sa1 08-direct-and-bending-stress

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TOPIC :-

SUBJECT :-structural Analysis - 1

GROUP MEMBERS:-

1.Bhatti Anisahmed A.130853106003

2.Jethwa Kashyapsinh M.130853106005

3.Keshariya Mitesh D.130853106007

4.Patel Ashish C.130853106010

5.Ramjiyani Dhaval J.130853106013

(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 = Eccentric load produces both direct stress (σo) and

bending stress (σb) in column.

Direct stress = σo =

Bending stress = σb = = .y Z =

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

= +

= (1+ )

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

= -

= (1- )

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, will be compressive. (+ve)

When load is acting at the point of e limit, will be zero.

When load is acting beyond e limit, will be tensile (-ve)

Example :-A column of T-section shown in figure is

subjected to a load of 100kN at a point on the centroidal axis, 40mm below the centroidal x-x axis. Calculate the maximum stresses induced in the section.Solution: Data Given,

P = 100 kNe = 40 mm

Part-1a1 = 200 x 20 = 4000y1 = 180 + 10 = 190mm

Part-2a2 = 180 x 15 = 2700y2 = 90 = mm

y = = = 149.70 mm

Ixx1 = Ig + ah2

= + 4000 x (190 – 149.70)2

= 6.629 x 106 mm4

Ixx2 = Ig + ah2

= + 2700 x (149.70 – 90)2

= 16.91 x 106 mm4

= 6.692 x + 16.91 x

= 23.539 x

= = = 157241.15

= + = +

= 14.92 + 25.43

= 40.35 N/ ( compressive)

= =

= 14.92 25.43

= 10.51 N/(Tensile)

(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 x ᵟ

where ᵟ = density of dam material in kN/

(2) Total water pressure on Dam:

Total water pressure = Area of water pressure diagram

P = x wh x h Where, w = density of water P = = 1000 kg/

= 10 kN/

(3) Eccentricity (e) : Total water pressure (p) acts horizontally at height from the base of dam.

Total weight of dam (W) acts vertically downwards.

R is the resultant of P and W.

R =

Resultant (R) cut the base at point K.

distance JK = x = x

distance AJ =

d = AJ + JK

eccentricity = e = d

(4) Maximum and Minimum Pressure :

Maximum Pressure.

= (1 )

Minimum Pressure.

= (1 )

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 = x P = x () Where, = active earth pressure coefficient

= = Angle of repose of soil

Total earth pressure (P) acts at height from the bases of retaining wall.

(C). Chimney

Chimney And Wall Subjected to Wind Pressure:

Consider a chimney or wall having plan dimension b x d and height h. Let, weight of chimney or wall

Where, = A = base area w = x V w = x A x h direct stress = σo = σo = 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

Bending stress caused on the base due to moment,

=

The extreme stresses on the base are,

=

=

Example :- A masonry wall, 5m high, is of solid rectangular section, 3m wide and 1m thick. A horizontal wind pressure of 1.2 kN/acts on the 3m side. Find the maximum and minimum stresses induced on the base, if unit weight of masonry is 22.4 kN/

Solution :-

= 22.4 kN/h = 5.0mb = 3.0mp = 1.2 kN/

= = 22.4 x 5.0 = 112 kN/

M = P x P = total wind force = p x b x h = 1.2 x 3 x 5 = 18 kN

M = P x = 18 x = 45 kN.m

Z = = = 0.5

= = = 90 kN/

=

= 112 + 90 = 202 kN/ (compressive)

= = 112 90 = 22 kN/ (compressive)

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