1- Given Data Column Footing fé fy D = 350 Kips = 250 Kips = 3.25 ksi = 60 ksi 40 in x 30 in qa = 2.7 ksf fy fe = 60 ksi = 3.25 ksi Find: suitable rectangular footing design for L/B = 1.6. • There is no need to determine the development length. • Iksf = 47.5 kPa В X= 40 in L 2- If the footing of previous question is on the following soil layers, estimate the settlement at the corner of it. Soil is layered clays with on sand seam from footing base to silt layer at -19 m; footing at -1 m. (u = v = 0.3) a. Es from 1 m to 4 m = 35 Mpa b. Es from 4 m to 9 m = 60 Mpa c. Es from 9 m to 13 m = 75 Mpa d. Es from 13 m to 16 m = 80 Mpa e. Es from 16 m to 19 m 800 Mpa %3D Y=30 in

1- Given Data Column Footing fé fy D = 350 Kips = 250 Kips = 3.25 ksi = 60 ksi 40 in x 30 in qa = 2.7 ksf fy fe = 60 ksi = 3.25 ksi Find: suitable rectangular footing design for L/B = 1.6. • There is no need to determine the development length. • Iksf = 47.5 kPa В X= 40 in L 2- If the footing of previous question is on the following soil layers, estimate the settlement at the corner of it. Soil is layered clays with on sand seam from footing base to silt layer at -19 m; footing at -1 m. (u = v = 0.3) a. Es from 1 m to 4 m = 35 Mpa b. Es from 4 m to 9 m = 60 Mpa c. Es from 9 m to 13 m = 75 Mpa d. Es from 13 m to 16 m = 80 Mpa e. Es from 16 m to 19 m 800 Mpa %3D Y=30 in

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Load and resistance factor design

Load and resistance factors design (LRFD), also known as Limit state design (LSD), refers to the design method used in building engineering. A boundary condition is a condition of a structure where it can no longer meet the relevant design requirements. The condition may refer to the load level or other actions on the structure, while the terms refer to the integrity of the structure, suitability for use, durability, or other design requirements. A structure designed for LRFD is limited to keeping track of all possible actions during its design time and is always ready for use, with the appropriate level of reliability in each boundary condition. LRFD-based construction codes clearly define the appropriate levels of reliability in their instructions.

Question

1- Given Data
Footing
= 3.25 ksi
= 60 ksi
Column
= 350 Kips
= 250 Kips
D
L
fy
40 in x 30 in
qa
= 2.7 ksf
= 60 ksi
fy
fe
= 3.25 ksi
Find: suitable rectangular footing design for L/B = 1.6.
• There is no need to determine the development length.
• Iksf = 47.5 kPa
B
X= 40 in
L
2- If the footing of previous question is on the following soil layers, estimate the settlement
at the corner of it. Soil is layered clays with on sand seam from footing base to silt
layer at -19 m; footing at -1 m. (u = v = 0.3)
a. Es from 1 m to 4 m = 35 Mpa
b. Es from 4 m to 9 m = 60 Mpa
c. Es from 9 m to 13 m 75 Mpa
d. Es from 13 m to 16 m 80 Mpa
e. Es from 16 m to 19 m = 800 Mpa
Y=30 in

1. Find Pus l4D+ 17h (w asually absent)
2 Detenmime Band L; B =Q+L4 Fer
unique Solutiom, B orL is fixed.
8. Find u-Pa/BL
4. Dutemime ve fr diagamal tension
5 Determmine d Vria appličable Egs. 6-6,
6-6a g6-6b. Fora rectamgular fohings cdeck
wide - beam Shear; Hue darger d gowems
6. Compute Hhe mequired flexune steelydeubp-
ment lemg thyetc, oneach direction. The mom-
ent is Computed at the emhcal sechions as
Specified by Code
7. Compute a desgn drawing all details
showing

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Determine the elastic settlement using the Schmertmann method of the 10'x 10' footing as shown below. Estimate the elastic modulus using Es = 10(N + 15), where Es is in ksf and N is the corrected SPT value. Loose sand Medium dense sand Loose sand Dense sand Very loose sand DEPTH 5° 10' 15' 20' 25' SPT 10 21 11 15 41 4 Q=250 kips 10' x 10' 0 0.2 0.4 0.6 STRAIN & Layer 1 Layer 2 Layer 3 Layer 4 Laver 5
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Foundation example
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