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 = F x d(Nm) = (N) x (m)

Note: take car with the perpendicular distance. Also the same as the shortest distance!

Always make clockwise positive.1.We can add up moments of each force to get a total moment.2.

Rules:

"Taking clockwise as positive, the sum of all moments around point A is zero"

The moments all have to be around the SAME POINT.1.You can pick any point you like!2.Pick a sensible point! 3.

Few things to note:

Example: 30kg pushes on 170mm crank. What is moment at axle?

M = F x d(Nm) = (N) x (m)M = (30 * 9.81) * (170/1000) = 50.031 Nm

Moment Equlibrium

Moments (Ch 6)

Moments Page 1

Moments Page 2

PCD = Pitch Circle DiameterRadius = D = 70.5mmEach bolt = 7.5 kNMoment for each bolt:M = FD= 7500 * 0.0705 = 528.75 NmTotal 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.

Examples

Moments Page 3

+ (5 × 0·50) - ( F x 0.25) = 0So 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 anywhereMonday, 12 March 20126:35 PM

Moments Page 4

Q19: A force at an angle of A = 54° has a magnitude of 1140 N. The mass of the block is 91 kg. Find moment at the pin joint. (Clockwise as +)

Weight: W = mass * gravity = 91*9.81 = 892.71 NAlpha = 90-54 = 36o

Theta = 180-36 = 144o

Draw this up in Autocad…Perp distance = 349.2mm

Now do moment equation:Ma = +(892.7*0.125) - (1140*0.3492) = -286.5 Nm

Moments 1 - graphical Monday, 9 March 20156:16 PM

Moments Page 5

Q19: A force at an angle of A = 54° has a magnitude of 1140 N. The mass of the block is 91 kg. Find moment at the pin joint. (Clockwise as +)

Weight: W = mass * gravity = 91*9.81 = 892.71 NAngle = 180-(90-54) = 144o

Split force B into x & y components: Bx = 1140*cos(144) = -922.28 NBy = 1140*sin(144) = 670.08 N

Now do the moment equation:Ma = +(W*0.125) - (Bx*0.25) - (By*0.25)

= -286.5 Nm (Same answer as last time!) = +(892.71*0.125) - (922.28*0.25) - (670.08*0.25)

Note: Ignore negative values inside the moment brackets.A negative moment is only decided by clockwise/anticlockwise.

Moments 1 - components Monday, 9 March 20156:16 PM

Moments Page 6

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 NmMoment C: M = FD= 2900*0 = 0 NmMoment B: Find perp distance D...

D = 341.5064mm or 0.3415mBy 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 2 - graphical

Moments Page 7

Make a moment equation for pivot point. Bx = 8400*cos(150) = -7274.6 NBy = 8400*sin(150) = 4200 N

= +(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) = -1472.7 Nm (Same as last time)

Remember: No negative signs inside the moment (brackets)

No negative forces in MomentEquation because +/- signs decidedby CW and CCW only

Moments 2 - components (alternative)

Moments Page 8

Force downhill = 30%*115*9.81 = 338.445 NThis 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=Frr = 0.120/2 = 0.06 mMoment equation...+(485*0.06)- (354*0.06) = 7.86 Nm

More Examples...Monday, 12 March 20127:19 PM

Moments Page 9

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 10

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 * 0.12Nm)+ (118*9.81* 0.125) + (522) + 144.6975 + 522 = 666.6975 Nm

Moveable Moments!Monday, 12 March 20128:18 PM

Moments Page 11

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 NmNow find F at latch: M=Fr so F = M/r... etc

Moments Page 12

Moment of beam due to weight.W = 0.18 * 9.81 = 1.7658 NM = F*r = 1.7658 * (.304/2) = .268402 NmSo Moment equation (equilibrium):+(1.7658 *0.152) - (.268402) = 0

------------------------------------------------------EXTRA: Now calculate torsion in wire...

Unknown wireWire diameter = 2.4mm, so r =1.2mmWire length = 530mmJ = Pi*2.4^4/32 = 3.2572 mm4

= 90*pi/180 = 1.57079 rad

T/J = G/L

= 27803 MpaE = 27803/0.4 = 69507.5 MPa

G =TL/J = 268.4*530/(3.2572*1.57079)

Note: This wire is not steel. It must be some silver brazing wire!

Here is proof that a moment can be applied anywhere and it still has the same effect!

Torque experimentThursday, 14 March 20139:43 AM

Moments Page 13

Note: This wire is not steel. It must be some silver brazing wire!

Stainless Steel wireD = 1.575 mm, L = 540mmExpect E = 195 so G = 194*0.4 = 77.6 GPaJ = Pi*1.575^4/32 = 0.6041 mm4

= 180*pi/180 = 3.1416 rad

T/J = G/L

G =TL/J = 268.4*540/(0.6041*3.1416)

E = 76369/0.4 = 190922.5 Mpa = 76369 Mpa

Note: Yes, this looks like stainless steel.

Moments Page 14