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Subject name : structure analysis-1
Subject code : 2140608
Guided by :- Prof. Pritesh Rathod
Prof. Rajan lad
Name Enrollment Number
Gandhi Harshil R. 141100106018
Deshmukh Bhavik H. 151103106002
Kotila Jayveer V. 151103106008
Misrty Aditya P. 151103106009
Pandya Dhrumil D. 151103106010
(A). Column :-
Axial load : When load is acting along the longitudinal
axis of column. It produces compressive stress in column.
Eccentric load : A load whose line of action does not
coincide with the axis of a column. It produces direct
and bending stress in column.
Eccentricity (e) : The horizontal distance between the
longitudinal axis of column and line of action of load.
In axially loaded column e = 0
Effect of Axial and Eccentric load on Column:
 When short column is subjected to axial compressive
force, only direct stress(σo) is produced in the column.
Direct stress = σo = P/A
 Eccentric load produces both direct stress (σo) and
bending stress (σb) in column.
Direct stress = σo = P/A
Bending stress = σb =M/z= M/I .y ∴ Z =I/Y
Maximum and Minimum Stresses :-
When a column is subjected to eccentric load, the edge of
column towards the eccentricity will be subjected to maximum
stress (σmax) and the opposite edge will be subjected to
maximum stress.
Maximum Stress (σmax) Minimum Stress (σmin)
σmax = direct stress + bending stress
= σo + σb
= P/A + M/Z
= P/A (1+ 6e/b)
σo = direct stress, σb = bending stress, M = Moment = P . e, e = eccentricity
Z = section modulus = , I = moment of Inertia, y = distance of extreme fibre
from c.g. of column.
σmin = direct stress - bending stress
= σo - σb
= P/A -M/Z
= P/A (1- 6e/b)
Stress distribution in Column :-
Stress distribution in column as the load (P) moves from
centre of column to the edge of column as shown in fig.
Limit of eccentricity (e limit):
The maximum distance of load from the centre of
column, such that if load acts within this distance there
is no tension in the column. The maximum distance is
called Limit of eccentricity.
When load is acting within e limit,
𝜎min will be compressive. (+ve)
When load is acting at the point of e limit,
𝜎min will be zero.
When load is acting beyond e limit,
𝜎min will be tensile (-ve)
(B). Dams and Retaining Wall :-
Maximum and Minimum pressure at the base of Dam :
(1) Weight of Dam:
Weight = cross sectional area of dam x density of
dam material
∴ W = (a + b) x (H/2)x ᵟ
where ᵟ = density of dam material in kN/m3
(2) Total water pressure on Dam:
Total water pressure = Area of water pressure diagram
∴ P =
1
2
x wh x h Where, w = density of water
∴ p=wh²/2 = 1000 kg/m3
= 10 kN/m3
(3) Eccentricity (e) :
Total water pressure (p) acts horizontally at height
h
3
from
the base of dam.
Total weight of dam (W) acts vertically downwards.
R is the resultant of P and W.
R = P2 + W2
Resultant (R) cut the base at point K.
distance JK = x = (p/w)x(h/3)
distance AJ =
a2+ab+b2
3 (a+b)
∴ d = AJ + JK
∴ eccentricity = e = d −(𝑏/2)
(4) Maximum and Minimum Pressure :
Maximum Pressure.
𝜎max = w/b(1 + 6e/b)
Minimum Pressure.
𝜎min = w/b(1 − 6e/b)
Retaining Wall :
A retaining wall is a structure used to retain soil(earth).
The basic difference between dam and retaining wall is that, a
dam retain water and subjected to water pressure while, a
retaining wall retain earth and subjected to earth pressure.
Total earth pressure :
P =
wh2
2
x Ka
∴ P =
wh2
2
x (
1 −sin∅
1+sin∅
)
Where,
Ka = active earth pressure coefficient
=
1 −sin∅
1+sin∅
∅ = Angle of repose of soil
Total earth pressure (P) acts at height
h
3
from the bases of
retaining wall.
