13 Select the lightest W section for the situation shown in

13. Select the lightest W section for the situation shown in the accompanying figure using A572 Grade 50 steel, assuming lateral support at the ends and at point A only. Determine the Cb factor as per AISC-FI. 4.2 k/ft (does not include weight of beam) 35 kips 35 kips t. 10\' 10\' 10\' 10\' 20\' 30\"

Solution

step 1:

Given data:

A572 Steel

Grade 50

step 2:

To select the lightest possible W section :

First calculate the maximum bending moment on the section and then using Z_x tables select the adequate one as per design strength requirement of shear force and bending moment

step 3:

Analysis of B.M and S.F:

consider the left end point as B and right end point as C

Taking the equilibrium in y direction:

R_b+R_c=35+35+4.2(30)

R_b+R_c=196 kips

taking moment about the point B

R_b(30)-35(20)-35(10)-4.2x30x15=0

R_b=98 kips

R_a=98 kips

For finding out the maximum bending moment consider a section at x from left side and equate it to zero, so that we can find out the maximum bending moment :

Taking equilibrium of SF from left side:

98-35-4.2x=0

x=15 feet

now calculate the BM at x= 15 feet

M_max=R_b(x)-4.2x²/2-35(x-10)

M_max=98x4-4.2x15x15/2-35(5)

M_max=94.5 ft-kips

The maximum shear force occurs at supports so consider the reaction as the maximum shear force:

V_max=98 kips

step 4:

selection of W section:

Consider W14x22

The design strength is given by M_p

M_p=F_yZ_x=50x29/12

M_p=120.83 ft-kipsx0.9

M_p=108.75 ft-kips

step 5:

Calculation of C_b:

C_b=12.5M_max/(2.5M_max+3M_a+4M_b+3M_c)

M_a= Moment at first quarter length

M_b= Moment at mid span

M_c= Moment at 3/4th of the span

due to symmetry

M_a=M_c

M_b= M_max

M_a=M_c

M_c=98x22.5-35(12.5)-35(2.5)-4.2x30x22.5/2-98(7.5)

M_a=472.5 ft-kips

C_b=12.5x94.5/(2.5x94.5+3x472.5+4x94.5+3x472.5)

C_b=0.34