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Tutorial, Step Stool, Step 8, extrude/remove or pocket leaves a thin wall

ali_e280ali_e280 Member Posts: 10
Hi,
I'm trying to complete the the step stool exercise.  The instructions are to create on offset on the face and use extrude/remove to create a 5 mm indent.  When I apply the remove extrude, I get an unexpected thin wall.  This thin wall prevents the shell step following this step.



SKETCH





EXTRUDE OPTIONS






PART WITH THIN WALL



What step am I missing here?

Thanks,
Ali

Best Answer

  • _anton_anton Member, Onshape Employees Posts: 366
    Answer ✓
    Because the sides of the part are drafted (at a nonzero angle with respect to the vertical), and you're extruding straight down, you'll get a really thin wall with a triangular cross-section.

Answers

  • robert_scott_jr_robert_scott_jr_ Member Posts: 418 ✭✭✭
    Hello. It looks like you've selected the wrong sketch area for the extrude. I read the intent in the design is to create a depression on the step surface in order to have a corrugated surface for traction. The sketch area you selected is the 5mm perimeter. I believe you need to select the large inside area of that perimeter. - Scotty
  • ali_e280ali_e280 Member Posts: 10
    Thank you.  Looks like you were right about the intent of the exercise, but the question still remains, why is there a thin wall if I remove the 5mm area around the part?
  • _anton_anton Member, Onshape Employees Posts: 366
    Answer ✓
    Because the sides of the part are drafted (at a nonzero angle with respect to the vertical), and you're extruding straight down, you'll get a really thin wall with a triangular cross-section.
  • ali_e280ali_e280 Member Posts: 10
    Thank you.  Looks like you were right about the intent of the exercise, but the question still remains, why is there a thin wall if I remove the 5mm area around the part?
    _anton said:
    Because the sides of the part are drafted (at a nonzero angle with respect to the vertical), and you're extruding straight down, you'll get a really thin wall with a triangular cross-section.

    Thank you for the explanation.
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