Frames - Engineering Mechanics Statics - Lecture Slides, Slides of Mechanical Engineering

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Engineering 36

Chp 6:

Frames

Introduction: MultiPiece Structures

• For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered.
• In the interaction between connected parts, Newton’s 3 rd^ Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.
• Three categories of engineering structures are considered: - Frames : contain at least one multi-force member , i.e., a member acted upon by 3 or more forces. - Trusses: formed from two-force members, i.e., straight members with end point connections - Machines: structures containing moving parts designed to transmit and modify forces.

Not Fully Rigid Frames

• Some frames may collapse if removed from their supports. Such frames can NOT be treated as rigid bodies.
• A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions.
• The frame must be considered as two distinct, but related , rigid bodies.
• With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components.
• Equilibrium requirements for the two rigid bodies yield 6 independent equations.

Example: Pin & Roller Frame

 Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force(s) in link DE and the components of the force exerted by the Pin at C on member BCD.

 SOLUTION PLAN

• Create a free-body diagram for the complete frame and solve for the support reactions. (won’t collapse)
• Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude (2-frc member); determined by summing moments about C.
• With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C.
• With member ACE as a free-body, check the solution by summing moments about A

Example: Pin & Roller Frame

• Define a free-body diagram for member BCD. The force exerted by the 2-force link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. ( )( ) ( )( ) ( )( ) 561 N

0 sin 250 mm 300 N 80 mm 480 N 100 mm = −

∑ = = − − − DE

C DE F

M F α

FDE = 561 N C

• Use the Sum of forces in the x and y directions to find the force components at C.

0 ( 561 N) cos 300 N

0 cos 300 N = − − +

α

α x

x x DE C

F C F =− 795 N Cx

0 ( 561 N) sin 480 N

0 sin 480 N = − − −

α

α

y

y y DE C

F C F Cy = 216 N

(DE in Compression)

Example: Pin & Roller Frame

( )( ) ( )( ) ( ) ( 561 cos )( 300 mm) ( 561 sin )( 100 mm) ( 795 )( 220 mm) 0

cos 300 mm sin 100 mm 220 mm

∑ = + −

MA FDE α FDE α Cx



 With member ACE as a free-

body, check the solution by

summing moments about A

Example: Pin & Pin Frame

• FBD for Member BC (^) • For AB take ΣM B = 0

( )( ) ( ) ( 450 ft lb) ( 6 ft) 75. 0 lb

0 150 lb 3 ft 6 ft

∑ = = − Ay

B Ay

F

M F

FAy = 75. 0 lb

Example: Pin & Pin Frame

• For AB take ΣF y = 0
  • Recall FAy = 75 lbs
    • For AB take ΣF x = 0
      • Will use Later:

F Bx = FAx

FBy = 75. 0 lb

( ) ( 150 lb) ( 75 lb) 75 lb

150 lb

0 150 lb

∑ = = + −

By

By Ay

y Ay By

F

F F

F F F

Bx Ax

x Ax Bx

F F

F F F

⇒ =

∑ =^0 = −

Example: Pin & Pin Frame

• For BC take ΣF y = 0
  • Recall from Before: FAx = F (^) Bx , and F (^) Bx = 86.6 lbs - Thus

75 lb

0

⇒ = =

∑ = = −

Cy By

y Cy By

F F

F F F

FCy = 75. 0 lb

FAx = 86. 6 lb

Example: Pin & Pin Frame

• Now Find SHEAR FORCE on

Pins at

A & C

• The Forces at A & C
  • Find the Shear Force

Magnitude by

• In this Case
• Thus the connecting pins

must resist a shear force of

115 lbs

F F i F j

F F i F j

C Cx Cy

A Ax Ay

2 2 F = F (^) x + F y

( ) ( )

( 86. 6 lb) ( 7 5 lb) 115 lb

86. 6 lb 75 lb 115 lb

2 2

2 2

C

A

F

F