Lecture Notes on Problem 1-23 from Mechanics of Materials

Overview

• Topic: Solving Problem 1-23 from Chapter 1 (Stress) of "Mechanics of Materials" by R.C. Hibbler.
• Previous problem: Problem 1-22 reviewed for context.
• Focus: Finding resultant internal loading at cross sections E and B.

Problem Setup

• A load of 120 Newton is applied to the handle.
• Aim: Determine internal loading at cross section E and in link BC.

Step 1: Analyzing Link BC

• When link is removed at point B, replace with a reaction force FBC.

• Remove support at point A; apply horizontal reaction AX and vertical reaction AY.

• Focus on calculating FBC using equilibrium equations:

  • Sum of Moments about Point A = 0 (Counterclockwise as positive):

    • Clockwise Moment (due to 120N load):
      • Calculation: 120 N * 500 mm (negative)
    • Counterclockwise Moment (due to FBC):
      • Calculation: FBC * cos(30°) * 50 mm (positive)

  • Equation:

    120 * 500 - FBC * cos(30°) * 50 = 0

• Solving for FBC:

  • Result: FBC = 1385.6 N or 1.3856 kN

Step 2: Internal Loading in Short Link BC

• After cutting link BC:
  • Resultant internal loading includes:
    • Shear Force V
    • Normal Force N
    • Moment M

Equilibrium Equations

1. Sum of Forces along Y = 0:

   • FBC - Normal Force = 0
   • Normal Force N = 1385.6 N or 1.3856 kN

2. Sum of Forces along X = 0:

   • Only shear force V present.
   • Result: V = 0

3. Sum of Moments about Point 1 = 0:

   • Only external moments produce a result of zero.
   • Thus, Moment = 0

Step 3: Analyzing Handle at Point E

• Free Body Diagram drawn for handle at point E with:
  • Load of 120 N coming down.
  • Forces include:
    • Normal Force NE
    • Shear Force VE
    • Moment ME

Equilibrium Equations for Point E

1. Sum of Forces along X' = 0:

   • Result: NE = 0

2. Sum of Forces along Y' = 0:

   • VE - 120 N = 0
   • Result: VE = 120 N

3. Sum of Moments about Point E = 0:

   • Moments considered:
     • Counterclockwise Moment (ME)
     • Clockwise Moment (120 N * 400 mm)
   • Resulting equation:
     • ME - (120 N * 0.4 m) = 0
     • Result: ME = 48 N*m

Summary of Results

• Resultant internal loadings found:
  • At Cross Section E:
    • NE = 0
    • VE = 120 N
    • ME = 48 N*m
  • In Short Link BC:
    • Normal Force N = 1.3856 kN
    • Shear Force V = 0
    • Moment M = 0

Conclusion

• Successfully solved Problem 1-23 and reviewed Problem 1-22.
• Encouragement to ask questions and subscribe for future content.