Made 2 copies of cover page
PHYSICS 23 LABORATORY E3-SHEAR MODULUS
NAME: GRADE: ________________ Partner’s Name: Laboratory Instructor: Laboratory Section: Recitation Section: _____ S 07
Recitation Instructor _____________________________
OBJECTIVES: To determine the shear modulus of a rod. Introduction: In this experiment we wish to get a feeling for what happens to a “rigid” body when we apply forces to it. These forces (and torques) sum to zero so that the body remains in static equilibrium after the forces are applied. This is a problem of fundamental concern to many engineers. Will the bridge collapse when a heavy truck is driven over it? Will the engine break loose from the wing during takeoff? Will the roof of the arena fall under heavy wind and water loads? As good scientists and engineers, we shall begin by trying very simple approximations and testing them against our measurements in the laboratory to see if they are sufficient. We know that rigid bodies are not really rigid. They distort under applied forces. Our simple assumption will be that they distort like springs. That is, if we pull on a rod, the change in length is proportional to how hard we pull. In other words, we expect a behavior like F = -kx for a spring (remember x in this equation is the change in length of the spring). Let’s consider the rotational equivalent of stretching a spring. We take a rod and try twisting it instead of stretching it. (continued next page)
F R r
∆θ
The rod, of length L and radius r, is clamped on the left hand side so that the left hand end cannot rotate. The right hand end is supported by bearings so it can rotate but not translate. None of this is shown in the drawing. Note that when we apply F at the circumference of the large wheel of radius R and call the twisting effect of this force a torque,τ = RF, the rod twists through an angleθ. That little arrow on the end of the rod is supposed to indicate a pointer so we can measure the angle θ. The applied torque is proportional to the angle θ
E r 4 2L
Eq (1)
where E is the shear modulus sometimes called the modulus of rigidity.
2τL E= πθr 4
In our experiment, we apply F, not at the radius of the rod, r, but at a larger radius, R, so τ = FR. Apparatus: Torsion lathe and specimens; Vernier caliper; weight pan and weights Procedure: 1.
Measure the dimensions of the specimen and place it in the torsion lathe, setting the Vernier on the large wheel to zero with the weight pan in place.
Add loads to the pan up to the maximum load which will produce a twist of no more than about 60 degrees (you need only use four masses ranging up to 500 gms). Record the angle of twist both for the loading, ∆θ load, and the unloading process, ∆θ unload. To help minimize friction effects use the 1 average value of θ avg = ( θ load + θ unload). 2 3. Take the necessary measurements to enable you to convert load to torque. Remember torque = τ = RF = Rmg where R is the radius of the large wheel and m is the hanging mass including the weight hanger. 4. Repeat steps 1-3 with a second specimen. 2.
Measurements: Your may wish to create your own table in EXCEL. R= r= L= τ = mgR
m
ΘLOAD
ΘUNLOAD
ΘAVE
Analysis: Make a table of torque in Newton-meters vs angle in rad. Plot a graph of torque in Newton-meters vs angle in radians. From the slope of this graph, determine E and compare this result with accepted values. Hint: Use eq. (1) to find the relation between the slope of τ vs. θ and the constant E. Compute the percent deviation between your measured value and the handbook value for the same material. Your report should include this plot and the slope. 2. Repeat for the second specimen. 1.
Conclusions: Material
Shear Modulus 1010 N m −2
Aluminum Brass Copper Iron Lead Magnesium Molybdenum Nickel Steel Tungsten From CRC Handbook of Chemistry and Physics
2.37 3.53 4.24 7.0 0.54 1.67 14.7 7.06-7.55 7.79 14.81