Solving Oneaxial Shear Test: Find Shear Force

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The discussion centers on calculating the shear force from a oneaxial shear test with a specimen height of 100mm and diameter of 54mm. The maximum force applied is 144.6 N, resulting in a deformation of 5mm. The user calculates the strain (ε) as 0.05 and the stress (σz) as 0.025 N/mm², leading to a shear force of 0.0127 N/mm², which is significantly lower than the expected 30 N/mm². Participants emphasize the need for additional details regarding the specimen's shape, force application points, and deformation measurement locations to clarify the calculations. Accurate context is essential for resolving discrepancies in the results.
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I´m really struggling to find the shear force here.
1. Homework Statement

Hight h of a specimen = 100mm and diameter is 54mm.
Max force F=144,6 N and deformation is δ=5mm

Homework Equations


I have to find ε=δ/h
Area of specimen before deformation=54mm*100mm=5400mm
σz =(F(1-ε))/A
Shear force = σz /2

The Attempt at a Solution


ε=δ/h=5/100=0.05
A=5400mm
σz =(F(1-ε))/A=(144,6(1-0.05))/5400=0,025N/mm2
Shear Force=0,025/2=0,0127N/mm2 which gives 12,7kN/m2. The right answer is 30kN/m2.

Any thoughts?
 
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I, for one, would need more description. What is the shape of the specimen? Where are the forces applied? Where is the deformation measured?
 
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