Stability conditions for retaining
wall or Dam
 A retaining wall or a dam is checked for the following conditions of
stability:-
 No overturning
two major forces acting on retaining wall/dam are:-
(a) total earth/water pressure(P)
(b) weight of wall/ dam(W)
for no overturning
resisting moment > overturning moment
JB > JK
 No tension at base
To avoid tension in the masonary at the base,
minimum stress should not be(-ve).
for no tension at base eccentricity should be less
than (b/6).
 No sliding
total frictional force at the base of wall/dam
= Fmax = 𝜇. 𝑊
for no sliding = 𝜇. 𝑊 > P
if factor of safety = 1.5
(𝜇. 𝑊/P) > 1.5
 No crushing at base
the material of wall/dam should be safe against
crushing at the base.
∴ 𝜎 max < 𝜎𝑐
𝜎𝑐 = permissible crushing stress
𝜎 max = maximum pressure at the base
Minimum width of base for no
tension
(C). Chimney
Chimney And Wall Subjected to Wind Pressure:
Consider a chimney or wall having plan dimension b x d and
height h.
Let, 𝛿 = unit weight of chimney or wall
𝛿 =
total weight
volume
Where,
∴ 𝛿 = w/v A = base area
∴ w = 𝛿 x V
w = 𝛿 x A x h
∴ direct stress = σo =w/A
=(𝛿 ∗ 𝐴 ∗ 𝐻)/𝐴
σo = 𝛿 x h ….(i)
If there is a uniform horizontal wind pressure (p) acting
on a side of width b,
wind force = P = p x b x h
This force will induce a bending moment on the base.
This force will induce a bending moment on the base.
∴ M = P x (h/2)
Bending stress caused on the base due to moment,
σb =M/Z
The extreme stresses on the base are,
𝜎max = σo + σb
𝜎min = σo − σb
Direct and bending stress
Direct and bending stress
Direct and bending stress

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Direct and bending stress

  • 1. Subject name : structure analysis-1 Subject code : 2140608 Guided by :- Prof. Pritesh Rathod Prof. Rajan lad
  • 2. Name Enrollment Number Gandhi Harshil R. 141100106018 Deshmukh Bhavik H. 151103106002 Kotila Jayveer V. 151103106008 Misrty Aditya P. 151103106009 Pandya Dhrumil D. 151103106010
  • 3. (A). Column :- Axial load : When load is acting along the longitudinal axis of column. It produces compressive stress in column. Eccentric load : A load whose line of action does not coincide with the axis of a column. It produces direct and bending stress in column. Eccentricity (e) : The horizontal distance between the longitudinal axis of column and line of action of load. In axially loaded column e = 0
  • 4. Effect of Axial and Eccentric load on Column:
  • 5.  When short column is subjected to axial compressive force, only direct stress(σo) is produced in the column. Direct stress = σo = P/A  Eccentric load produces both direct stress (σo) and bending stress (σb) in column. Direct stress = σo = P/A Bending stress = σb =M/z= M/I .y ∴ Z =I/Y
  • 6. Maximum and Minimum Stresses :- When a column is subjected to eccentric load, the edge of column towards the eccentricity will be subjected to maximum stress (σmax) and the opposite edge will be subjected to maximum stress. Maximum Stress (σmax) Minimum Stress (σmin) σmax = direct stress + bending stress = σo + σb = P/A + M/Z = P/A (1+ 6e/b) σo = direct stress, σb = bending stress, M = Moment = P . e, e = eccentricity Z = section modulus = , I = moment of Inertia, y = distance of extreme fibre from c.g. of column. σmin = direct stress - bending stress = σo - σb = P/A -M/Z = P/A (1- 6e/b)
  • 7. Stress distribution in Column :- Stress distribution in column as the load (P) moves from centre of column to the edge of column as shown in fig.
