Torsion of circular sections lab report

CIVE 302 – Lab #4 Torsion and Principal Stresses T. Johnson Spring 2014 Types of Deformation So far, we have only dealt with axial deformations and briefly discussed flexure. However, there are several types of deformation that we must consider: V P M M P V Axial Deformation Shear Deformation Flexural Deformation Each of these types of deformation occur under different loading scenarios and different structural layouts – always try to identify which are dominant in your system so you know your analysis is going to be accurate to the actual physics of your problem! Torsion: Applying the Load Let’s consider a circular bar in torsion. How is torsion physically defined in the 3-D space? T A B A T T B T Torsion: Deformation Characteristics Now that we have applied the load, how does this body deform? For circular members, we see points rotate on a plane: A’ B’ A’ A B B’ Deriving Relationships Now let’s break this down mathematically: we have a pure rotation and we know several dimensions. L γ L’ s r r θ s Deriving Relationships Let’s assume these angles are small (which they are for most materials) and say the arc length, s, is essentially linear: L’ ~r s γ θ L   tan   s L r &   tan   s  L  r  r L &  L r s r Now what? Now we’ve established the geometry of how our applied torque deforms the bar, but how do actually relate this to the torque itself? Well, what is the definition of torque from physics: it’s a moment – a force “F” applied at a certain distance “r” that causes rotation about an axis.  dT   r  dF   r  dF We want this in terms of stress, though, not force. However, stress is simply a force divided by an area so we can draw the relationship: dF    dA So:  dT   r  dF   r   dA A What pieces are missing? We’re almost there, but we still don’t know exactly what this stress τ is. Let’s look back one more time at the deformed shape of the rod and see if we can figure it out by tracing the region two points rotate: Stress element with area approaching zero Stress is related to area, so let’s zoom in on this region and break down this behavior both logically and chronologically. Shear Deformation (1/2) So: γ γ V γ V γ V (a) Undeformed object. (b) Right face shears down by angle γ. This is caused by a force – let’s call it V. (c) ∑Fx = 0 has a balancing shear “V” develop on other face. However, they form a couple that causes the system to want to rotate. Shear Deformation (2/2) Wrapping this up we have: o V τ V V γ V V (d) ∑Mo = 0 yields equal but opposite couple on perpendicular faces. V (e) Redrawing these forces on the undeformed shape shows how this couple works. τ (f) Lastly, let’s put these in terms of shear stresses.