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:


                         

INTERIOR BAYS: DESIGN FOR STRENGTH I POSITIVE FLEXURE - EXAMPLE

Using the previous example, determine the bar spacing for both the positive and negative moments in the interior bays using the following:

               Use:#25 bars

                     f'c   =  27.58 MPa
                     fy   =  413.69 MPa

   

To determine the properties of the rebar, check the values from the table below:

harrissupplysolutions.com

So with #25 bar: 

                       diameter = 25.4mm
                                 

DESIGN FOR STRENGTH I POSITIVE FLEXURE

Solve for the required area of steel, , and the required bar spacing is found as follows:

Effective depth (from previous computation) : 238mm

Solve for the required amount of reinforcing steel for a typical section with width b=1m. The formula for the nominal resistance is:

                                    

where the resistance force can be taken from the AASHTO table found in 5.5.4.2.1.

So,

                                                                                                                                                                    

From the general reinforced concrete design equations:

                                    

                               
                                  

Compare with :


                                 

Thus, use  .


The required are is then computed:

                             
                                  

The required bar spacing can be computed as follows:

                                 

CHECK FOR POSITIVE MOMENT FLEXURE CRACKING

For flexure, load factors considered are:

                         
                         

                          

                          


Compute for the positive service moment. Values for moments have already been previously computed.








AASHTO specifies  bar spacings (s) from the formula 5.7.3.4.-1


                      

in which

                     

where: 


  exposure factor
    =  1.0 for Class 1 exposure condition
    =  0.75 for Class 2 exposure condition

  thickness of concrete cover measured from extreme tension fiber to center of the flexural reinforcement located closest thereto (mm)

 tensile stress in steel reinforcement at the service limit state (MPa)

  overall thickness or depth of the component (mm)


So, using:

                  class I exposure condition:   

                     
                          

                          

for
                         


                         

where:
                       

                       

                       

                       

Therefore, for n = 8,
                 


                       

                      

So                 


From AASHTO 5.7.3.4.               

                        

                       


Compare the two spacing values. The selected spacing for strength 1 positive moment 500mm with 136mm. Thus the flexure cracking value should be used.

**OK for Service I positive moment in interior bays.