Columns lecture#6

Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns

LECTURE # 6

CASE 8: e IS GIVEN AND e < e b:

If e < eb , then we are in compression

failure region. For this case we know that

fs ≠ fy , but whether fs′ = fy or not, this is not

known.

In figure comparing a b c and c f g,

we get,

As, fs = Es. εs , so,

……………… (1)

Now, in figure comparing c f g and c d e, we get,

……………… (2)

As, fs′ = Es. εs′ , so,

……………… (3)

Now,

Assuming, fs′ = fy and using equation (1) in above load equation we get,

……………… (4)

Now we know that,

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LOAD EQUATION


Using value of Pn from equation (4) in above equation we get,

……………… (5)

Also,

………… (6)

We compare equation (5) and (6) to find the value of ‘a’ and using equation (2) we check

our assumption,

If εs′ ≥ εy , then our assumption is correct and we compute Pn and Mn using load

equation and moment equation respectively.

If εs′ < εy , then fs′ = Es. εs′ and we proceed as follows,

Using equation (3) in load equation we get,

………… (7)

Now we know that,

Using value of Pn from equation (4) in above equation we get,

……………… (8)

Also,

We compare equation (8) and (9) to find the value of ‘a’ and using this value of ‘a’ we

compute fs′ from equation (3). Finally using theses values of a and fs′ in load and moment

equation, we find out Pn and Mn respectively.

CASE 9: Mn IS GIVEN AND AT FAILURE STEEL IS YIELDING:

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……. (9)


Here fs = fy but whether fs′ = fy or not, this is not known. We know that,

……………… (1)

Now assuming As = As′ and fs′ = fy , we get,

……………… (2)

Also,

……………… (3)

Solving equation (3) yields the value of ‘a’ and using this value of ‘a’ in equation (1) we check

our assumption,

If εs′ ≥ εy , then our assumption is correct and we compute Pn using load equation.


……………… (4)

Using equation (4) in (3) we get,

………… (5)

Solving equation (5) yields the value of ‘a’ and using this value of ‘a’ in equation (4) we find

value of fs′. Now using theses values of a and fs′ in load equation, we find out Pn

CASE 10: Mn IS GIVEN AND FAILURE IS DUE TO CRUSHING OF CONCRETE:

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LOAD EQUATION

MOMENT

EQUATION

LOAD EQUATION


In this case we know that,

……………… (1)

and

……………… (2)

Now,

Assuming, fs′ = fy and using equation (2) in above load equation we get,

……………… (3)

Also,

………… (4)

We solve equation (4) for the value of ‘a’ and using this value of ‘a’ in equation (1) we check

our assumption,

If εs′ ≥ εy , then our assumption is correct and we compute Pn using equation (3).


……………… (5)

Using equation (5) in (4), we get,

Now we solve above equation for the value of ‘a’ and using this value of ‘a’ in equation (5) we

find value of fs′. Similarly using ‘a’ value in equation (2) we find value of fs. Now using

theses values of a, fs and fs′ in load equation, we find out Pn.

CASE 11: DEPTH OF N.A. IS GIVEN ( c OR a ):

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LOAD EQUATION


If ‘c’ is given, then compute ‘a’ using,

Compute εs′ using,

If εs′ ≥ εy , then fs′ = fy.

If εs′ < εy , then fs′ = Es. εs′

Now, compute εs′ using,

If εs ≥ εy , then fs = fy.

If εs < εy , then fs = Es. εs

Now, using theses values of a, fs and fs′ in load equation and moment equation, we find out Pn

and Mn respectively.

PROBLEM:

fc = 25 MPa , fy = 300 MPa

Analyze the column shown in figure

for the following conditions;

(a) Pure Axial Case.

(b) Balanced Condition.

(c) Pu = 1300 kN.

(d) e = 300 mm.

(e) Mu = 200 kN-m.

