Your SlideShare is downloading. ×
MECHANICS ENGINEERING - Equilibrium
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

MECHANICS ENGINEERING - Equilibrium

4,009

Published on

Published in: Education
0 Comments
3 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
4,009
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
54
Comments
0
Likes
3
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. 2-1
  • 2. • For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.• The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero,     ∑ F = 0 ∑ M O = ∑ (r × F ) = 0• Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium, ∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0 ∑Mx = 0 ∑M y = 0 ∑Mz = 0 4-2
  • 3. First step in the static equilibrium analysis of arigid body is identification of all forces acting onthe body with a free-body diagram. • Select the extent of the free-body and detach it from the ground and all other bodies.• Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.• Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.• Include the dimensions necessary to compute the moments of the forces. 4-3
  • 4. SOLUTION: • Create a free-body diagram for the crane. • Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A. • Determine the reactions at A byA fixed crane has a mass of 1000 kg solving the equations for the sum ofand is used to lift a 2400 kg crate. It all horizontal force components andis held in place by a pin at A and a all vertical force components.rocker at B. The center of gravity ofthe crane is located at G. • Check the values obtained for the reactions by verifying that the sum ofDetermine the components of the the moments about B of all forces isreactions at A and B. zero. 4-4
  • 5. • Determine B by solving the equation for the sum of the moments of all forces about A. ∑ M A = 0 : + B(1.5m ) − 9.81 kN( 2m ) − 23.5 kN( 6m ) = 0 B = +107.1 kN • Determine the reactions at A by solving the equations for the sum of all horizontal forces• Create the free-body diagram. and all vertical forces. ∑ Fx = 0 : Ax + B = 0 Ax = −107.1 kN ∑ Fy = 0 : Ay − 9.81 kN − 23.5 kN = 0 Ay = +33.3 kN • Check the values obtained. 4-5
  • 6. SOLUTION: • Create a free-body diagram for the car with the coordinate system aligned with the track. • Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle. • Determine the cable tension byA loading car is at rest on an inclined solving the equation for the sum oftrack. The gross weight of the car and force components parallel to the track.its load is 5500 lb, and it is applied atat G. The cart is held in position by • Check the values obtained by verifyingthe cable. that the sum of force components perpendicular to the track are zero.Determine the tension in the cable andthe reaction at each pair of wheels. 4-6
  • 7. • Determine the reactions at the wheels. ∑ M A = 0 : − ( 2320 lb ) 25in. − ( 4980 lb ) 6in. + R2 ( 50in.) = 0 R2 = 1758 lb ∑ M B = 0 : + ( 2320 lb ) 25in. − ( 4980 lb ) 6in. − R1 ( 50in.) = 0 R1 = 562 lb• Create a free-body diagram W x = +( 5500 lb ) cos 25 • Determine the cable tension. = +4980 lb ∑ Fx = 0 : + 4980 lb − T = 0 T = +4980 lb W y = −( 5500 lb ) sin 25 = −2320 lb 4-7
  • 8. SOLUTION: • Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A. • The three forces must be concurrent for static equilibrium. Therefore, the reactionA man raises a 10 kg joist, of R must pass through the intersection of thelength 4 m, by pulling on a lines of action of the weight and roperope. forces. Determine the direction of theFind the tension in the rope and reaction force R.the reaction at A. • Utilize a force triangle to determine the magnitude of the reaction force R. 4-8
  • 9. • Create a free-body diagram of the joist.• Determine the direction of the reaction force R. AF = AB cos 45 = ( 4 m ) cos 45 = 2.828 m CD = AE = 1 AF = 1.414 m 2 BD = CD cot(45 + 20) = (1.414 m ) tan 20 = 0.515 m CE = BF − BD = ( 2.828 − 0.515) m = 2.313 m CE 2.313 tanα = = = 1.636 AE 1.414 α = 58.6  4-9
  • 10. • Determine the magnitude of the reaction force R. T R 98.1 N = = sin 31.4 sin 110  sin 38.6  T = 81.9 N R = 147.8 N 4 - 10
  • 11. THANK YOU………. 2 - 11

×