Question

***NOTE*** you *MUST draw complete FBD, and complete Shear-Force and Bending-Moment diagrams Select the lightest-weight steel wide-flange (W) overhanging beam from Appendix B that will safely support the loading. Assume the support at A is a pin and the support at B is a roller. The allowable bending stress is allow (see email data) and the allowable shear stress is "Zaitow -- 14 ksi. 8 ft + Non 4ft af Selected beam section is: W X P (kip) = 3 Kip Ksi = 24 Kri (see email data for value of P)

Answer 1

Given data:

The maximum allowable bending stress, $\sigma$ = 24 ksi

The maximum allowable shear stress, $\tau$ = 24 ksi

P = 3 kip

Solution:

Draw the free body diagram of the beam.

8 ft 2 ft 4 ft B C D B 3 kip 3 kip

Write the equilibrium equation of forces along the x-axis.

ΣF = 0 Α, = 0

Take the moment of all the forces at point A.

ΣΜΑ = 0 3x 14 + 3 x 10 - By X8 = 0 By = 9 kip

Write the equilibrium equation of forces along the x-axis.

$\\\sum F_{y}=0\\ A_{y}+B_{y}-3-3=0\\ A_{y}+9-3-3=0\\ A_{y}=-3 \text{ kip}$

The shear force details at x = 0 to 14 ft

Shear force at x = 0 is -3 kip.

Shear force at x =8 ft is -3+9 = 6 kip.

Shear force at x =10 ft is 6-3 = 3 kip

Shear force at x =14 ft is 3-3 = 0

The bending moment details at X = 0 to 14 ft.

Bending moment at x =0 is 0

Bending moment at x =8 ft is -3x8 = -24 kip-ft

Bending moment at x =10 ft is -3x10 + 9x2 = -12 kip-ft

Bending moment at x =14 ft is (-3x14 +9x6 - 3x4 )=0

The shear force and bending moment diagrams are as follows.

8 ft 21 ft 4 ft A B C D X BFD By 3 kip 3 kip 6 kip 3 kip +ve 0 SFD -ve - 3 kip BMD 0 -ve 12 kip-ft -24 kip-ft

From the moment diagram, the maximum bending moment is

$M_{max}= 24 \text{ kip}\cdot \text{ft}$

The maximum bending stress formula is

$\sigma _{max}=\frac{M_{max}}{S}$

Here = 24 ksi,

mui

s = section modulus

substituting the respective values

$\\24 = \frac{24*12}{S}\\ S = 12 \text{ inch}^{3}$

Select W 6 x 20 which has S = 13.4 inch³ d = 6.28 inch and $t_{w}$ = 0.26 inch.

From the shear force diagram,

kip

Now, provide the shear stress check for W 6 x 20

$\tau _{max} = \frac{V_{max}}{t_{w}\times d}$

$\tau _{max} = \frac{6}{0.26\times 6.28}$

$\tau _{max} = 3.67 ksi < \tau _{allow }=14 \text{ ksi}$

so the selected wide flange beam W 6 x 20 is acceptable.