springs & loads

8/2/2019 Springs & Loads

1/52

Chapter 2

SRIDHAR CONDOOR


2/52

Review - Springs

A spring is an elastic elementwhich deforms under theaction of forces.

Springs come in various shapesand sizes.

Even seemingly rigid structuralelements (such as columns andbeams) can be considered

springs as they deform (to amuch lesser extent) under theaction of forces.

For the ease of sketching, the schematic iscommonly used to denote a spring.


3/52

Springs

When a tensile (pulling) force is applied, a springstretches or extends.

A compressive (squeezing) force decreases thelength. When the spring is unloaded, it returns to itsoriginal length.

This property of returning to its original shape is

known as elasticity. A spring that returns to itsoriginal shape is an elastic spring.


4/52

Springs


5/52

Spring Rate

The spring rate or spring

constant defines the

stiffness of a spring.

It is the slope of the force

versus the deflection

curve.

A linear spring has a

constant slope.


6/52

Spring Rate

For extension and compression springs, obtain the springrate (k) by dividing the applied force (F) by the deflection().

The spring rate uses the units N/m or lb/in.

A stiff spring has a high spring constant and requires largerforce per unit deflection compared to a flexible spring.

k

F


7/52

Spring Arrangements

Springs can be arranged in

series

parallel

a combination of series and parallel

arrangements.


8/52

Parallel Arrangement

The force is shared between the springs.

The deflections are the same for all the springs.

The effective spring constant for a parallel arrangement is given by

eff 1 2k k k


9/52

Parallel Arrangement

When springs act in parallel, the assemblybecomes stiffer and, therefore, the effectivespring constant is larger than the individual

spring constants.

Use this property to qualitatively verify the

numerical calculations.


10/52

Series Arrangement

The individual springs experience the same force.

The total deflection is equal to the sum of individual springdeflections.

The effective or equivalent spring constant for the springassembly can be computed using

1 2

eff 1 2

1 1 1

k k k


11/52

Series Arrangement


12/52

Series Arrangement

The effective spring constant is smaller than

the spring constants of the individual springs.

Use this property to qualitatively verify the

numerical calculations.


13/52

SERIES OR PARALLEL?

Load is shared

Same deflection

Therefore, the springs

are in PARALLEL


14/52

SERIES OR PARALLEL?

Load is shared

Same deflection

Therefore, the springs

are in PARALLEL


15/52


16/52

DEFLECTION & STIFFNESS

A short bar with a large

cross-sectional area

made of a material with

high Youngs moduluswill have high stiffness.


17/52

Analysis of a Prismatic Bar Subjected

to Several Axial Loads

Compute the stresses and the change in its

length in this prismatic bar.


18/52



STEP 1 - Draw the free-body diagram using the

method of sections and compute the loads in

each section using the equilibrium equations.


19/52



STEP 2 Compute the stresses and deflections

in each segment.

Previous Result


20/52



STEP 3 Determine the total deflection by

algebraically summing individual deflections.


21/52

PROBLEM

Determine the maximum stress and the

deflection of a circular bar hanging due to its

self-weight. The weigh density is and the

length of the bar is L.


22/52

SOLUTION

Consider a small element

whose thickness is dx at a

distance x from the lower

end.

The force experienced by

this particular element

will be equal to theweight of the rod hanging

below the element.


23/52

SOLUTION

The maximum stressoccurs at the top of thebar.

All rods of the samematerial and length willhave the same stressesirrespective of their cross-sectional area.


24/52

SOLUTION

Deflection of the disc isequal to

The total deflection is

All rods of of the same

material and length willstretch by the sameamount irrespective oftheir cross-sectional area.


25/52

Statically Indeterminate Structures

Statically indeterminate structuresrequire the consideration of bothequilibrium equations and thedeformations to determine the internalforces and reactions.

If the block is guided so that it movesvertically down under the action ofapplied force, the displacement of thetwo springs will be equal.

The equilibrium equation yields

with two unknown forces. To solve forthe unknown forces, we need to knowthe stiffness of the springs.

FFF 21


26/52

Steps for Analyzing Statically Indeterminate

Structures

Apply the equilibrium equations

Establish geometric compatibility

Relate forces and deformations

Solve for the unknowns


27/52

PROBLEM

Force is applied on an

aluminum bar. The

diameter of the bar is

10 mm. If themagnitude of is 50 kN,

determine the stresses

in each section of the

bar .


28/52

SOLUTION

STEP 1: Apply the

equilibrium equations


29/52

SOLUTION

STEP 2: Establish

geometric compatibility


30/52

SOLUTION

STEP 3: Relate forces and deformations


31/52

SOLUTION

STEP 4: Solve for the unknown forces, stresses, anddeflections.

The results make sense because: The sum of the two forces is equal to the applied force

The stiffer spring is carrying more load (because its deflection isthe same as that of the weaker spring).


32/52

SOLUTION (ALTERNATIVE APPROACH)

First compute the stiffnesses of individual

members.


33/52


Visualize thedeformation

Node C moves down.

The extension in

segment AC will be equalto the compression insegment BC.

Further, the two

segments AC and BCresist this deformationand share the appliedload.


34/52


The effective stiffness for this assembly


35/52

PROBLEM

Determine

the forces carried by the

bar and the casing

the individual

deflections


36/52

PROBLEM

Steel bar Diameter = 0.5 in.

Length = 10 in.

Casing Inner diameter = 0.6 in

Outer diameter = 0.75 in

The casing is shorter thanthe steel bar by 0.01 in.

Force = 10,000 lb


37/52

SOLUTION

STEP 1: Apply the

equilibrium equations


38/52

SOLUTION

STEP 2: Establish

geometric compatibility


39/52

SOLUTION

STEP 3: Relate forces and deformations


40/52

SOLUTION

STEP 4: Solve for the unknown forces, stresses, anddeflections.

The results make sense because: The sum of the two forces is equal to the applied force

The stiffer spring is carrying more load (because its deflection isthe same as that of the weaker spring).


41/52

PROBLEM

The cross-sections of thepost are square, withdimensions bXb at the topand 1.5bX1.5b at the base.

Derive a formula for theshortening of the post dueto the compressive load Pacting at the top.

Assume The angle of taper is small

Disregard the self-weight


42/52

SOLUTION


43/52

SOLUTION


44/52

PROBLEM

The line of action of theloads has an eccentricity eof such magnitude thateach part of the bar isstressed uniformly.

Determine the axial forces.

Determine the eccentricitye of the loads.

Determine the ratio of thestresses.


45/52

SOLUTION


46/52

PROBLEM

The angle of rotation of

the bar due to the

action of the load P is

limited to 3.

What is the maximum

permissible load Pmax?


47/52

SOLUTION


48/52

SOLUTION


49/52

PROBLEM Determine the tensile

stresses In the wires dueto the load P = 340 lb.

Find the downward

displacementB at end B

of the bar.


50/52

SOLUTION


51/52

PROBLEM

Determine the

reactions RA and RD at

the fixed supports.

Determine the

compressive axial force

FBC in the middle

segment of the bar.


52/52

SOLUTION

springs & loads

Documents