Tuesday, November 19, 2019
Ground Engineering Essay Example | Topics and Well Written Essays - 1500 words
Ground Engineering - Essay Example Pile foundations are chosen for the situations where structures transmit huge loads and suitable soil layer capable of supporting these loads are available only at greater depths. The piles are long and slender members capable of transmitting the loads of the structure into the hard soil layer or the rocks successfully even if very poor or soft soils that are considered to be inferior to meet this purpose is present in the upper layers. Based on the load carrying function piles are classified as end bearing pile, friction pile or combination pile that undertakes both the functions appropriately. The friction piles are suitable for the locations where granular soil is present and frictional resistance between the pile and the soil could be utilised in the resistance. While the end bearing piles helps to transmit the entire load a hard stratum through the tip resistance. A sketch of the load action on the end bearing pile and friction pile are shown in the figure 1. The basic structura l components of any pile are pile cap and the body of the pile. And (iii) All piles carry only vertical force. ... Figure 1(a) : End bearing pile Figure 1 (b) : Friction pile (Abebe and Smith, 2005) (Abebe and Smith, 2005) Pile load computation Following assumptions are incorporated in the load estimation on piles (i)All piles are considered as rigid (ii) The pile is pinned at the top and bottom And (iii) All piles carry only vertical force. The different loads that the pile need to carry would be direct loads from the columns and any other imposed loads on the structure. The direct loads from the columns to the pile are already given in the problem. The other loads are the load from the glass facades and the imposed loads. The glass faade load is the load along the edges of the floor and it could be computed by multiplying the unit load of the faade by the distance between the columns. The details of computation is as follows Glass faade load between the columns 1 & 8 Distance between the columns 1 and 8 = 20 metre Unit load of facde between the column 1 and 8 = 35 KN/m The load from the faade is assumed to be transferred equally to both the columns . So the loads are computed as 35 20 / 2 = 350 KN So the columns 1 and 8 would carry load of 350 KN each from the glass facades. Glass faade load between the columns 4 & 5 Distance between the columns 4 and 5 = 20 metre Unit load of facde between the column 4 and 5 = 35 KN/m The load from the faade is assumed to be transferred equally to both the columns . So the loads are computed as 35 20 / 2 = 350 KN So the columns 4 and 5 would carry load of 350 KN each from the glass facades placed between the columns 4 and 5. Glass faade load between the columns 1 and 4 The columns in this span are 1,2,3
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.