Question

328 CHAPTER 11 Elasticity and Periodic Motion EXAMPLE 11.3 Minor earthquake damage Video Tutor Now let's look at the shear stress that can develop in an earthquake. Suppose the brass baseplate of an out- door sculpture experiences shear forces as a result of a mild earthquake. The plate is a square, 0.80 m on a side and 0.50 cm thick. How large a force Fi must be exerted on each of its edges if the displacement x in Figure 11.10 is 0.016 cm?The shear modulus of brass is listed in Table 11.3. SOLUTION SET UP Figure 11.11 shows our sketch. The shear stress at each edge → 0016 cm of the plate is the force divided by the area of the edge: Fi Shear stress thickness 4 (0.80 m) (0.0050 m) (250 m)F The shear strain is Shear strain =-= 1.6 × 10-4 m 2.0 x 10" 0.80 m 0.80 m SOLVE From Table 11.3, the shear modulus S is ▲ FIGURE 11.11 Our sketch for this problem. Ip(250 m-2)F Shear stress = 0.36 × 1011 Pa = 2.0×10-4 Practice Problem: We replace the plate with another bras has the same horizontal dimensions (0.80 m on a side) b ent thickness. If the shear forces have magnitude 1.2 the shear strain is the same as before, what is the plate Answer: 2.1 cm. Shear strain 2.0 × 10-4 Solving for Fi, we obtain Fi = 2.9 × 104 N. REFLECT This force is nearly 7000 lb, or roughly twice the weight of a medium-size car

For the Shear stress, I’m wondering how in the world you get 250m^2 from (.80m)(.0050m)?

Answer 1

Answer:

Shear stress = F₁ / A = F₁ / (0.80 m) (0.0050 m) = F₁ / (0.004 m²) = (250 m^-2) F₁ .

[ Since 1 / 0.004 m² = 250 m^-2 ]

I hope that you would understand.