lecture 3 d_equilibrium [compatibility mode]

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Equilibrium Equations• Six scalar equations are required to express the

conditions for the equilibrium of a rigid body in the general three dimensional case.

000000

zyx

zyxMMMFFF

• These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections.

• The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium,

00 FrMF O

Engineers Mechanics- 3D equilibrium

Reactions at Supports and Connections for a Three-Dimensional Structure


Ball and Socket

Roller on Bridge

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Reactions at Supports and Connections for a Three-Dimensional Structure


Universal Joint

Bearing

Engineers Mechanics- 3D-equilibriumProblem 1 :

SOLUTION KEY:A and B are hinged.

Statically indeterminate

Take moment about AB tosolve for FCD

A window is temporarily held open in the 50o position shown by a woodenprop CD until a crank-type opening mechanism can be installed. If a=0.8mand b=1.2 m and the mass of the window is 50 kg with mass center at itsgeometric center, determine the compressive force FCD in the prop

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Engineers Mechanics- 3D-equilibriumProblem 1 :

SOLUTION:

The co-ordinates of point D:

(0.8*sin (50), -0.8*cos(50),0)

= (0.613,-0.514,0)

FCD = FCD [ 0.828 i + 0.386 j + 0.406 k ]

Engineers Mechanics- 3D-equilibriumProblem 1:

1 - 6

Summation of moments about Z axis passing through line AB:

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SOLUTION KEY:

A rectangular sign over a store has a mass of 100kg, with the center of mass inthe center of the rectangle. The support against the wall at point C may betreated as a ball-and-socket joint. At corner D support is provided in the y-direction only. Calculate the tensions T1 and T2 in the supporting wires, the totalforce supported at C, and the lateral force R

6 unknowns

Statically determinate

No force contribute to the moment about x axis except Cy


1 - 8

SOLUTION:

∑MX = 0; ∑MAB = 0 ;

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1 - 9

On solving, we get-


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The two bars AB and OD, pinned together at C, form the diagonals of ahorizontal square AOBD. The ends A and O are attached to a vertical wallby ball and socket joint, point B is supported by a cable BE, and a verticalload P is applied at D. Find the components of the reactions at A and O,and tension in the cable

SOLUTION KEY:

total unknown reactions = 12

total equations = 12 (from 2 rigid bodies)

Therefore, the problem is statically determinate.

The components of reactions can be solved by taking force and moment equilibrium.


∑MOX = -AY.a + Pa = 0 => AY = P

∑MOY = 0 => AX.a = 0=> AX = 0

SOLUTION:

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∑FX= 0

=> AZ + OZ = 0

=> AZ = -OZ = -P

AX = 0

AY = PAZ = -POX

= POY = -POZ = P

AX = 0AY = PAZ = -P

OX = POY = -POZ = P


Three identical steel balls, each of mass m, are placed in the cylindricalring which rests on a horitontal surface and whose height is slightlygreater than the radius of the balls. The diameter of the ring is such thatthe balls are virtually touching one another. A fourth identical ball is thenplaced on top of the three balls. Determine the force P exerted by the ringon each of the three lower balls.

SOLUTION KEY:

Consider the equilibrium of the top ball. Three equal contact reactions balances the weight

Take the equilibrium of one of the lower balls to find the force exerted by the ring

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푟 2 − (√3

) = 2푟 2⁄

Equilibrium of top ball

훴퐹푍 = 3푅 cos휃 −푚푔 = 0

3푅 2푟 2 3⁄2푟

= 푚푔 => 푅 = 푚푔 √6⁄

Lower ball

훴퐹푋 = 0;푃 − 푅sin 휃 = 0

푃 =푚푔√6

2 푟 √3⁄2푟 =

푚푔3√2

AO =

AO = 푟cos 30표

= 2푟√3

OC = 푟 22 − ( 2√3

)2 = 2푟 2 3⁄


A uniform 0.5mx0.75m steel plate ABCD has a mass of 40 kg and is attached to ball-and-socket joints at A and B. Knowing that the plate leans against a frictionless vertical wall at D, determine (a) the location of D, (b) the reaction at D.

SOLUTION KEY:

Find the location of point D from the geometry of the problem

Take moment of all the forces about axis AB and set it zero to find the unknown reaction at D. Vector mechanics could be easier!!

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/ / 0D A B A r r

(a) The location of D follows fromthe geometry of the problem

Denoting the coordinates of D by (0, y, z):

/ 0.1 m 0.7 mD A y z r i j k

/ 0.3 m 0.4 mB A r i k

Thus, / / 0.03 0.4 0.28 0D A B A z r r

Or 0.625 m.z

2 22/ 0.1 m 0.625 m 0.7 m 0.75 mD A y r 0.73951 my


2 2

0.3 0.4 0.6 0.80.3 0.4

ABABAB

i k i k

/ 0.1 m 0.73951 m 0.075 mD A r i j k

/ 0.4 m 0.73951 m 0.625 m 0.3 mD B r i j k

D DNN i

240 kg 9.81 m/s 392.4 Nmg W j j j

Note:

rG/B = 0.5 rD/B

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/ /0: 0AB AB D A D AB G BM r N r W


0.6 0 0.8 0.6 0 0.80.1 0.73951 0.075 0.2 0.36976 0.1625 0

0 0 0 392.4 0DN

0.59161 24.525 0DN 41.455 NDN

41.455 ND N i

lecture 3 d_equilibrium [compatibility mode]

Engineering