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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
The force F located at the perpendicular distance d1 from point A, causes the point to rotate clockwise, its moment is M = Fd1
d2
Q F A The force Q located at the perpendicular distance d2 from point A causes the point to rotate counterclockwise, its magnitude is d1 M = Qd2 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 0 0N X Y EXAMPLE 4
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
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
4m EXAMPLE 6
Z
dz 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 EXAMPLE  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 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. EXAMPLE BALCONY (continued)
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
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 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 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 along the beam. 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
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
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. 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 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
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
Notes: 1) A x B B x A but A x B =  B x A 2) Magnitude of A x B = AB sin 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
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
Given = 10i + 19j 8k Determine B x A : 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) MOMENT OF A FORCE
(VECTOR FORMULATION) The magnitude Mo of M o = r F can be obtained as Mo = (r)(F)sin
Where 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 3m
3m 5m 4m 4m 20 N Y X Mx = 0 Nm My = 0 Nm Mz = 48 Nm EXAMPLE 8  continued
Determine the moment of the 20 N force about the Z axis. Z F = 20 N
F
X 3 m O 4 mF
20m 0N 5 2 F=N F =F +F
X
Y Y Y X 3 4 F = 20 i + 20 j + 0k 5 5 F = 12i + 16 j + 0k EXAMPLE 8  continued
Z O
X 4m F F = 12i + 16 j + 0k r1 = 3i
Y m Y F=20N 5m X F 3 i j k M Z = r1 F = 3 0 0 = 0i + 0 j + 48k  12 16 0 M Z = 48 k N  m EXAMPLE 8  continued
Z F = 12i + 16 j + 0k
Y O
X 4m F m r2 = 4 j
Y F=20N 5m X F 3 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
Z Take r1 = 3 i
i j k 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
X 4m F m Y F=20N 5m X F 3 From scalar formation, Mz = 48 Nm
The results are the same for both position vectors and the scalar formulation. 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 :
r1 5 12 F1 = 260 i + 260 j 13 13 =100i + 240 j r2 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 SOLUTION 9  continued r1 Alternatively, geometry can be used to calculate 1 = 123.69o and 2 = 134.04o and then calculate the magnitude of the total moment MP as follows:
1 = 123.69o M P = r1 F1 sin 123.69
2 2 r2 + r2 F2 sin 134.04
MP = = 3151.2 N  m
2 = 134.04o ( 13 )( (100 + 240 ) ) sin 123.69 + ( 68 )( ( 346.4 + 200 ) ) sin 134.04
2 2 EXAMPLE 10 Determine the resultant moments of the force about points O and P in vector format SOLUTION 10  continued
r1 Draw r1 from point O : r1 = 3i  7 j + 4k
r2 Draw r2 from point P : 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
r1 Determine moment about point P : r2 r2 i M P = r2 F =  7 60 j  13 k 6 = ( 260 + 180 ) i  (140  360 ) j + ( 210 + 780 ) k = { 440i + 220 j + 990k} N  m  30  20 ...
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This note was uploaded on 02/03/2012 for the course CE 221 taught by Professor Buch during the Summer '08 term at Michigan State University.
 Summer '08
 Buch

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