### Cup Drawing: Determining Drawing and Blank-Holding ForcesFor the drawing operation of Problem 3 (Cup Drawing), the sheet metal used is low-carbon steel with the following properties:- **Tensile strength** = 300 MPa- **Yield strength** = 175 MPa- **Die corner radius** = 5 mmTasks:1. Determine the **drawing force**.2. Determine the **blank-holding force**.Given:- **Blank diameter** = 150 mm- **Punch diameter** = 100 mm### Equations Used#### Drawing Force:\[ F = \pi D_p t (TS) \left( \frac{D_b}{D_p} - 0.7 \right) \]Where:- \( F \) = Drawing Force- \( D_p \) = Punch Diameter- \( t \) = Thickness of the sheet- \( TS \) = Tensile Strength- \( D_b \) = Blank Diameter#### Blank-Holding Force:\[ F_h = 0.015 Y \pi \left( D_b^2 - \left( D_p + 2.2t + 2R_d \right)^2 \right) \]Where:- \( F_h \) = Blank-Holding Force- \( Y \) = Yield Strength- \( R_d \) = Die Corner RadiusThese equations will allow the calculation of forces required in the cup drawing process, ensuring proper material formation and avoiding defects.

Answered: Image /qna-images/answer/44cae035-c1f9-4674-ad05-a42efc88b6ff.jpg

For the drawing operation of Problem 3 (Cup Drawing), the tensile strength of the sheet metal (low-carbon steel) = 300 MPa and yield strength = 175 MPa. The die corner radius = 5 mm. Determine: a. the drawing force and, b. the blank-holding force blank diameter =150mm punch diameter= 100mm Equations used Ꭰ F = nD₂t (TS) (Dr – 0.7) Dp F₁ = 0.015Yπ{D² – (D₂ + 2.2t + 2Ra)

For the drawing operation of Problem 3 (Cup Drawing), the tensile strength of the sheet metal (low-carbon steel) = 300 MPa and yield strength = 175 MPa. The die corner radius = 5 mm. Determine: a. the drawing force and, b. the blank-holding force blank diameter =150mm punch diameter= 100mm Equations used Ꭰ F = nD₂t (TS) (Dr – 0.7) Dp F₁ = 0.015Yπ{D² – (D₂ + 2.2t + 2Ra)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: Three-hole punches are commonly used. The three punch heads' diameter is 6mm. Assume the force one…

A: let N be thumber of shears that can be sheared. shear area = π d t d = diameter of hole =0.006 mt…

Q: 1. A rod made from a material with a constitutive law of the form σ = 100 + 75€ 0.2 is drawn from a…

Q: Q1: Find a relation between true strain () and reduction in area (r)?

Q: Give an example of the Absolute Maximum Shear Strain?

Q: No wrong answer please , i could downvote The piece of suture is tested for its stress relaxation…

Q: A compression test is performed on a 2” diameter by 5” long bar of an unknown metal. At the end of…

A: Given:d1=2 inl1=5 inl2=3.5 inK=250 MPan=0.4a)Volume before compression=volume after compression…

Q: . For the following diagram, calculate the force needed to is A-36 steel, wire diameter is 0.125…

Q: 5. Consider forging a 2024 aluminum slab from the initial and final "brick" shapes shown below. The…

A: For the solution refer below images.

Q: What is the linear variation in the shear strain?

A: The change in shape obtained due to shear stress is defined by shear strain.

Q: You have built the majority of the tension testing machine, but much of the instrumentation is still…

A: The diameter is

Q: 8. For the following diagram, calculate the force needed to stretch the wire AB by 5%. Wire is A-36…

A: For the following diagram, calculate the force needed to stretch the wire AB by 5%. Wire is A-36…

Q: 3XPL@!N THE PROCESS @ND PROVIDE EXAMPLES FORMING OPERATIONS 1. Forging is mechanically working or…

A: Forging ,rolling , extrusion , drawing

Q: Determine the single uniformly distributed load in the x direction that wo produce the same…

Q: You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior…

A: Given:To find:Yield strength

Concept explainers

Forming and Shaping

Metal forming is a category of metalworking process through which the material can be given the desired shape and size by subjecting it to plastic deformation, where the component undergoes bulk deformation process upon application of loads. Unlike the shearing and machining process, the material does not undergo up to a failure point as indicated in the stress-strain curve. Hence, the stresses induced during the forming process are greater than yield strength but less than the failure strength of the material. The process does not lead to the loss of material, so it can be said to be an economical process. During forming, the material is subjected to tensile forces, compressive forces, bending, and shearing loading conditions. The material should possess the property of ductility to undergo the forming process.

Question

### Cup Drawing: Determining Drawing and Blank-Holding Forces

For the drawing operation of Problem 3 (Cup Drawing), the sheet metal used is low-carbon steel with the following properties:
- **Tensile strength** = 300 MPa
- **Yield strength** = 175 MPa
- **Die corner radius** = 5 mm

Tasks:
1. Determine the **drawing force**.
2. Determine the **blank-holding force**.

Given:
- **Blank diameter** = 150 mm
- **Punch diameter** = 100 mm

### Equations Used

#### Drawing Force:
\[
F = \pi D_p t (TS) \left( \frac{D_b}{D_p} - 0.7 \right)
\]

Where:
- \( F \) = Drawing Force
- \( D_p \) = Punch Diameter
- \( t \) = Thickness of the sheet
- \( TS \) = Tensile Strength
- \( D_b \) = Blank Diameter

#### Blank-Holding Force:
\[
F_h = 0.015 Y \pi \left( D_b^2 - \left( D_p + 2.2t + 2R_d \right)^2 \right)
\]

Where:
- \( F_h \) = Blank-Holding Force
- \( Y \) = Yield Strength
- \( R_d \) = Die Corner Radius

These equations will allow the calculation of forces required in the cup drawing process, ensuring proper material formation and avoiding defects.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

In a slab rolling operation, the maximum thickness reduction (del hmax) is given by del hmax = μ²R, where R is the radius of the roll and u is the coefficient of friction between the roll and the sheet. If µ = 0.1, find the maximum angle subtended by the deformation zone at the centre of the roll.
Please solve d through g
...). A 450 mm-long sheet with a cross sectional area of 320 mm² is stretched with a force, F, until α = 22°. The material has a true-stress-true-strain curve o = 750 x 106 0.35 (a) Find the total work done, ignoring end effects and bending. (b) What is a max before necking begins? F 300 mm - 150 mm
what are t three components of shear strain?
A 250 mm diameter circle is etched on a 400 mm x 400 mm x 12 mm piece of sheet metal. The elastic modulus is 190 GPa and Poisson's ratio is 0.3 for the material If stresses of 40 MPa and 75 MPa are applied in the x and y directions respectively, calculate the following: a) Change in sheet thickness, i.e. find At. Hint engineering and true strain are equal in the elastic regime. b) Sketch the shape of the circle after deformation on the figure below c) What is the elastic strain energy in the thickness direction only?
Given: Flat rolled sheets are produced in wide rolling mills, but many products are manufactured from strip stock. The below figure is a free-body diagram of a shaft for a short slitting line where a single blade is placed in the center of the shaft and a motor drives the shaft through a pulley at the far right end. Find: The shaft diameter If the maximum shear stress is 40 MPa and the largest gage sheet causes a blade force of 2000 N,