MECHANICS  CASE STUDY SOLUTION

Strain Gage Rosette on Bracket 

In order to determine the location and weight of a particular hanging load P, a strain rosette is attached to the top of the bracket near the wall (assume strain gages are at the wall). The rosette has three gages, each are 120^{o} apart as shown in the diagram. The bracket material is steel, E = 29,000 ksi and ν = 0.29.
When the weight P is placed at a distance s on the rectangular bracket arm, the three gages measure the following strains,
ε_{a} = 224.8 μ
ε_{b} = 118.3 μ
ε_{c} = 132.9 μ
From these three strains, the load P and the distance s need to be determined. 





Transform Gage Strains to Strain Element



The first step is to transform the strain gage strains to strains that can be represented in a normal xy strain element. This can be done by rotating the three unknown strains, ε_{x}, ε_{y} and γ_{xy}. into the three known stains, ε_{a}, ε_{b} and ε_{c}. using the basic strain rotation equation,
ε_{x'} = (ε_{x} + ε_{y})/2 + (ε_{x} + ε_{y})/2 cos2θ + γ_{xy}/2 sin2θ
Applying this equation three times, once for each gage where θ_{a} = 0^{o}, or θ_{b} = 120^{o}, or θ_{c} = 240^{o}, gives,
These simplify to
Substituting the first equation into the second two and solving for ε_{x} and ε_{y} gives




Gage Direction Strains Transformed
to xy Strain Element 

Substituting the actual gage values gives the final strains in the xy coordinate system,
ε_{x} = 224.8 μ
ε_{y} = [2(118.3 μ) + 2(132.9 μ)  224.8 μ]/3
= 65.2 μ
γ_{xy} = [2(118.3 μ) + 2(132.9 μ)]/1.7321
= 290.0 μ
Notice, the ydirection normal strain is not zero due to Poisson's effect even though the stress will be zero. 





Stresses at Gage Rosette

Actual xy Stresses at Gage Location 

Hooke's law can be used to calculate the xdirection stress and the shear stress.






Bending and Twisting Moment

Bending Moment (M) and
Twisting Moment (T) and 

The load P will cause both a bending moment and a twisting moment at the wall where the strain gages are located. The bending moment is
M_{bending} = P (6 in)
This moment will cause a bending stress at the top of the circular bar at the gage location.
P = 45.0 lb 





The twisting moment is
T_{twisting} = Ps
This torque will cause a shear stress in the circular bar.
s = 6.0 in 


