j Y
i j M Z = r2 F = 0 4 12 16 k 0 = 48 N  m k 0 O
3X m 4 mY F F=20N 5m X The results are the same for both position vectors (also the same as the results obtained from scalar formulation). F EXAMPLE 8  continued
Determine Mx, My and Mz :
d = 4 x 3 = 2.4 m 5 3m 5m 4m Z 3m 4m 20 N Y X Mx = 0 Nm My = 0 Nm Mz = 48 Nm EXAMPLE 9 Determine the magnitude and direction of the resultant moment of the forces about point P SOLUTION 9
First, obtain F1 and F2 in vector format :
r 5 12 F1 = 260 i + 260 j 13 13 =100i + 240 j r F2 = 400 cos 30i + 400 sin 30 j = 346.4i + 200 j
Project r1 and r2 from point P : r1 = 2i  3 j r2 = i 8 j 2 SOLUTION 9  continued
Determine the resultant moment about point P : i M P = r1 F1 + r2 F2 = 2 j 3 k i 0 + 2 0 j 8 k 0 0 100 240 346.4 200 = ( 480 + 300 ) k + (  400 + 2771.2 ) k = 3151.2 k N  m The resultant moment is counterclockwise and has a magnitude of 3151.2 Nm EXAMPLE 10 Determine the resultant moments of the force about points O and P in vector format SOLUTION 10  continued
r Draw r1 from point O : r1 = 3i  7 j + 4k
2 Draw r2 from point P : r r2 = 7i 13 j + 6k (140 +120)i ( 60 240 ) j +(90 +420 )k ={260i +180 j +510k } N  m Determine the moment about point O i j k M O =r1 F = 3 7 4 = 60 30 20 SOLUTION 10  continued
r Determine moment about point P : r2
2 r i 60 j  13 k 6 M P = r2 F =  7 = ( 260 + 180 ) i  (140  360 ) j + ( 210 + 780 ) k = { 440i + 220 j + 990k} N  m  30  20 QUESTION
Based on the RightHand Rule, which of the followings is true? A) i x k = j C) j x i = k B) j x k = i D) k x k = 1
k i j QUESTION
Z Which of the followings is true? A) Mx = 60 Nm B) Mz = 60 Nm C) Mz = 140 Nm D) My = 0 Nm
X 20 N 3m 7m Y QUESTION
S R P Q
If a force of magnitude F can be applied in four different 2D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum torque values on the nut. (Max, Min) A) (Q, P) C) (P, R) B) (R, S) D) (Q, S) QUESTION
If M = r F, then what will be the value of M x r ? A) 0 C) r 2 x F B) 1 D) None of the above QUESTION
10 N 5N 3m P 2m The net moment of the two forces about point P is A) 10 Nm B) 20 Nm C)  20 Nm D) 40 Nm E)  40 Nm QUESTION
If r = { 3 i + 5 j } m and F = { 20 i 30 j + 10 k } N, the moment equals { _______ } Nm A) 50 i 30 j 190 k C) 50 i + 30 j 190 k
i j k B) 50 i 30 j 190 k D) 50 i + 30 j + 190 k M = r F = { 3 i + 5 j} { 20 i  30 j + 10 k } = 3 5 0 20  30 10 5 0 3 0 3 5 =i j +k  30 10 20 10 20  30 = { 50 i  30 j  190 k} N  m ...
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This note was uploaded on 07/26/2009 for the course CE 221 taught by Professor Buch during the Spring '08 term at Michigan State University.
 Spring '08
 Buch

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