Shearing a lead slabसीसे की पट्टिका का अपरूपण

NCERT Fig 8.6 · interactive diagram

Parts and their functions

• Lead slab 50 cm × 50 cm × 10 cm (सीसे की पट्टिका 50 cm × 50 cm × 10 cm) — The force acts on the 50 cm × 10 cm face: A = 0.05 m².
• Shearing force F (अपरूपण बल F) — Stress = F/A = 9 × 10⁴/0.05 = 1.8 × 10⁶ N m⁻².
• Displacement Δx (विस्थापन Δx) — Δx = (F/A)L/G = 1.8 × 10⁶ × 0.5/5.6 × 10⁹ = 1.6 × 10⁻⁴ m = 0.16 mm.
• Height L (ऊँचाई L) — Shear strain = Δx/L.
• Fixed lower edge (स्थिर निचला किनारा) — The floor supplies the equal and opposite force.

Original diagram drawn by Simple Learning Hub, based on NCERT Fig 8.6. Report a mistake