Unformatted text preview: MOMENT OF A FORCE MOMENT
SCALAR AND VECTOR SCALAR FORMULATION FORMULATION LEARNING OBJECTIVES LEARNING Be able to understand and define moment Be able to determine the moment of a force Be in 2D and 3D cases in PREREQUISITE KNOWLEDGE KNOWLEDGE Units of measurements Trigonometry concepts Vector concepts Rectangular components concepts MOMENT OF A FORCE MOMENT A moment of a force around a point or an moment axis is a measure of the tendency of the force to cause rotation of the point or axis. force The moment of a force around a point or The an axis can be calculated as the magnitude of the force times the distance between the force and the point or the axis. force The sign of a moment follows the righthand rule, clockwise is negative and hand counterclockwise is positive. counterclockwise MOMENT OF A FORCE AROUND A POINT POINT
The force F located at the perpendicular distance d1 from point A, causes the point to rotate clockwise, its moment is M = Fd
d2 F A Q The force Q located at the perpendicular distance d2 from point A causes the point to rotate counterclockwise, its magnitude is d1 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 EXAMPLE
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 EXAMPLE
Determine Mx, My and Mz :
Z Mx = My = Mz = 0 Nm 20 00 N X Y EXAMPLE 4 EXAMPLE
Determine Mx, My and Mz :
Z Mx = (20)(2) = 40 Nm My = 0 Nm
2m X 20 N
5m Y Mz = (20)(5) = 100 Nm EXAMPLE 5 EXAMPLE
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 EXAMPLE
Z
dz 4m 3 m 4m 20 N X 5m Y Mx = 0 Nm My = 0 Nm Mz = (20)(2.4) = 48 Nm DESIGN EXAMPLES USING MOMENT CALCULATION MOMENT EXAMPLE  BALCONY 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 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 50,000 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. 1 ton = 2000 lb EXAMPLE – BALCONY (continued) (continued) EXAMPLE
3.6 ft W1 = 3 tons W2 = 1 tons The total moment that can be applied is 50000/2 = 25000 lbft; where 2 is the factor of safety A● 2 ft
2 ft 1 ton = 2000 lb If the maximum number of students that can stand on the balcony is (S), 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)(S)(3.6) = 25000 lbft S = 8.2 8 Students EXAMPLE  SIMPLY SUPPORTED BEAM EXAMPLE
200 lb
5.2 ft 5.2 ft •
5 ft R = 100 lb 5 ft Shear = S = 100 lb MMax 5 ft •
• 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 shear and the maximum moment exerted on the beam. 1. Based on equilibrium, R = 100 lb. 2. Draw FBD of half the beam R = 100 lb 100 – S = 0.0, S = 100 lb. M Max = (100)(5) = 500 lbft EXAMPLE  SIMPLY SUPPORTED BEAM
200 lb
5.2 ft 5.2 ft •
5 ft R = 100 lb 5 ft •
• Plot the moment and shear diagram Plot along the beam. along Step 1 – draw a free body diagram of x feet of the beam where x < 5 ft R = 100 lb Shear = S = 100 lb M At x ft from the support, the moment X < 5 ft R = 100 lb and shear can be calculated as follows: M = 100x lbft (positive) S = 100 lb downward (positive) EXAMPLE  SIMPLY SUPPORTED BEAM EXAMPLE
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 = 100 lb 5 ft M X > 5 ft R = 100 lb At x ft from the support, the moment Shear = S and shear can be calculated as follows: M = {100x 200(x5)} lbft S = 200 – 100 = 100 lb (upward, negative) EXAMPLE  SIMPLY SUPPORTED BEAM EXAMPLE
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 CALCULATION So far, the moment has been calculated using scalar approach. convenient to determine the moment using a different approach, “Vector Formulation”. This requires crossproduct of vectors or forces. For a complex system, it is more CROSS PRODUCT OF VECTORS OR VECTOR PRODUCT OR 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 CROSS
