Question

A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maximum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.

Answer 1

W = 11,416.6879 N

L ≈ 64.417 cm

Step-by-step explanation:

The maximum shear stress,
`\tau_(max)`, is given by the following formula;

`\tau_(max) = (W)/(8 \cdot I_c \cdot t_w) * \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )`

`t_w` = 1 cm = 0.01

h = 29 cm = 0.29 m

`h_w` = 25 cm = 0.25 m

b = 15 cm = 0.15 m

`I_c` = The centroidal moment of inertia

`I_c = (1)/(12) \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )`

`I_c` = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

`I_c = (1)/(12) \cdot \left (0.15 * 0.29^3 - 0.15 * 0.25^3 + 0.01 * 0.25^3 \right ) = 1.2257083\bar 3 * 10^(-4)`

`I_c` = 1.2257083
`\bar 3` × 10⁻⁴ m⁴

From which we have;

`4,500,000 = (W)/(8 * 1.225708\bar 3 * 10 ^(-4)* 0.01) * \left (0.15 * 0.29^2 - 0.15 * 0.25^2 + 0.01 * 0.25^2 \right )`

Which gives;

W = 11,416.6879 N

`\sigma _(b.max) = (M_c)/(I_c)`

`\sigma _(b.max)` = 1500 N/cm² = 15,000,000 N/m²

`M_c` = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

`M_(max) = (W \cdot L)/(4)`

`L = (4 \cdot M_(max))/(W) = (4 * 1838.5625)/(11,416.6879) \approx 0.64417`

L ≈ 0.64417 m ≈ 64.417 cm.