490 Prof. C. A. Carus Wilson on the Influence of 



To find 0, take 



3(24+<9)=4(17-75 + 0), or 0=1. 



Correcting the distances, we have 



12-5 15-25 18-75 25-0 37-0 76'0, 

 and the products of the distances into the shears become 



75-0 76-25 75-0 75'0 74-0 7fr0. 



III. Same beam, &c, as before, roller and load readjusted. 

 Distance to successive fringes : — 



10-75 12-5 15-25 19-25 26-0 39*0 80*5. 

 To find 6 take 3(26 + 0) = 4(19'25 + 0), whence 0=1. 

 Correcting the distances, we have 



11-75 13-5 16-25 20'25 27-0 40'0 81-5, 

 and the products become 



82-25 81-0 81-25 81*0 81-0 80-0 81-5. 



The law of variation of shear along the normal is thus 

 shown to be hyperbolic. 



Experiment 3. The steel straining-frame was removed from 

 the instrument and — by a screw inserted in the place of the 

 straining-screw — hung from a balance, which could be loaded 

 with shot and had a leverage of 50 to 1 : a steel stirrup 

 was hung over the frame with two hardened points resting 

 on the two guiding-pins ; one lower end of the stirrup was 

 secured to the body of the balance, the beam inserted and 

 balanced, and shot put in the pan. This lifted the straining- 

 frame and loaded the beam. 



Beam [B] 56 millim. x 20 millim. x 6*5 millim. placed on 

 the base of the steel frame on a piece of thin paper : loaded by 

 a steel roller 2 millim. in diameter. Viewed through nicols 

 crossed and at 45° to the horizontal axis of the beam. 



The balance was loaded until the first blue fringe was 

 brought down to a given position on the beam, and the weight 

 of shot observed ; the same fringe was then brought down to 

 a lower given position, and the weight of shot again observed, 

 and so on for successive points. 



