I have to correct the Internal diameter of 2 half round cast piece of aluminuim against the actual mandrel. Because after casting the pieces are open when cooling.i will use chain block to correct or tie the cast piece with the Mandrel. could you please help me to find out what will be force to apply to make it correct. is there any particular formula? eg: the required I.D is 180mm the cast pieces are now 187mm. so i have to round it with the mandrel which is 180mm dia and will use chain block to tighten it with force. just i wnat to know what is the force will be there to correct it.?
Are the cast pieces cylindrical, or spherical? What are the thicknesses?
Usually the cast pieces are a little oversized so that when they cool, they will fit.
What is the temperature of the cast pieces when the ID measured 187mm?
The force required to "squeeze" them from 187mm to 180mm ID will depend on the thickness of the pieces. The thicker they are, more force will be required.
Are you studying in mechanical engineering, or is it a real-life problem? It sounds to me that some data are missing.
Q: I have to correct the Internal diameter of 2 half round cast piece of aluminuim against the actual mandrel. Because after casting the pieces are open when cooling.i will use chain block to correct or tie the cast piece with the Mandrel. could you please help me to find out what will be force to apply to make it correct. is there any particular formula? eg: the required I.D is 180mm the cast pieces are now 187mm. so i have to round it with the mandrel which is 180mm dia and will use chain block to tighten it with force. just i wnat to know what is the force will be there to correct it.?
Answer received:-
Are the cast pieces cylindrical, or spherical? What are the thicknesses?
Usually the cast pieces are a little oversized so that when they cool, they will fit.
What is the temperature of the cast pieces when the ID measured 187mm?
The force required to "squeeze" them from 187mm to 180mm ID will depend on the thickness of the pieces. The thicker they are, more force will be required.
Are you studying in mechanical engineering, or is it a real-life problem? It sounds to me that some data are missing.
Reply: thanks for the answer: more details as per your question. The thick ness of the cast piece is 50mm and the straight length is 400mm ; means the product dim is 400mm long x 50mm thk x 180mm i.d and 280mm O.D - in 2 halves. hope this data will satisfy to give more details to me against my question.
please assit me to solve the above said problem.
To determine the force required to correct the internal diameter of the cast piece, you need to consider the material properties of the aluminum and the desired amount of deformation.
One approach to calculate the force is by using the concept of stress and strain. The stress applied to a material is related to the force required to produce a certain deformation. In this case, the deformation is the difference between the actual diameter (187mm) and the desired diameter (180mm).
To calculate the force, you can use the equation:
Force = Stress × Area
First, you need to calculate the stress required to achieve the desired deformation. The stress can be calculated using the formula:
Stress = Young's Modulus × Strain
where Young's Modulus is a material property that represents its stiffness, and strain is the ratio of the deformation to the original dimension of the material.
For aluminum, the Young's Modulus is usually around 70 GPa (Gigapascal).
Next, calculate the strain:
Strain = (Actual diameter - Desired diameter) / Desired diameter
Once you have the strain value, you can calculate the stress. Then, use the stress value in the force equation along with the area of the half-round cast piece in contact with the mandrel to find the force required for correction.
However, it's important to note that this calculation provides an estimate and assumes linear elastic behavior, which might not be accurate for aluminum and complex shapes. It is always advisable to consult with a mechanical engineer or an expert in the field to ensure safety and accuracy in the correction process.