Question 7.20: An axle supported on two bearings carries a vertical load of......

Allowable strength

\sigma_{\mathrm{ult}}=630 \mathrm{~MPa}

\sigma_y=380 \mathrm{~MPa}

\sigma_{\mathrm{all}}=0.6 \times 380=228 \mathrm{~MPa}

or = 0.36 × 630 = 226.8 MPa

Hence, \sigma_{\mathrm{all}}=226.8 MPa

The loading on the axle is shown in Figure 7.23.

Reaction at the bearing supports

Let R_A \text { and } R_B are supports at two bearings.

From the force equilibrium,

R_A+R_B=80000+5 \times 1000 \times 1.9=89500 \mathrm{~N}

Taking moment about A

R_B \times 1900-80000 \times 500-(5000 \times 1.9) \times \frac{1900}{2}=0

R_B=25802.63 \mathrm{~N}

R_A=89500-25802.63=63697.37 \mathrm{~N}

Taking any section at a distance x from end A, the bending moment equation is

M_x=R_A \times[x]-80 \times 10^3[x-500]-(5000 \times 1.9) \times\left[x-\frac{1900}{2}\right] (7.73)

At x = 500 mm

M_1=63697.37 \times[500]=31848.685 \times 10^3 \mathrm{~N} \cdot \mathrm{mm}

At section at x = 950 mm

M_x=63697.37[950]-80 \times 10^3[950-500]

=24512.5 \times 10^3 \mathrm{~N} \cdot \mathrm{mm}

To obtain the maximum bending moment, the shear force equation can be used to locate the point of maximum bending moment.

The shear force (SF) equation from left is

\mathrm{SF}=63697.37-80 \times 1000-\left(5000 \times \frac{x}{1000}\right) \text {, where } x \text { is in } \mathrm{mm} .

The above equation should be evaluated stepwise.

Between 0 \leq x<500

\mathrm{SF}=63697.37-\left(5000 \times \frac{x}{1000}\right)

At point A, \mathrm{SF}=63697.37 \uparrow \mathrm{N}

At point C and x < 500 mm

\mathrm{SF}=63697.37-5000 \times \frac{500}{1000}=61197.37 \uparrow \mathrm{N}

At point C, \mathrm{SF}=61197.37-80000=-18802.63 \downarrow \mathrm{N}

At point B, \mathrm{SF}=-18802.63-\frac{5000 \times(1900-5000)}{1000}=-25802.63 \downarrow \mathrm{N}

Shear force diagram is shown in Figure 7.24.

The bending moment is maximum at x = 500 mm.

M_{\max }=R_A \times[x]-\left(5000 \times \frac{x}{1000}\right) \times\left[\frac{x}{2}\right]

M_{\max }=63697.37 \times[500]-\left(5000 \times \frac{500}{1000}\right) \times\left[\frac{500}{2}\right]

M_{\text {max }}=M=31223685 \mathrm{~N} \cdot \mathrm{mm}

d=\left(\frac{32 \times 31.22 \times 10^6}{\pi \times 226.8}\right)^{1 / 3}=111.9 \mathrm{~mm}

d = 112 mm