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

38 Chapter Three
Example 3-7
A floor system has supports at 20 ft in each direction that must support
an 8-in.-thick slab of concrete whose weight is 150 lb per cu ft.
Motorized buggies will be used to transport and place the concrete.
The vertical concentrated load acting on the support can be calculated
as follows:
Dead load of concrete = (150 lb per cu ft)[(20 ft × 20 ft) × (8/12 ft)]
= 40,000 lb
Estimated dead load of forms and hardware
= (10 lb per sq ft) (20 ft × 20 ft)
= 4,000 lb
Live load of workers using buggies = (75 lb per sq ft)(20 ft × 20 ft)
= 30,000 lb
Total design load = 44,000 lb dead load + 30,000 lb live load
= 74,000 lb
Example 3-8
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A continuous concrete beam is 16 in. wide and 24 in. deep, with 150
lb per cu ft concrete. The uniform vertical load per foot acting on the
bottom of the beam form can be calculated as:
Dead load of concrete = (150 lb per cu ft)(16/12 ft)(24/12 ft)
= 400 lb per lin ft
Estimated dead load of form material = (5 lb per sq ft)(16/12)
= 3 lb per lin ft
Live load of workers and tools = (50 lb per sq ft)(16/12)
= 67 lb per lin ft
Total design load = 403 dead load + 67 live load
= 470 lb per lin ft
Example 3-9
Concrete that weighs 150 lb per cu ft is placed for an elevated
6-in.-thick slab. Motorized buggies will not be used. The vertical pressure
on the slab forms can be calculated as:
Dead load of concrete = (150 lb per cu ft)(6/12 ft)
= 75 lb per sq ft
Estimated dead load of form material = 10 lb per sq ft
Live load of workers = 50 lb per sq ft
Total design load = 85 dead load + 50 live load
= 135 lb per sq ft
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