Moment (Structural)
Content
Moments (Ch 6)
Moment is another name for Torque. In engineering; Torque for something spinning (shaft) Moment for something still (torsion spring) Moment = Force x Perpendicular Distance
M=Fxd
(Nm) = (N) x (m) Note: take car with the perpendicular distance. Also the same as the shortest distance!
Rules: 1. Always make clockwise positive. 2. We can add up moments of each force to get a total moment.
Example: 30kg pushes on 170mm crank. What is moment at axle?
M=Fxd
(Nm) = (N) x (m) M = (30 * 9.81) * (170/1000) = 50.031 Nm
Moment Equlibrium
"Taking clockwise as positive, the sum of all moments around point A is zero"
Few things to note: 1. The moments all have to be around the SAME POINT. 2. You can pick any point you like! 3. Pick a sensible point!
Moments Page 1
Moments Page 2
Examples
PCD = Pitch Circle Diameter Radius = D = 70.5mm Each bolt = 7.5 kN Moment for each bolt: M = FD = 7500 * 0.0705 = 528.75 Nm Total torque: M = 528.75 * 4 = 2115 Nm
When you apply a moment to a body, you can apply it ANYWHERE and it has the same effect!!! Wow! So if you get a question with an applied MOMENT, simply move the moment to wherever you like - such as the pivot point for example.
Moments Page 3
Moments anywhere
Monday, 12 March 2012 6:35 PM
+ (5 × 0·50) - ( F x 0.25) = 0 So force F = 2·50 Nm ÷ 0·25 m = 10 N
Moments at B: +(5*0.25) + (15*0.25) - (F *0.5) = 0 (F *0.5) = 5 F = 5/0.5 = 10 N
The moments add up to zero no matter where you take them - point A or point B. We deliberately do moment on the fulcrum so that the 15N force does not create a moment (Since still unknown)
Moments Page 4
Equlibrium of Moments - graphical
Find moment around pin joint P Mp = Moment A + Moment B + Moment C + MomentWeight
= + (5000*0.25) - (8400*0.3415) + (2900*0) + (119*9.81*0.125) = -1472.6763
Moment A: M = F D = 5000* 0.25 = 1250 Nm Moment C: M = FD = 2900*0 = 0 Nm Moment B: Find perp distance D...
D = 341.5064mm or 0.3415m By geometry; D = 0.35355*cos(15) = 0.3415 m Moment B: M = FD = 8400 * 0.3415 = 2868.6 Nm
Total Moment: Mp = 1250 - 2868.6 + 0 = -1618.6 Nm Taking weight into account: M = FD = (119 * 9.81) * 0.125 = 145.9238 Nm Total Moment at P: -1618.6 + 145.9 = -1472.7 Nm
Moments Page 5
Equlibrium of Moments - components
Make a moment equation for pivot point. Bx = 8400*cos(150) = -7274.6 N By = 8400*sin(150) = 4200 N
No negative forces in Moment Equation because +/- signs driven by CW and CCW only
+(A*0.25) - (Bx * 0.25) - (By * 0.25) + (C*0) + (W*0.125) +(5000 * 0.25) - (7274.6 * 0.25) - (4200*0.25) + 0 + (119*9.81*0.125) = -1472.7263 No negative signs inside the moment (brackets)
Moments Page 6
More Examples...
Monday, 12 March 2012 7:19 PM
Force downhill = 30%*115*9.81 = 338.445 N This is the force at the perimeter of wheel. So from Moment definition; M = Fr = 338.445 * (0.211/2) = 35.706 Nm
Apply moment to drive pulley; M=Fr r = 0.120/2 = 0.06 m Moment equation... +(485*0.06)- (354*0.06) = 7.86 Nm
Moments Page 7
Apply moment at F3 (eliminates the MOMENT effect of F3) +(733*0.229) - (F2*0.1) = 0 (F2*0.1) = (733*0.229) = 167.857 F2 = 167.857 /0.1 = 1678.57 N
Moments Page 8
Moveable Moments!
Monday, 12 March 2012 8:18 PM
A pure moment (torque) has the same effect no matter where it is applied! So we can move the moment to somewhere convenient... Moment eqn: + (W * 0.125) + (522Nm) + (118*9.81* 0.125) + (522) + 144.6975 + 522 = 666.6975 Nm
Moments Page 9
At the spring: M = Fr, so F = M/r = 0.51/(51/1000) = 10.0 N At the lever: M=Fr = 10*(6.8/1000) = 0.068 Nm Now find F at latch: M=Fr so F = M/r... etc
Moments Page 10
Moment is another name for Torque. In engineering; Torque for something spinning (shaft) Moment for something still (torsion spring) Moment = Force x Perpendicular Distance
M=Fxd
(Nm) = (N) x (m) Note: take car with the perpendicular distance. Also the same as the shortest distance!
