Question 35.8: Design a concrete caisson with a W-section (steel) at the co...
Design a concrete caisson with a W-section (steel) at the core to carry a load of 1,000 tons. Assume the skin friction to be 150 psi and the end bearing to be 200 psi. See Fig. 35.13. The following parameters are given. The ultimate steel compressive strength is 36,000 psi, and the ultimate concrete compressive strength is 3,000 psi.
STEP 1: Structural design of the caisson.
Assume a diameter of 30 in. for the concrete caisson. Since E (the elastic modulus) of steel is much higher than for concrete, the major portion of the load is taken by the steel. Assume 90% of the load is carried by the steel.
load carried by steel = 0.9 × 1,000 tons = 900 tons
allowable steel compressive strength = 0.5 ×36,000 = 18,000 psi
(It should be noted that in the New York City Building Code, Table 11.3 of that code recommends a factor of safety of 0.5. Check your local building code for the factor of safety value.)
\text { steel area required }=\frac{(900 \times 2,000)}{18,000}=100 \text { sq in. }
Check the manual of steel construction for an appropriate W-section.
Use W 14 × 342. This section has an area of 101 sq in. The dimensions of this section are given in Fig. 35.14.
STEP 2: Check whether this section will fit inside a 30 in. hole.
The distance along the diagonal can be calculated using the Pythagorean theorem.
\left(16.36^{2}+17.54^{2}\right)^{1 / 2}=23.98 in
This value is smaller than 30 in. Hence the section can easily fit inside a 30 in. hole.
STEP 3: Compute the load carried by concrete.
concrete area = area of the hole – area of steel
= 706.8- 101 = 605.8
allowable concrete compressive strength = 0.25
× ultimate compressive strength
= 0.25 × 3,000 = 750 psi
(It should be noted that in the New York City Building Code, Table 11.3 of that code recommends a factor of safety of 0.25. Engineers should refer to local building codes for the relevant factor of safety values.)
load carried by concrete = concrete area × 750 psi
= 605.8 × 7501b
= 227.6 tons
load carried by steel = 900 tons (computed earlier)
total capacity of the caisson = 900 + 227.6 = 1,127.6 tons > 1,000 tons
Note that the designer can start with a smaller steel section to optimize the above value.
STEP 4: Compute the required length, L, of the caisson.
The skin friction is developed along the perimeter of the caisson.
total perimeter of the caisson = π × (diameter) × (length) = π × D × L
total skin friction = π × 30 × L × unit skin friction of rock
(in this case 150 psi)
total skin friction = π × 30 × L × 150 lb
(L should be in units of inches)
Design the caisson so that 95% of the load is carried by skin friction
0.95 × 1,000 tons = 950 tons
Hence, the total load carried through skin friction is 950 tons.
total skin friction =π × 30 × L × 150 = 950 tons = 950 × 2,000 lb
length of the caisson required, L = 134 in. = 11.1 ft
