Formwork for Concrete Structures by R.L.Peurifoy and G.D- By EasyEngineering.net

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Design of Wood Members for Formwork 101
For a single-span beam with a uniform load of w over its entire
length of L:
Total vertical load = wL
All loads within a distance d from the supports = w(2d/12)
Therefore, the net total vertical load = wL – w(2d/12)
For this condition, the shear force V is
V = (wL/2) – w(d/12)
= (wL/2)[1 – 2d/l] (5-16)
Substituting L = l/12, Eq. (5-16) for shear in a single-span beam
can be written in the following form:
V = wl[1–2d/l]/(24)
(5-16a)
From Eq. (5-13), the total shear for a beam is
V = 2f v
bd/3
(5-16b)
ww.EasyEngineering.n
Combining Eqs. (5-16a) and (5-16b) and solving for f v
gives:
2f v
bd/3 = wl[1 – 2d/l]/24
f v
= 3wl[1 – 2d/l]/[2 × 24bd]
= wl[1 – 2d/l]/16bd
= w[l – 2d]/16bd (5-17)
For multiple-span beams the shear force V can be calculated as
follows:
V= (5wL/8) – [(5w/8) × (2d/12)]
= (5wL/8)[1 – 2d/1] (5-18)
Substituting L = l/12, Eq. (5-18) for shear in a multiple-span beam
can be written in the following form:
V = 5wl[1 – 2d/l]/96
(5-18a)
From Eq. (5-13), the total shear in a beam is
V = 2f v
bd/3
(5-18b)
Combining Eqs. (5-18a) and (5-18b) and solving for f v
gives:
2f v
bd/3 = 5wl[1 – 2d/l]/96
f v
= 15wl[1 – 2d/l]/192bd
= 15w[l – 2d]/192bd (5-19)
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