# Same E for axial and rotational stiffness?

When we solve frames, we assume that the materials are infinitely rigid in axial direction, so all deformations are due to bending. So this means, we assume infinitely rigid modulus of elasticity, E, from formula:

$$\text{axial deformation} = \frac{\text{force} \cdot \text{length}}{ \text{Area} \cdot E}$$

And for bending the stiffness is EI. Is this E not the same E as the one in the formula above? And if so, if we assumed that E to be infinity for axial, then the EI will also be infinity. Or we can take E as differently for axial and rotation, so there is the axial E and rotational E?

• Could you elaborate you statement "*When we solve frames, we assume that the materials are infinitely rigid in axial direction, so all deformations are due to bending. *"? I would like to see an example of what you are discussing. Somehow, I get the impression that that you are talking about analysis of trusses (pin jointed structures), which are a simple form of frames.
– NMech
Feb 6, 2021 at 16:53
• No. You don't assume they are infinitely rigid in axial direction. You reason that the axial deflections are small enough that they can be ignored for the case at hand. Feb 7, 2021 at 1:19
• @PhilSweet I think what you say answers my question. Thanks.... Feb 7, 2021 at 7:08
• @NMech no not at all.... in structural engineering, trusses and frames are two entirely distinct type of structures. In trusses everything is axial because they are connected by pins and moment cannot be transferred. In frames (columns and beams) moment is transferred and bending is the key. Phil answered my question with his comment.My assumption about E was wrong. Even if we neglect axial deformations Feb 7, 2021 at 7:27
• Ok. Apologies from getting confused from the phrasing. As pete and kamran replied ignoring axial its not a necessity. Its convenience in the calculations. Usually the bending deformation is much greater.
– NMech
Feb 7, 2021 at 8:44