Question

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. Determine the maximum shearing stress in shaft BC. (Round the final answer to one decimal place.) The maximum shearing stress in shaft BC is 56 MPa.

Answer 1

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

Answer 2

Shearing stress is a stress that acts coplanar to the surface. The maximum shearing stress on the BC part of the shaft is 25.37 MPa.

Shearing stress:

It the defined as the ratio of the force to the cross-sectional area where the force is applied.

$\tau_{max} = {\tau _{bc} \times \dfrac {r(bc)} { J(bc)}$

Given here,

$\tou_{bc}$= 273.374 Nm

$r_{bc}$= 19 mm

$J_{bc}$ = $\dfrac \pi 2 \times 19^4$

Put the values in the equation,

$\tau_{max} = 73.374\rm \ Nm \times \dfrac {19 \ mm} { \dfrac \pi 2 \times 19^4}\\\tau_{max} = 25.37\ MPa$

Therefore, the maximum shearing stress on the BC part of the shaft is 25.37 MPa.

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