The case where there is no net effect of the forces …roneducate.weebly.com/uploads/6/2/3/8/6238184/statics_lecture_5_f... · forces acting on a rigid body Outline: Introduction

The case where there is no net effect of the

forces acting on a rigid body

Outline:

Introduction and Definition of Equilibrium

Equilibrium in Two-Dimensions

Special cases

Equilibrium in Three-Dimensions

Constraints and Determinacy

ENGR 1205 1 Chapter 7

We will now use concepts from previous chapters to

solve static equilibrium problems

Requirements for equilibrium:

For a RB in equilibrium, the external forces impart no

translation or rotation motion

So, the RB (Rigid Body) is not moving, or is moving at

a constant velocity (translational and rotational)


2D, is relatively straightforward since Fz = 0

so, Mx = 0, My = 0 therefore Mz = Mo

We use:

since the location of the origin is arbitrary, we can

also say ∑ Mo = ∑ MA = ∑ MB = ∑ MC = etc = 0

we have multiple ∑ M equations that we can use

in general, we have 3 equations in total, and we can

solve for up to 3 unknowns

1 fixed support, or 2 rollers and a cable, or 1 roller

and 1 pin in a fitted hole


For example:

Truss loaded as shown

Given P, Q, S, find reactions

at A and B

Draw FBD, include known forces, weight and reaction

forces at A and B

y

x


we know P, Q and S, so pick the

equations that will isolate the

unknowns at A and B

Case 1 (always works)

use ∑ MA = 0 (to get 𝐵𝑦)

use ∑ Fx = 0 (to get 𝐴𝑥)

use ∑ Fy = 0 (to get 𝐴𝑦)

You can get the same answers by using different

equations in different orders, but sometimes the

equations will be more complicated to solve.

For example - in this case solving Σ𝐹𝑦 = 0 first give 2

equations and 2 unknowns

y

x


you can use 3 other equations if you wanted …

Case 2: ∑ Fx = 0 ∑ MA = 0 ∑ MB = 0

- valid so long as A and B aren’t on the same line parallel

to the y-axis

Case 3: ∑ Fy = 0 ∑ MA = 0 ∑ MB = 0

- valid so long as A and B aren’t on the same line parallel

to the x-axis

note: since A and B are often where ground rxn forces

apply, these 3 eqns are generally not a good choice

Case 4: ∑ MA = 0 ∑ MB = 0 ∑ MC = 0

- valid so long as A, B and C are not in a line i.e. collinear

Using an inappropriate equation will not give you the

wrong answer, but you will not be able to solve (0 = 0).


In general, choose equations of equilibrium that have

1 unknown each (where possible) by:

i] Summing moments at points of

intersection of unknown forces

ii] Summing components in a direction

perpendicular to their common

direction, if they’re parallel

Figure at right

- use (┴ to FA and FB, to get FDx)

- use (intersection for FA and FD,

to get FBy)

- use (intersection for FB and FD,

to get Fay)


A fixed crane has a mass of 1000 kg and it is used to

lift a 2400 kg crate. It is held in place by a pin at A

and a rocker at B. The center of gravity of the crane

is located at G. Determine the components of the

reactions at A and B.


ENGR 1205 Chapter 7 9


The uniform 100 kg I-beam is supported initially by its

end roller and pin on the horizontal surface at A and

B. By means of the cable at C it is desired to elevate

end B to a position 3 m above end A. Determine the

required tension P and the reaction at A, when end B

is 3 m higher than A and C is directly below the

pulley.


2 m

P

6 m

A B

C


TWO-FORCE ELEMENT

a special case where forces are only applied at 2

points on a Rigid Body (RB)

if a two-force element is in equilibrium then the two

forces have

e.g. to satisfy: ∑ Fx = 0

∑ Fy = 0

∑ M = 0

FA

FB


if they didn’t have the same line of

action (seen by joining the two

points of application for the forces)

then sum of moment equal zero.

we can treat small (or thin) RBs such as links, where

their weight is small compared to external forces, as

2 force elements

in general, if we can reduce forces acting on an RB to

F’s at only 2 points, we can create a convenient two-

force element problem (this helps in Chap 9)

FA

FB


THREE-FORCE ELEMENT

another special case is where the RB is subjected to forces

acting at 3 points

if a RB is in equilibrium, then the lines of action will be

either concurrent (act through the same point) or parallel

D is the point of intersection of F1 & F2

To satisfy ∑MD=0, the LOA of F3 must act

through D.

this situation can make finding the directions of all

unknown forces easier

the only exception is when all forces are parallel (no

intersection)

F1

F2

D


F3 acts at dot

Four types important elements

i) two-force RB is subject to forces at 2 points

~ Resultants of forces acting at each point have the same

magnitude, line of action and opposite sense

~ The direction is along the 2 points of application

ii) three-force RB is subject to forces at 3 points

~ Resultants of forces acting at each point must have

lines of action that are either concurrent or parallel

~ Use intersection of LOAs of 2 force to get the 3rd

iii) Particle is subjected to forces at only 1 point

~ All forces are concurrent

~ Moment condition is automatically satisfied

iv) Frictionless pulleys only change the direction of a cable


The lever ABC is pin supported at A and connected to

a short link BD as shown. If the weight of the

members is negligible, determine the force of the pin

on the lever at A.