Given
A = AX i + AY j + AZ k B = BX i + BY j + BZ k i A × = AX B BX
i AX BX j AY BY k i AZ − AX BZ BX j AY BY k i AZ + AX BZ BX j AY BY
j AY BY k AZ BZ k AZ BZ Recall CROSS PRODUCT OF VECTORS CROSS
Notes: 1) A x B ≠ B x A but A x B =  B x A 2) Magnitude of A x B = AB sin θ i A × B = Ax Bx A × B = ( Ay Bz − B y Az ) i − ( Ax Bz − Bx Az ) j + ( Ax B y − Bx Ay )k j Ay By k Az Bz EXAMPLE 7 EXAMPLE
A = 3i + 2 j + 1k and B = 4i − 5k
Determine A x B :
ijk A × B = 3 2 1 = ( − 10 − 0 ) i − ( − 15 − 4 ) j + ( 0 − 8) k 4 0 −5
Given = 10i + 19j – 8k Determine B x A : B × A = − A × B = 10i  19j + 8k MOMENT OF A FORCE MOMENT
(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) MOMENT OF A FORCE MOMENT
(VECTOR FORMULATION) The magnitude Mo of M o = r × F can be obtained as Mo = (r)(F)sinθ
Where θ is the angle between r and F as shown in Figure b above Determine Mx, My and Mz using scalar:
Z
d = 4 x 3 = 2.4 m 5 EXAMPLE 8 EXAMPLE 3m
3m 5m 4m 4m 20 N Y X Mx = 0 Nm My = 0 Nm Mz = 48 Nm EXAMPLE 8  continued continued
Determine the moment of the 20 N force about the Z axis. Z F = 20 N
O 4m
m
F
X
Y F =F +F
X
Y 3 Y X 20m 0N 52 F=N 3 4 F = −20 i + 20 j + 0k 5 5 F = −12i + 16 j + 0k F EXAMPLE 8  continued continued EXAMPLE
Z O
3X m 4 mY F = −12i + 16 j + 0k r1 = 3i
Y F F=20N 5m X F i jk M Z = r1 × F = 3 0 0 = 0i + 0 j + 48k − 12 16 0 M Z = 48 k N  m EXAMPLE 8  continued continued EXAMPLE
Z F = −12i + 16 j + 0k
4 mY O
3X m r2 = 4 j
Y F F=20N 5m X F i M Z = r2 × F = 0 j 4 k 0 = 48 k N  m − 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 continued
Z Take r1 = 3 i
i jk M Z = r1 × F = 3 0 0 = 48 k N  m − 12 16 0
Now, take r2 = 4 j
i j Y M Z = r2 × F = 0 4 −12 16 k 0 = 48k N  m 0 O
3X m 4 mY F F=20N 5m X F From scalar formation, Mz = 48 Nm
The results are the same for both position vectors and the scalar formulation. EXAMPLE 9 EXAMPLE Determine the magnitude and direction of the Determine resultant moment of the forces about point P resultant SOLUTION 9 SOLUTION
First, obtain F1 and F2 in vector format : format
r 5 12 F1 = 260 i + 260 j 13 13 =100i + 240 j r F2 = 400 cos 30°i + 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 continued SOLUTION
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 The and has a magnitude of 3151.2 Nm and SOLUTION 9  continued r Alternatively, geometry can be Alternatively, used to calculate θ1 = 123.69o and θ2 = 134.04o and then calculate the magnitude of the total moment MP as follows: as
θ = 123.69 M P = r1 F1 sin 123.69
2 2 r + r2 F2 sin 134.04
MP = = 3151.2 N − m
θ = 134.04 ( 13 )( (100 + 240 ) ) sin 123.69 + ( 68 )( ( 346.4 + 200 ) ) sin 134.04
2 2 EXAMPLE 10 EXAMPLE Determine the resultant moments of the force about Determine points O and P in vector format points SOLUTION 10  continued continued
r Draw r1 from point O : r1 = −3i − 7 j + 4k
2 Draw r2 from point P : r r2 = −7i −13 j + 6k MO (140 +120)i −( 60 − 240) j + ( 90 + 420)k = {260i +180 j +510k } N  m Determine the moment about point O i j k = r1 ×F = −3 −7 4= 60 −30 − 20 SOLUTION 10  continued continued
r Determine moment about point Determine 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 30 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 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.
 Fall '10
 Baladi

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