ASSIGNMENT: Part c, d and e ( To be submitted on Thursday, 19/11/2009 )

1. NOMINAL & DESIGN INTERACTION CURVE:

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Nominal curve is one on which no reduction factor on material’s strength. Design curves

include reduction factor for the material strength. Reduction factor for various cases is as

follows;

For compression controlled failure, φ = 0.65

For Tension controlled failure i.e., εs ≥ 0.005 , φ = 0.90

For Transition controlled failure i.e., εy < εs < 0.005

For Ties,

For Spirals,

Fig. Nominal Vs Design Interaction Curve

2. DESIGN OF SHORT COLUMN FOR UNI-AXIAL ECCENTRICITY WHEN STEEL IS

ON OPPOSITE FACES:

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In this region design curve is obtained by multiplying nominal curve with φ = 0.65 (for Ties)

φ = 0.70 (for Spirals)

In this region design curve is obtained by multiplying nominal curve with

φ = 0.9

(φMnb ,φPnb)

(Mnb , Pnb)

(Mn ,0)

( 0, Pn)

(φMn ,0)

(0, φPn)

(0, φPn)

This point is obtained by

using,φ= 0.65 x 0.8 (for Ties)

φ= 0.7 x 0.85 (for Spiral)

This point is obtained by using,φ= 0.65 (for Ties)

φ= 0.70 (for Spiral)

Interpolation is required


GIVEN:

Pu and Mu

fc and fy

Cross-sectional size (not always given)

REQUIRED:

Ast

Ties/ Spirals

Step 1:

Assume yielding of compression steel at balance failure point i.e, fs′ = fy. and also assume that

As = As = .

Compute ab using,

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Step 2:

In load equation use fs′ = fy. and As = As

Using above equation we can find φPnb

Step 3:

Check εs′ using,

If εs′ ≥ εy , then our assumption is correct and we will use φPnb value computed in Step 2.

If εs′ < εy , then fs′ = Es. εs′ and we proceed as follows;

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LOAD EQUATION

For Spirals,

φ = 0.7 x 0.85


……………… (1)

Using equation (1) in load equation, we get,

Above equation yields the value of φPnb .

CASE 1: Pu ≥ Pnb

Step 1:

For this case we know that tension steel is not yielding. So,

Assume yielding of compression steel i.e, fs′ = fy. and also assume that As = As = .

Step 2:

Above equation results as,

Ast = f1 (a) ……………… (2)

Step 3:


Ast = f2 (a) ……………… (3)

Step 4:

Compare equation (2) and (3) and find out values of a and Ast .

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For Spirals,

φ = 0.7 x 0.85


Step 5:

Now using the value of ‘a’ computed from Step 4, check the assumption made in Step 1 using,

If εs′ ≥ εy , then our assumption is correct and we will use values of a and Ast computed

in Step 4.

If εs′ < εy , then fs′ = Es. εs′ . So take,

Using the above value of fs′ in load and moment equation, repeat Step 2,3 and 4.

CASE 2: Pu < Pnb

Step 1:

For this case we know that tension steel is yielding. Assume yielding of compression steel i.e,

fs′ = fy and also assume that As = As = .

Step 2:

Assume φ value using an empirical formula as under,

Step 3:

Using load equation and applying the assumptions made in Step 1 to it, we get,

Using the above equation compute value of ‘a’.

Step 4:

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Now using the value of ‘a’ computed from Step 3 check the assumption made in Step 1 using,

If εs′ ≥ εy , then our assumption is correct and we will use values of a computed in Step 4

in moment equation to find Ast.

If εs′ < εy , then fs′ = Es. εs′. So take,

Using the above value of fs′ in load equation, we get,


Ast = f1 (a) ……………… (4)

Also,


Ast = f2 (a) ……………… (5)

Compare equation (4) and (5) and find out values of a and Ast .

Step 5:

Check the assumed value of φ computed in Step 2 using,

If εs ≥ 0.005 , then φactual = 0.9

If εs ≥ 0.005 , then

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For Ties,

For Spirals,

If φactual is greater than of less than φassumed with in the limit of 10 % then we can use φassumed for

the design otherwise we will use φactual for the design and repeat Step 3 and 4

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Columns lecture#6

Design

Transcript of Columns lecture#6