Engineering Questions with Answers - Multiple Choice Questions
Stress Distribution – Newmark’s Influence Chart
1 - Question
_________ chart is used to find the vertical stress on Westergaard’s equation.
a) Influence chart
b) Isocurve chart
c) Isobar chart
d) Fenske’s chart
View Answer
Answer: a
Explanation: The influence chart is used to find the vertical stress on Westergaard’s equation. Whereas, an isobar is a curve or contour connecting all points below the ground surface or ground level of equal vertical pressure. In an isobar, at each and every point the vertical pressure is the same.
2 - Question
_________ is more accurate method of determining the vertical stress at any point.
a) Isobar chart
b) equivalent point load method
c) Influence chart
d) Fenske’s chart
View Answer
Answer: c
Explanation: The Influence chart or the influence diagram is a more accurate method of determining the vertical stress at any point under a uniformly loaded area of any shape than the equivalent point load method.
3 - Question
The Newmark’s influence chart consists of _________
a) a single circle only
b) a number of circles and radiating lines
c) bar diagram
d) small rectangular unit areas
View Answer
Answer: b
Explanation: The Newmark’s influence chart consists of a number of circles and radiating lines, is so prepared that the influence of each area unit is same at the centre of the circle.
4 - Question
If a uniformly loaded circular area is divided into 20 sectors, then the influence value if is given by ___________
a) 120[1−[11+(az)2]32]
b) 20[1−[11+(az)2]32]
c) 20[1−[11+(az)2]52]
d) 120[1−[11+(az)2]52]
View Answer
Answer: a
Explanation: Let a uniformly loaded circular area is divided into 20 sectors.
q=load
σz=vertical stress at depth z
The stress at each unit area will be σz20
∴ σz20=q20[1−[11+(az)2]32]
The influence factor is given by if=120[1−[11+(az)2]32].
5 - Question
If the influence value if=135[1−[11+(az)2]32] for a uniformly loaded circular area, then the circular area is divided into _________ sectors.
a) 20
b) 35
c) 7
d) 14
View Answer
Answer: b
Explanation: let the vertical stress be σz.
The Boussinesq’s vertical pressure σz under a uniformly loaded circular area is given by,
σz=q[1−[11+(az)2]32]
If the area is divided into 35 area units, then the stress in each unit is given by,
σz35=q35[1−[11+(az)2]32]
Therefore for if=135[1−[11+(az)2]32],
The circular area is divided into 35 sectors.
6 - Question
If a uniformly loaded circular area is divided into 44 sectors, then the influence value if is given by ___________
a) 144[1−[11+(az)2]32]
b) 44[1−[11+(az)2]32]
c) 44[1−[11+(az)2]52]
d) 144[1−[11+(az)2]52]
View Answer
Answer: a
Explanation: Let a uniformly loaded circular area is divided into 44 sectors.
q=load
σz=vertical stress at depth z
The stress at each unit area will be σz44
∴ σz44=q44[1−[11+(az)2]32]
The influence factor is given by if=144[1−[11+(az)2]32].
7 - Question
In Newmark’s influence chart method, the point below which pressure is required should lie within the loaded area.
a) True
b) False
View Answer
Answer: b
Explanation: In Newmark’s influence chart method, the point below which pressure is required may lie within the loaded area or outside the loaded area as pressure distributes uniformly in all directions in a homogeneous soil mass.
8 - Question
The vertical stress under the corner of a uniformly loaded rectangular area of size a, b at depth z and m=a/z, n=b/z is given by ___________
a) σz=2q′πz1[1+(xz)2]2
b) σz=q4π[2mn(m2+n2+1)√m2+n2+m2n2+1]
c) σz=q4π[2mn(m2+n2+1)√m2+n2+m2n2+1∗m2+n2+2m2+n2+1+tan−12mn(m2+n2+1)√m2+n2+m2n2+1]
d) σz=q[1−[11+(az)2]32]
View Answer
Answer: c
Explanation: The vertical stress under the corner of a uniformly loaded rectangular area of size a, b at depth z and m=a/z, n=b/z is given by,
σz=q4π[2mn(m2+n2+1)√m2+n2+m2n2+1∗m2+n2+2m2+n2+1+tan−12mn(m2+n2+1)√m2+n2+m2n2+1]
m and n are interchangeable terms. The above form of solution is after Newmark(1935).
