Question

Information P Flag question HORIZONTAL SHEAR STRESS Find fy max for the beam shown below. Use actual dimensions shown For the built-up section Ix = 136 in4 W-200 lb ft L-101 6 in 2 in NA 6 in 5 in 2 in

Answer 1

given a simply supported beam with

uniformly distrubuted load (w) = 200 lb/ft

length of beam (L) = 10 ft

The cross section is given as T-beam with dimensions:

Width of flange (B_f) = 6 in

Thickness of flange (T_f) = 2 in

Thickness of web (T_w) = 2 in

Height of web (H) = 6 in

the distance of neutral axis is given as 5 inches from bottom of beam and (8-5)=3 inches from top of beam

• Question 12

V is shear force. Shear force is computed as:

The two vertical support reaction will be equal as the structure is symmetrical

So, vertical support reaction at each end (V) = (WL)/2

V = (200 * 10) / 2

therefore, V = 1000 lbs (ANSWER)

This is the maximum value of shear force because in simply supported beam with UDL , maximum shear force is at the supports. It can be shown below :

Equation for shear force :-

SF = 1000 - 200x [ x = distance from left end ]

putting x = 0 gives maximum value of shear force = 1000 lbs

• Question 13

Q is the moment of area about neutral axis.

6 ( 2 t = 2 Tees y -NA

moment of area of flange part = 6 * 2 * (2/2 + 1) = 24 in³

moment of area of web part = 1 * 2 * (1/2) = 1 in³

Total moment of area (Q) = 24 + 1 = 25 in³ (ANSWER)

• Question 14

f_{v max} is the maximum horizontal shear stress

Maximum horizontal shear stress will occur at neutral axis of the beam

Horizontal shear stress (f) is given by

  $f = \frac{VQ}{Ib}$

where, V = shear force

For maximum horizontal shear stress,  the maximum value of shear force is choosen

maximum value of shear force (V) = 1000 lbs

  Q = moment of area about neutral axis = 25 in³ [ calculated before ]

I = moment of inertia about neutral axis = 136 in⁴ [ given ]

b = width of section = 2 in

since maximum horizontal shear stress occur at neutral axis(NA) and neutral axis lies in web of beam, so width of section is taken as thickness of web

putting the values in above equation, we get:

$f_{v\: max}= \frac{1000 \times 25}{136 \times 2}$

f_{v max} = 91.91 lb/in²  (ANSWER)