Please see the sketch attached. Generally Fig: A is the loading pattern (or teasing mechanism) of shear wall. 'P' is the vertical force, 'H' is the horizontal force, and 'R' is the resultant force, for which a wall is seen to be cracked (Fig: B). If 'R' is given by UTM (Fig C), what other difference in the result could occur, than in the real case (Fig: A)?
I've already seen that inclined setup for squat masonry panels, but those studies belong to decades ago. It is clear that such inclined setup would fail to resemble the real boundary condition of a real vertical wall. Just look at the lower right corner of the vertical wall in Fig. A. This point would experience significant tension forces and all longitudinal reinforcements would start to yield from this point. I don't see such condition in the inclined setup (Fig. C).
I'm not sure, but the idea might be applicable for squat walls which their failures are governed by diagonal cracks. In any case you can perform two separate analysis (one for Fig. A and one for Fig.C) and decide based on the obtained results.
It seems you are very interested in UTMs…! Just kidding.
Actually due to lack of instruments required to make test for shear wall I thought for an alternative using UTM.
By the way thanks for your comment. I got your point.
The fig.C setup is usual for masonry squat walls' analysis.
The crack line, maybe, is in the opposite diagonal.
Dear Pierantonio Barizza:
Your information is very interesting to me. Can you please shear the results, test setup, material property used at that test, also the reason for the crack in vertical way not being formed in any other way?
Dear Mahzuz,
this isn't my field of research, but I can share a link useful to understand the aim of this test.
http://amipp.rutgers.edu/assets/documents/ASTM/DiagShearCerE519.pdf
We note a vertical crack because of a tensile stress, orthogonal to the compression stress (due to the load path). You can use the Mohr circle to try a first approach. Good luck.
The Aim of the test is more important. In the test set up discussed by P. Barizza, the aim is evaluation of diagonal tension for shear panels.
As per the testing requirement of H. Muhuz, if the vertical force P in the set up is constant and monotonic increment of the horizontal force is required to be simulated, one way to do using UTM is a pulley arrangement. A rope with a sufficiently high elasticity modulus and strength, pulley and its support compatible to be held in the lower head of the UTM and rigid enough to avoid any local deformations during the test will be sufficient for carrying out the required test. Conventionally such a test is done using a servo hydraulic actuator. However, with the pulley system attached to the UTM lower head at the elevation of the horizontal force H, and the rope connected to the upper movable head and the point of application of "H" through the pulley, the test can be carried out.
One point of caution however. In the above technique, a horizontal force will be acting on the lower head of the UTM. There is a possibility that the head may not be able to take the force and some additional load transfer mechanism will be required for catering this horizontal force. Technical support from the UTM company may be sought in this regard.
Thanks V. Mahadik for your Idea.
I want to add that in Fig A moment will be generated at bottom. This moment in no sense can be generated by Fig C. Therefore the result of Fig A must differ widely than Fig C.
No doubt that moments will be generated at the bottom in case of Fig-A...Fig C in no way simulates the boundary conditions of Fig-A and hence the results will be different.
Here, we are talking about shear panels/walls. In the in-plane behaviour, the span to depth ratio for a shear wall will determine whether it is moment critical or shear critical. Those having large span to depth ratios typically fail with flexure cracks, and it is required that the foundation of the specimen is pinned down to avoid any lift up for having the expected behaviour. For squat shear walls, span to depth ratio is small and they predominantly fail with diagonal cracking---a shear failure.
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