  • 8. Limit of eccentricity (e limit): The maximum distance of load from the centre of column, such that if load acts within this distance there is no tension in the column. The maximum distance is called Limit of eccentricity. When load is acting within e limit, 𝜎min will be compressive. (+ve) When load is acting at the point of e limit, 𝜎min will be zero. When load is acting beyond e limit, 𝜎min will be tensile (-ve)
  • 9.
  • 10.
  • 11. (B). Dams and Retaining Wall :- Maximum and Minimum pressure at the base of Dam :
  • 12. (1) Weight of Dam: Weight = cross sectional area of dam x density of dam material ∴ W = (a + b) x (H/2)x ᵟ where ᵟ = density of dam material in kN/m3 (2) Total water pressure on Dam: Total water pressure = Area of water pressure diagram ∴ P = 1 2 x wh x h Where, w = density of water ∴ p=wh²/2 = 1000 kg/m3 = 10 kN/m3
  • 13. (3) Eccentricity (e) : Total water pressure (p) acts horizontally at height h 3 from the base of dam. Total weight of dam (W) acts vertically downwards. R is the resultant of P and W. R = P2 + W2 Resultant (R) cut the base at point K. distance JK = x = (p/w)x(h/3) distance AJ = a2+ab+b2 3 (a+b) ∴ d = AJ + JK ∴ eccentricity = e = d −(𝑏/2)
  • 14. (4) Maximum and Minimum Pressure : Maximum Pressure. 𝜎max = w/b(1 + 6e/b) Minimum Pressure. 𝜎min = w/b(1 − 6e/b)
  • 15. Retaining Wall : A retaining wall is a structure used to retain soil(earth). The basic difference between dam and retaining wall is that, a dam retain water and subjected to water pressure while, a retaining wall retain earth and subjected to earth pressure.
  • 16. Total earth pressure : P = wh2 2 x Ka ∴ P = wh2 2 x ( 1 −sin∅ 1+sin∅ ) Where, Ka = active earth pressure coefficient = 1 −sin∅ 1+sin∅ ∅ = Angle of repose of soil Total earth pressure (P) acts at height h 3 from the bases of retaining wall.
  • 17. Stability conditions for retaining wall or Dam  A retaining wall or a dam is checked for the following conditions of stability:-  No overturning two major forces acting on retaining wall/dam are:- (a) total earth/water pressure(P) (b) weight of wall/ dam(W) for no overturning resisting moment > overturning moment JB > JK
  • 18.  No tension at base To avoid tension in the masonary at the base, minimum stress should not be(-ve). for no tension at base eccentricity should be less than (b/6).
  • 19.  No sliding total frictional force at the base of wall/dam = Fmax = 𝜇. 𝑊 for no sliding = 𝜇. 𝑊 > P if factor of safety = 1.5 (𝜇. 𝑊/P) > 1.5  No crushing at base the material of wall/dam should be safe against crushing at the base. ∴ 𝜎 max < 𝜎𝑐 𝜎𝑐 = permissible crushing stress 𝜎 max = maximum pressure at the base
  • 20. Minimum width of base for no tension
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  • 25. (C). Chimney Chimney And Wall Subjected to Wind Pressure:
  • 26. Consider a chimney or wall having plan dimension b x d and height h. Let, 𝛿 = unit weight of chimney or wall 𝛿 = total weight volume Where, ∴ 𝛿 = w/v A = base area ∴ w = 𝛿 x V w = 𝛿 x A x h ∴ direct stress = σo =w/A =(𝛿 ∗ 𝐴 ∗ 𝐻)/𝐴 σo = 𝛿 x h ….(i) If there is a uniform horizontal wind pressure (p) acting on a side of width b, wind force = P = p x b x h This force will induce a bending moment on the base.
  • 27. This force will induce a bending moment on the base. ∴ M = P x (h/2) Bending stress caused on the base due to moment, σb =M/Z The extreme stresses on the base are, 𝜎max = σo + σb 𝜎min = σo − σb