A wide-flange beam (see figure) is subjected to a shear force V. t- 0 Using the following dimensions of the cross section, calculate the moment b = 6 in., t = 0.5 in., h = 10 in., h₁ = 8.4 in., V = 28 kips Note: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles. (a) the maximum shear stress max (in psi) in the web 6496.51✔✔✔ psi max aver (b) the minimum shear stress min (in psi) in the web 5414.99✔ psi Taver (c) the average shear stress Taver (obtained by dividing the shear force by the area of the web) and the ratio max (Enter your answer for Taver in psi.) Taver = 6666.67✔✔✔ psi Ⓡ = 0.974✔ = (d) the shear force Vweb (in kips) carried in the web and the ratio web V Vweb = x kips Vweb inertia and then determine the following quantities.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Please answer part d. Thank you

A wide-flange beam (see figure) is subjected to a shear force V.
t-
0
Using the following dimensions of the cross section, calculate the moment
b = 6 in., t = 0.5 in., h = 10 in., h₁ = 8.4 in., V = 28 kips
Note: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles.
(a) the maximum shear stress max (in psi) in the web
6496.51✔✔✔ psi
max
aver
(b) the minimum shear stress min (in psi) in the web
5414.99✔
psi
Taver
(c) the average shear stress Taver (obtained by dividing the shear force by the area of the web) and the ratio max (Enter your answer for Taver in psi.)
Taver = 6666.67✔✔✔ psi
Ⓡ
= 0.974✔
=
(d) the shear force Vweb (in kips) carried in the web and the ratio web
V
Vweb = x kips
Vweb
inertia and then determine the following quantities.
Transcribed Image Text:A wide-flange beam (see figure) is subjected to a shear force V. t- 0 Using the following dimensions of the cross section, calculate the moment b = 6 in., t = 0.5 in., h = 10 in., h₁ = 8.4 in., V = 28 kips Note: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles. (a) the maximum shear stress max (in psi) in the web 6496.51✔✔✔ psi max aver (b) the minimum shear stress min (in psi) in the web 5414.99✔ psi Taver (c) the average shear stress Taver (obtained by dividing the shear force by the area of the web) and the ratio max (Enter your answer for Taver in psi.) Taver = 6666.67✔✔✔ psi Ⓡ = 0.974✔ = (d) the shear force Vweb (in kips) carried in the web and the ratio web V Vweb = x kips Vweb inertia and then determine the following quantities.
Step 1 Find Moment of inertia
Note: As per our guidelines we can answer 1st 3 sub-parts only.
- Moment of inertia about neutral axis is given by :
3
1/₂2 (bh²³ - bh₁³ + the ³)
I = (6 × 10³6 × 8.4³ + 0.5 × 8.4³)
228.344 inª
I
=
I =
Step 2 Find Maximum shear stress in web
Maximum shear stress in web is given by;
2
(bh² - bh₁² + th₁²)
max =
Smax
Smax
5min
Smin
5min
Savg
=
=
Step 3 Find Minimum shear stress in web
Vh (h² h²1)
Same
=
=
Smur
Save
=
=
V
8 11
Step 4 Find Average shear stress in web
Average shear stress in web is given by :
Save
=
28 x 10³
8 x 228.344 x 0.5
6496.51 psi
28 x 10 x6
8 x 228.344 x 0.5
5414.99 psi
=
th₂
Savg
6666.67 psi
Find ratio for max to average;
28 x 10³
0.5 x 8.4
x (6 x 10²6 x 8.4² +0.5 x 8.4²)
6491.51
6666.67
x (10² - 8.4²)
= 0.974
Transcribed Image Text:Step 1 Find Moment of inertia Note: As per our guidelines we can answer 1st 3 sub-parts only. - Moment of inertia about neutral axis is given by : 3 1/₂2 (bh²³ - bh₁³ + the ³) I = (6 × 10³6 × 8.4³ + 0.5 × 8.4³) 228.344 inª I = I = Step 2 Find Maximum shear stress in web Maximum shear stress in web is given by; 2 (bh² - bh₁² + th₁²) max = Smax Smax 5min Smin 5min Savg = = Step 3 Find Minimum shear stress in web Vh (h² h²1) Same = = Smur Save = = V 8 11 Step 4 Find Average shear stress in web Average shear stress in web is given by : Save = 28 x 10³ 8 x 228.344 x 0.5 6496.51 psi 28 x 10 x6 8 x 228.344 x 0.5 5414.99 psi = th₂ Savg 6666.67 psi Find ratio for max to average; 28 x 10³ 0.5 x 8.4 x (6 x 10²6 x 8.4² +0.5 x 8.4²) 6491.51 6666.67 x (10² - 8.4²) = 0.974
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