Question

A B C D E F G H 1 J K L M N 0 P Q Exercise Single shear test of timber Insert any discussions/comments on this page For the data provided in the table below, determine the single shear strength for each test regime. Also, comment on the respective test results, whether they are satisfactory or not. Note: The cells for 'Area of cross section', and 'Single shear stress' are not locked for editing Failure load, N Single shear stress Width of the Thickness of Area of Sample failure plane the failure cross Ref (w), mm plane (d), section 1 50.1 51.3 50.1 50.3 50.1 51.0 18500 19000 18800 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 50.0 49.9 50.4 56.4 50.0 50.3 24000 24500 24300 3 50.6 49.8 48.0 50.3 50.6 49.9 28700 29000 28800 4 50.9 51.0 50.2 50.5 54.4 50.4 4000 7500 4400 5 50.5 50.3 50.2 50.3 50.4 149.9 18000 26500 20000 Single Shear - Theory Exercises +

Answer 1

The single shear stress on a plate is given by:

$\tau=\frac{F}{A}$

where F is the Failure Load (in N) and A is the Area of cross-section(in mm²).

Also, A=w*d.

The table turns out to be as given below:

Sample Ref Width of the failure plane (w),mr Thickness of the failure plane (d),mn Area of cross section,mm^2 Failure Load, N Single Shear Stress, MPa 50.1 51.3 2570.13 18500 7.198079475 50.1 50.3 2520.03 19000 7.539592783 50.1 51 2555.1 18800 7.357833353 2 50 56.4 2820 49.9 50.4 50 50.3 2495 2535.12 24000 24500 24300 8.510638298 9.819639279 9.585345072 3 50.6 28700 50.3 50.6 49.9 49.8 48 2545.18 2519.88 2395.2 29000 11.27621622 11.50848453 12.0240481 28800 4 4000 50.9 51 50.5 54.4 50.4 2570.45 2774.4 2530.08 7500 4400 1.556147756 2.703287197 1.739075444 50.2 5 50.3 50.5 50.3 50.2 50.4 49.9 2540.15 2535.12 2504.98 18000 26500 20000 7.086195697 10.45315409 7.984095681

The test results are satisfactory.