• Pins at A & D 4 unknowns

• Look at BD and AB separately


START WITH BD:

Pins at B & D direction and magnitude of force is unknown

But only 2 forces so for equilibrium, LOA must run through B & D and

forces must be equal yet opposite


NOW LOOK AT AB:

Pins at A & B unknowns

But 𝑭𝑩 acts at 𝟒𝟓° from BD

AB is a 3-force member so all the forces meet at 1 point

We can find ∅ (for direction of 𝑭𝑨 )



∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0

∑ Mx = 0 ∑ My = 0 ∑ Mz = 0

these 6 equations can be solved for up to 6

unknowns, usually for rxns at supports or

connections

generally use vectors to solve 3D problems

∑M = r x F = 0, set each of the components = 0

∑ F = 0 and do the same for its components

choose the points around which you calculate

moments carefully (can also take moments about

axes)


1. draw the FBD - for each support note the 1-6 rxns

(any prevented movements are rxns … go through

carefully)

2. ∑ F = 0 and ∑ M = 0 (about any point) will give

you

3. seek equations involving as few unknowns as

possible

e.g. sum moments about ball and socket joints or

hinges, or draw an axis through points of

application of all but 1 unknown rxn, and then

solve for it using mixed triple products


Rod AB has a weight of 200 N that acts as shown.

Determine the reactions at the ball-and-socket joint A

and the tension in the cables BD and BE.





A 20-kg ladder used to reach

high shelves is supported by

two wheels A and B mounted

on a rail and by an unflanged

wheel (treat like a ball) at C

resting against a rail fixed to

the wall. An 80-kg man

stands on the ladder and

leans to the right such that

the line of action of the

combined weight of the

ladder and man intersects

the floor at point D.

Determine the reactions at A,

B, and C.





A uniform pipe cover of

radius 240 mm and mass

30 kg is held in a

horizontal position by

cable CD. Assuming that

the bearing at B does not

exert any axial thrust,

determine the tension in

the cable and the

reactions at A and B.





The tension in the cable is 343N, the

reaction at A is 𝟒𝟗𝒊 + 𝟕𝟑. 𝟔𝒋 + 𝟗𝟖. 𝟏𝒌 𝑵 and the reaction at B is 𝟐𝟒𝟓𝒊 + 𝟕𝟑. 𝟔𝒋 𝑵

The bar ABC is supported by ball and socket

supports at A and C and the cable BD. The

suspended mass is 1800 kg. Determine the

magnitude of the tension in the cable.


E


A 450-lb load hangs from the corner C of a rigid piece of pipe which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D, and by a cable attached at the midpoint BC to the wall at G. Determine where G should be located if the tension in the cable is to be minimum, and the corresponding value of the tension.


“Completely Constrained”

means the rigid body can’t move under the given

loads (or under any loads)

“Statically Determinate”

means the values of the unknowns can be determined

under static equilibrium (in 2D 3 unknowns & 3

equations, in 3D 6 unknowns & 6 equations

This case is both

statically determinate

and completely

constrained.


“Statically Indeterminate”

means there are more constraints

than needed

4 unknowns (𝐴𝑥 , 𝐴𝑦 , 𝐵𝑥 , 𝐵𝑦) but

only 3 independent equations

of equilibrium

Can’t determine 𝐴𝑥 and 𝐵𝑥

separately.

∑ MA = 0 gets us By and

∑ MB = 0 gets us Ay

but ∑ Fx = 0 only gets us 𝐴𝑥 and 𝐵𝑥


“Partially Constrained”

means the rigid body has fewer constraints

than necessary (it can move and is

unstable)

2 unknowns and 3 equations

one equation will usually not be

satisfied

∑ MA = 0 gets us By, ∑ MB = 0 gets us Ay

but Σ𝐹𝑥 ≠ 0

Occasionally in these cases we get lucky

with applied forces and ∑ Fx=0, but we

can’t ensure that it equals 0 because we

can’t control the x-direction with a

reaction force. ENGR 1205 41 Chapter 7

In order to be statically determinate and completely

constrained, we need an equal number of unknowns

and equations of equilibrium.

If not, the system will be partially constrained or

statically indeterminate or both.

Having an equal number of unknowns and equations is

necessary (but not sufficient ) for static determinacy


Example 1

3 unknowns and

Statically indeterminatecan’t

determine the values of A,B & E

Improperly constrained


Example 2

4 unknowns and Σ𝑀𝐴 ≠ 0

Statically determinate 4 unknowns

and 3 independent equations

Improperly constrained Σ𝑀𝐴 = 0

doesn’t hold since the LOA of the rxn

forces all go through the A


In 3D:

if rxns involve > 6 unknowns, some rxns will be

statically indeterminate

if rxns involve < 6 unknowns, the RB will only be

partially constrained (although it may still be in

equilibrium depending on specific loads)

even with 6+ unknowns, some equations may not be

satisfied (such as when rxns are parallel or intersect

the same line) in which case the RB is then improperly

constrained


A system is improperly constrained whenever the

supports (even if providing enough rxns) are arranged

such that the rxns are concurrent (as above in

example 2) or in parallel (as in example 1).

To get static determinancy, make sure the rxns involve

3 unknowns and that the supports don’t require

concurrent or parallel rxns

supports involving statically indeterminate rxns can

be dangerous, so be careful using them when

designing things


“Completely Constrained” means the rigid body can’t

move under any loads

“Statically Determinate” means the values of the

unknowns can be determined under static equilibrium

“Statically Indeterminate” means there are more

constraints than needed

“Partially Constrained” means the rigid body has fewer

constraints than necessary

In order to be statically determinate and completely

constrained, we need an equal number of unknowns and

equations of equilibrium and that the supports don’t

require concurrent or parallel rxns


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The case where there is no net effect of the forces …roneducate.weebly.com/uploads/6/2/3/8/6238184/statics_lecture_5_f... · forces acting on a rigid body Outline: Introduction