Rules: 1. Always make clockwise positive. 2. We can add up moments of each force to get a total moment.
Example: 30kg pushes on 170mm crank. What is moment at axle?
M=Fxd
(Nm) = (N) x (m) M = (30 * 9.81) * (170/1000) = 50.031 Nm
Moment Equlibrium
"Taking clockwise as positive, the sum of all moments around point A is zero"
Few things to note: 1. The moments all have to be around the SAME POINT. 2. You can pick any point you like! 3. Pick a sensible point!
Moments Page 1
Moments Page 2
Examples
PCD = Pitch Circle Diameter Radius = D = 70.5mm Each bolt = 7.5 kN Moment for each bolt: M = FD = 7500 * 0.0705 = 528.75 Nm Total torque: M = 528.75 * 4 = 2115 Nm
When you apply a moment to a body, you can apply it ANYWHERE and it has the same effect!!! Wow! So if you get a question with an applied MOMENT, simply move the moment to wherever you like - such as the pivot point for example.
Moments Page 3
Moments anywhere
Monday, 12 March 2012 6:35 PM
+ (5 × 0·50) - ( F x 0.25) = 0 So force F = 2·50 Nm ÷ 0·25 m = 10 N
Moments at B: +(5*0.25) + (15*0.25) - (F *0.5) = 0 (F *0.5) = 5 F = 5/0.5 = 10 N
The moments add up to zero no matter where you take them - point A or point B. We deliberately do moment on the fulcrum so that the 15N force does not create a moment (Since still unknown)
Moments Page 4
Equlibrium of Moments - graphical
Find moment around pin joint P Mp = Moment A + Moment B + Moment C + MomentWeight
= + (5000*0.25) - (8400*0.3415) + (2900*0) + (119*9.81*0.125) = -1472.6763
Moment A: M = F D = 5000* 0.25 = 1250 Nm Moment C: M = FD = 2900*0 = 0 Nm Moment B: Find perp distance D...
D = 341.5064mm or 0.3415m By geometry; D = 0.35355*cos(15) = 0.3415 m Moment B: M = FD = 8400 * 0.3415 = 2868.6 Nm
Total Moment: Mp = 1250 - 2868.6 + 0 = -1618.6 Nm Taking weight into account: M = FD = (119 * 9.81) * 0.125 = 145.9238 Nm Total Moment at P: -1618.6 + 145.9 = -1472.7 Nm
Moments Page 5
Equlibrium of Moments - components
Make a moment equation for pivot point. Bx = 8400*cos(150) = -7274.6 N By = 8400*sin(150) = 4200 N
No negative forces in Moment Equation because +/- signs driven by CW and CCW only
+(A*0.25) - (Bx * 0.25) - (By * 0.25) + (C*0) + (W*0.125) +(5000 * 0.25) - (7274.6 * 0.25) - (4200*0.25) + 0 + (119*9.81*0.125) = -1472.7263 No negative signs inside the moment (brackets)
Moments Page 6
More Examples...
Monday, 12 March 2012 7:19 PM
Force downhill = 30%*115*9.81 = 338.445 N This is the force at the perimeter of wheel. So from Moment definition; M = Fr = 338.445 * (0.211/2) = 35.706 Nm
Apply moment to drive pulley; M=Fr r = 0.120/2 = 0.06 m Moment equation... +(485*0.06)- (354*0.06) = 7.86 Nm
Moments Page 7
Apply moment at F3 (eliminates the MOMENT effect of F3) +(733*0.229) - (F2*0.1) = 0 (F2*0.1) = (733*0.229) = 167.857 F2 = 167.857 /0.1 = 1678.57 N
Moments Page 8
Moveable Moments!
Monday, 12 March 2012 8:18 PM
A pure moment (torque) has the same effect no matter where it is applied! So we can move the moment to somewhere convenient... Moment eqn: + (W * 0.125) + (522Nm) + (118*9.81* 0.125) + (522) + 144.6975 + 522 = 666.6975 Nm
Moments Page 9
At the spring: M = Fr, so F = M/r = 0.51/(51/1000) = 10.0 N At the lever: M=Fr = 10*(6.8/1000) = 0.068 Nm Now find F at latch: M=Fr so F = M/r... etc
Moments Page 10
