Question

1.8 Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 mm², determine the magnitude of the load P for which the normal stress in that portion of BD is 50 MPa. 1.9 Knowing that the link DE is 3 mm thick and 25 mm wide, determine the normal stress in the central portion of that link when (a) ? = 0, (b) ? = 90°.

          1.8 Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 mm², determine the magnitude of the load P for which the normal stress in that portion of BD is 50 MPa.

1.9 Knowing that the link DE is 3 mm thick and 25 mm wide, determine the normal stress in the central portion of that link when (a) ? = 0, (b) ? = 90°.

1.8 Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 mm², determine the magnitude of the load P for which the normal stress in that portion of BD is 50 MPa.

1.9 Knowing that the link DE is 3 mm thick and 25 mm wide, determine the normal stress in the central portion of that link when (a) ? = 0, (b) ? = 90°.

Added by Jose C.

Mechanics of Materials

R. C. Hibbeler 10th Edition

Chapter 1

Instant Answer

Solved by Expert Dominador Tan

06/30/2023

Step 1

Given: - Cross-sectional area \( A = 800 \, \text{mm}^2 \) - Normal stress \( \sigma = 50 \, \text{MPa} \) The formula for normal stress is: \[ \sigma = \frac{P}{A} \] Rearrange to solve for \( P \): \[ P = \sigma \times A \] Substitute the given values: \[ P Show more…

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