Sample number 1a 1b 2a 2b 3a 3b 4a 4b 5a 5b ming of ni...

Sample Number 1A 1B 2A 2B 3A 3B 4A 4B 5A 5B ming of ni beritanama Length den Load Type Compression Perpendicular to Grain Compression Parallel to Grain 48 5B originates of carela1 par to grasizishnu na 3B 18 2B 100 width 1180 Thickness Weight (gm) Strength in Compression and Tension Sample # 1A 2A Specific Gravity Measurements 3A 5A 4A Cross-section Area (in²) Contact Area (in²) Max Load (lbs.) Max Specific Stress Gravity (psi) Specific Gravity Specific Strength (Stress/Sg)

Transcript

00:01 Hello students in this question given as a the diameter is equal to 12 millimeter of all three samples a b and c so for this the radius will be equal to the 6 into 10 to the power minus 3 meter so from the eskimo law we have to write the eskimo law we have to write the critically resolved shear stress critically resolved shear stress is equal to the sigma applied stress into cos of phi and cos of lambda so now first of all we have to find the critically resolved shear state stress which is which is same for all the materials so now this will be equal to the sigma stress is equal to load that means force upon area cos of phi and cos of lambda so this is same for all the materials so now we can take the data anyone so now we will take the data from the sample one the sample one applied force is given as 77 point 1 newton and area is phi r square r means 6 into 10 to the power minus 3 whole square and cos of phi means 70 .5 all these data given in the table which is given in question so and cos of lambda lambda value is for sample one is 29 so after calculation we will get the critically resolved shear stress is equal to the 196 kilo newton 196 kilo newton per meter square now, this is same for all the materials now we can find the load load for the material c for the material c is equal to the tau into this from this equation from equation number one or equation number two so from equation number two we can see that f is equal to tau into area upon cos of phi into cos of lambda now we will substitute all the values tau is critically resolved shear stress 196 10 to the power 3 newton per meter square into pi r square means 6 into 10 to the power minus 3 whole square upon cos of phi for the material c is 13 and cos of lambda for materials is 78 so after calculation we will get the load is equal to for the material c is equal to the 109 point 37 newton now we have to find the stress so controlling stress for the material c is equal to fc upon area substitute all the values 109 in point seven three seven newton upon the area is pi into 6 into 10 to the power minus 3 whole square so after calculation, we will get the controlling stress for material c is equal to 967 point 53 kilo newton per meter square...

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