4.5. AREA METHOD: SHEAR AND MOMENT OF BEAMS

The area method is a shorter way in determining the shear and moment of beams; however, the procedures still entails the computation of the reactions. The process starts with the shear diagram considering the forces and reactions observed (depending on the kind of loading).

EXAMPLE: 3.3.1 TORQUE FOR A GIVEN ANGLE OF TWIST

What torque should be applied to the end of the shaft to produce a twist of 5 degrees? The modulus of rigidity for steel is G=77GPa.

From the formula of angle of twist, we will be able to solve for the torque, T.

                          

                                                  

where:

              

              

              

For J, the same formula for the polar moment of inertia is use. The shaft is hollow so:


             

             

With all parameters determined, we can compute for the required torque, T:

EXAMPLE: 3.1.3. MULTIPLE GEARS

Shaft BC is hollow with inner and outer diameters of 80mm and 120mm, respectively. Shafts AB and CD are solid and of diameter d. With the loading shown, determine:

a. The maximum and minimum shearing stress in shaft BC;

b. The required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 60 MPa.

Mechanics of Materials

Solution A:

The shearing stress in the shafts are measured with the transfer of stress from gears. Still using the same formula:

                                                

we should determine the values for the parameters T, c, and J.

For T, provide a cutting plane at shaft BC and determine the torque either starting from A or from D.

with both torque rotating the same.


If we consider the other end (starting at D)




with both torque rotating against each other.

For c, given the diameter of 120mm, radius will be computed as 60mm.

For J, take note that shaft BC is a hollow shaft.

                       

                       
                     

With all values already determined, we can solve the maximum shearing stress on shaft BC.

      

The minimum shearing stress will be solved using ratio and proportion.

                   
             
                   

                    

Solution B:

Using the same principle and process with the previous solution, use the general formula of the shearing stress to solve for the diameter. 

**Diameter = 2 x radius = 2r

For T:  Placing the cutting plane on either shaft AB or CD, 

With summation of torques:

Take note that for shafts AB and CD, the shafts are solid.

                          
                          

                         

                            

                             

Thus, the diameter can be determined:

                             

EXAMPLE 3.1.2. HOLLOW SHAFT

1. Determine the torque T that causes a maximum shearing stress of 50 MPa in the hollow cylindrical steel shaft shown.

2. Determine the maximum shearing stress caused by the same torque T in a solid cylindrical shaft of the same cross-sectional area.

Problem 1: 

The general formula relating shearing stress with torque is:

                                         where shearing stress is already given as 50 MPa


The value of c to be considered will be the radius which can give the maximum shearing stress, thus, consider the outermost radius.

                             c= 45mm

The polar moment of inertia is computed as:

                                      

                               
                                      

Thus:


                     

                                      

Problem 2: 

With the computed valued of T=5.74 kN-m (5,743,225.07N-mm) and a cross-sectional area of 45mm radius, we are asked to determine the shearing stress.

Before solving the shearing stress, we need to consider another value for the polar moment of inertia, J, since it is transformed to a solid shaft.

                          

Then, we can solve for the shearing stress:


                         

EXAMPLE 3.1.1. SOLID SHAFT

1. Determine the torque, T, that causes a maximum shearing stress of 100 MPa in the steel cylindrical shaft shown.

2. For the cylindrical shaft shown, determine the maximum shearing stress caused by a torque of magnitude T=1kN-m.

1. Required: Torque, T:

The general formula to be used is as follows, where all are given except the torque, T, and the polar moment of inertia, J so compute for J first.
                    

For polar moment of inertia, use the solid shaft:


                          where c will be taken as the radius to have the maximum shearing stress.

So,


                  

Thus,

               

             

2. Required: Shearing stress:

              

           
             
         
                       

EXAMPLE 1.4.3. HOOP STRESS

A large pipe called a penstock in hydraulic work is 1.5m in diameter. It is composed of wooden staves bound together by steel hoops, each  in cross-sectional area, and is used to conduct water from a reservoir to a powerhouse. If the maximum tensile stress permitted in the hoops is 130MPa, what is the maximum spacing between hoops under a head of water of 30m? (Use mass density of water = .

For one hoop, the FBD shows two F for the hoop against the hydrostatic force caused by pressure.

And the force of the hoops is computed as:





From the FBD, use statics to attain equilibrium (prevent the staves to explode).







where pressure,




So, the required length becomes: