Introduction:

A compression test was performed to analyze a statically indeterminate system. This system

consisted of two cylinders: one larger steel cylinder and another inner smaller brass cylinder. The

brass cylinder is also slightly taller than the steel cylinder. The system is indeterminate because

there are more parts than necessary to obtain equilibrium. Properties of the system that are of

interest include: the relation of stress vs. absolute value of average strain for when only brass is

carrying the load as well as after the point where both cylinders carry the load, the point where

both cylinders begin carrying the load, the force at the point where both cylinders begin sharing

the load. The slopes of the graphs are useful for analyzing internal forces as well as stress.

The experiment is to be done by axially loading the said brass and steel specimen. This was done

by applying a load with a hydraulic controlled machine. Stresses were measured at 3 points on

the specimen using strain gauges. By knowing the total load and stress at incremental points a

graph was constructed to aid in analyzing internal forces of the two separate specimens.

Theory:

For a problem to be statically indeterminate one or more forces may not be solved from statics

alone. This occurs in situations such as when adding additional unnecessary supports for

stability. Additional equations can be used to solve such problems called compatibility equations

which describe the deformation of a system. In this particular case an axially loaded system of

two cylinders, one larger steel cylinder and one smaller brass cylinder, is indeterminate through

statics. The brass cylinder is slightly longer than the steel so it will initially bear the full load

until it is compressed down to the height of the sleet cylinder where both will then support the

load. When compressed to this point a FBD (Free Body Diagram) of the system looks like:

Figure 1: Free-body diagram of the system

According to convention a tensile stress is a positive value. By looking at this diagram we know

that Ns and Nb are going to be negative as expected since the system is undergoing compression.

Looking at figure 1 the sum of forces in equilibrium can be expressed as:

∑ F y=0=F tot+N b+N s (1)

Nb is the normal force due to brass, Ns is the Normal force due to steel, and F tot is the applied

load. However this is not enough information to solve for the stress in each cylinder (σ=N/A)

In this case compatibility equations can be used to obtain additional equations necessary to solve.

This will be done by examining figure 2 as shown below:

Figure 2: Diagram of system before loading

Examining figure 2 it can be seen that the brass cylinder is slightly longer than the steel one.

When the system is initially compressed only the brass cylinder will bear the load until the length

of the brass cylinder is equal to the length of the steel cylinder. Knowing this a compatibility

equation can be obtained:

δ b=δs+(Ls−Lb) (2)

Where δ is a change in length, L is the initial length of the specimen, b and s refer to brass and

steel respectively. δ is defined to be positive when the length increases, this also indicates

tension. Knowing this it can be seen that δb and δs will be negative since the cylinders are to be in

compression. For an axially loaded specimen, δ in the elastic region can be expressed by:

δ= NLEA (3)

Where N is internal force of a specimen, L is original length, E is Young’s Modulus, and A is

original cross-sectional area. Substituting (3) into (2) gives:

N b LbEb Ab

=N s LsEs A s

+(L s−Lb) (4)

Where Lb and Ls are original cylinder length, Ab and As are original cylinder cross sections, Eb

and Es are Young’s Modulus for materials, and Nb and Ns are internal forces in the cylinder.

Assuming all information listed above is known including Ftot (Ftot will be the applied load), Nb

and Ns can be solved for.

Strain is to be recorded by 3 strain gauges attached to the system and an average should be taken

for an active strain value.

When analyzing measured data a plot of force vs. strain can be useful as the slope, k, can be

equated to

k1=kb Lb (5)

In the first area of the graph where only brass supports the load. Kb is a constant that is described

as

k b=Eb AbLb

(6)

The same constant can be found for steel in the same manner which is useful in finding k2:

k 2=(kb+k s)Lb (7)

On a force vs. strain graph of the system, the point where both cylinders bear the load together

the equation of the line for k1 should be equal to k2. Strain of the system at this point (ε¿¿¿)¿

should be

ε ¿=Lb−L sLb

(8)

At the point ε ¿ a force F* will be present and can be calculated by relating kb to the change in

length of the brass specimen.

F ¿=kb∆ Lb (9)