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b) Channel Section: (4-Stringer Problem) • Example 5: V 2 3 1 4 F2 F1 F4 Given: A1 = A2 = A3 = A4 = 300mm² V = 4.8 KN Required: -F1, F2, F3, F4, e -Shear flow in all webs Solution: -Section is symmetric w.r.t neutral axis; -Similar to the 2-stringer problem, it's easy to prove through the compatibility relation that, since $\frac{F_i}{A_i\delta_i}$ = constant Then, F1 = F2 = F3 = F4

          b) Channel Section: (4-Stringer Problem)
• Example 5:
V
2
3
1
4
F2
F1
F4
Given:
A1 = A2 = A3 = A4 = 300mm²
V = 4.8 KN
Required:
-F1, F2, F3, F4, e
-Shear flow in all webs
Solution:
-Section is symmetric w.r.t neutral axis;
-Similar to the 2-stringer problem, it's easy to prove through the compatibility
relation that, since
$\frac{F_i}{A_i\delta_i}$ = constant
Then,
F1 = F2 = F3 = F4
        
Show more…
b) Channel Section: (4-Stringer Problem)
• Example 5:
V
2
3
1
4
F2
F1
F4
Given:
A1 = A2 = A3 = A4 = 300mm²
V = 4.8 KN
Required:
-F1, F2, F3, F4, e
-Shear flow in all webs
Solution:
-Section is symmetric w.r.t neutral axis;
-Similar to the 2-stringer problem, it's easy to prove through the compatibility
relation that, since
(Fi)/(Ai) = constant
Then,
F1 = F2 = F3 = F4

Added by Francisco Javier M.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Solve it quickly. Channel Section: (4-Stringer Problem) Exam 15: Given: A = 300mm, V = 4.8KN Required: - Shear flow in all webs Solution: - The section is symmetric with respect to the neutral axis. - Similar to the 2-stringer problem, it's easy to prove through the compatibility relation that, since Then, F = F = F = 16.
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Transcript

-
00:01 Shear stress is given by tau equals to mu times du by dy.
00:09 Now this u is the equation of the center line where u as a function of y is given in this problem as q naught into 1 minus y square by b square...
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