At a point in a strained material the principal stresses are 60MPa and 40MPa. Find the position of the plane across which the resultant stress is most inclined to the normal and determine the value of this stress.
Google "Mohr's circle".
for example:
http://www.youtube.com/watch?annotation_id=annotation_851576&feature=iv&src_vid=1plaFAWN8rk&v=qqgfKHBpj1k
To find the position of the plane across which the resultant stress is most inclined to the normal, we need to determine the direction of the maximum shear stress. The maximum shear stress occurs on a plane that is 45 degrees to the principal stress planes.
To determine the value of this stress, let's assume the two principal stresses are σ1 = 60 MPa and σ2 = 40 MPa. The maximum shear stress, τ_max, can be calculated using the formula:
τ_max = (σ1 - σ2) / 2
Substituting the values, we have:
τ_max = (60 MPa - 40 MPa) / 2
= 20 MPa / 2
= 10 MPa
So, the maximum shear stress is 10 MPa.
Now, to find the position of the plane across which the resultant stress is most inclined to the normal, we consider that this plane is at 45 degrees to the principal stress planes. Since we have two principal stress planes at 60 degrees apart (σ1 and σ2), the plane with the maximum inclination (45 degrees) will be at the average of these two angles.
The inclination angle, θ, can be calculated using the formula:
θ = (θ1 + θ2) / 2
where θ1 and θ2 are the angles of the principal stresses with respect to a reference plane (usually the x-axis).
Let's assume that σ1 is inclined at an angle of 30 degrees and σ2 is inclined at an angle of 90 degrees. The inclination angle, θ, can be calculated as:
θ = (30 degrees + 90 degrees) / 2
= 120 degrees / 2
= 60 degrees
So, the position of the plane across which the resultant stress is most inclined to the normal is at an angle of 60 degrees with respect to the reference plane.
Additionally, if you want to determine the value of the resultant stress across this inclined plane, you can use the formula of the resolved shear stress:
τ = |(σ1 - σ2) / 2| * cos(2θ)
Substituting the values, we have:
τ = |(60 MPa - 40 MPa) / 2| * cos(2 * 60 degrees)
= 10 MPa * cos(120 degrees)
= 10 MPa * (-0.5)
= -5 MPa (Note: the negative sign indicates a shear stress in the opposite direction)
Therefore, the value of the resultant stress across the plane with the maximum inclination is -5 MPa.