9 - Question
The vertical stress under the corner of a uniformly loaded rectangular area of size 2m*4m at depth 5m and load of 80 kN/m2 is given by ___________
a) 6.22 kN/m2
b) 7.45 kN/m2
c) 8.12 kN/m2
d) 9.23 kN/m2
View Answer
Answer: b
Explanation: Given,
Load q=80 kN/m2
Size of rectangle=2m*4m
Depth z=5m
Therefore m=2/5=0.4
n=4/5=0.8
∴ the vertical stress at the corner of rectangle is given by,
σz=q4π[2mn(m2+n2+1)√m2+n2+m2n2+1∗m2+n2+2m2+n2+1+tan−12mn(m2+n2+1)√m2+n2+m2n2+1]
∴ σz=804π[2∗0.4∗0.8(0.42+0.82+1)√0.42+0.82+0.420.82+1∗0.42+0.82+20.42+0.82+1+tan−12∗0.4∗0.8(0.42+0.82+1)√0.42+0.82−0.420.82+1]
∴ σz=7.45 kN/m2.
10 - Question
The influence factor for the vertical stress under the corner of a uniformly loaded rectangular area of size 1m*2m at depth 5m and load of 80 kN/m2 is given by ___________
a) 0.6212
b) 0.7465
c) 0.0328
d) 0.0624
View Answer
Answer: c
Explanation: Given,
Load q=80 kN/m2
Size of rectangle=1m*2m
Depth z=5m
Therefore m=1/5=0.2
n=2/5=0.4
∴ the vertical stress at the corner of rectangle is given by,
σz=q4π[2mn(m2+n2+1)√m2+n2+m2n2+1∗m2+n2+2m2+n2+1+tan−12mn(m2+n2+1)√m2+n2+m2n2+1]
∴ σz=kq
The influence factor k=14π[2mn(m2+n2+1)√m2+n2+m2n2+1∗m2+n2+2m2+n2+1+tan−12mn(m2+n2+1)√m2+n2+m2n2+1]
k=14π[2∗0.2∗0.(0.22+0.42+1)√0.22+0.42+0.220.42+1∗0.22+0.42+20.22+0.42+1+tan−120.20.4(0.22+0.42+1)√0.22+0.42−0.220.42+1]
∴k=0.0328.
11 - Question
The Westergaard’s equation is given by ___________
a) σz=1[1+2(rz)2]32
b) σz=12[1+2(rz)2]32
c) σz=1π[1+2(rz)2]32Qz2
d) σz=1π[1+2(rz)2]32
View Answer
Answer: c
Explanation: The Westergaard’s equation is given by,
σz=1π[1+2(rz)2]32Qz2
Where Q=load
Z= depth of point at which load is found
r=radial distance.
12 - Question
The Westergaard’s influence factor is given by _____________
a) KW=1π[1+2(rz)2]32
b) KW=Qz2
c) KW=1π[1+2(rz)2]52Qz2
d) KW=1π[1+2(rz)2]3Qz2
View Answer
Answer: a
Explanation: The Westergaard’s equation is given by,
σz=1[1+2(rz)2]32Qz2
∴ σz=KWQz2.
Where KW is the Westergaard’s influence factor given by,
KW=1π[1+2(rz)2]32.
13 - Question
For a uniformly loaded rectangular area, the Newmark’s influence factor given by ___________
a) K=[20.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1∗0.22+0.42+20.22+0.42+1+tan−120.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1]
b) K=14π[20.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1∗0.22+0.42+20.22+0.42+1+tan−120.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1]
c) K=14π
d) K=q4π[20.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1∗0.22+0.42+20.22+0.42+1+tan−120.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1]
View Answer
Answer: b
Explanation: The Newmark’s influence factor for a uniformly loaded rectangular area of size a, b at depth z is given by,
K=14π[20.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1∗0.22+0.42+20.22+0.42+1+tan−120.20.4(0.22+0.42+1)√0.22+0.42+0.220.42+1]
Where m=a/z
n=b/z.
m and n are interchangeable.
14 - Question
Find the vertical pressure at depth 5m for a uniformly loaded circular area of 80 kN/m2 load and radius of 5m.
a) 51.72 kN/m2
b) 54.12 kN/m2
c) 78.325 kN/m2
d) 12.24 kN/m2
View Answer
Answer: a
Explanation: Given,
uniformly loaded circular area of radius a=5m
depth z=5m
load q=80 kN/m2
The Boussinesq’s vertical pressure σz under a uniformly loaded circular area is given by,
σz=q[1−[1(1+(az)2]32]
∴ σz=80[1−[11+(55)2]32]
∴ σz=51.72 kN/m2.
15 - Question
Find the influence factor for the vertical pressure at depth 5m for a uniformly loaded circular area of 80 kN/m2 load and radius of 1m.
a) 0.6212
b) 0.0571
c) 0.0328
d) 0.0624
View Answer
Answer: b
Explanation: Given,
uniformly loaded circular area of radius a=1m
depth z=5m
load q=80 kN/m2
The Boussinesq’s vertical pressure σz under a uniformly loaded circular area is given by,
σz=q[1−[11+(az)2]32]
∴ σz=kq
∴ k=[1−[11+(az)2]32]
K=[1−[11+(15)2]32]
k=0.0571.