Cee 312(2)

CEE-312

Structural Analysis and Design Sessional-I

(1.0 credit)Lecture: 2

Bijit Kumar Banik

Assistant Professor, CEE, SUSTRoom No.: 115 (“C” building)

[email protected]

Department of Civil and Environmental Engineering

Design philosophy

A general statement assuming safety in engineering design

Resistance (of material & x-section) >= Effect of a pplied load

It is essential in the above equation that both sid es are evaluated in same conditions; e.g. if of load to produce comp ressive stress

on soil, then it should be compared with the bearin g capacity of soil

Design methods

1. ASD (Allowable Stress Design)

2. LRFD (Load & Resistance Factor Design)

ASD

Safety in design is obtained by specifying that the effects of the loads should produce stresses that were the fraction of yield stress, F y

This is equivalent to:

FOS = Resistance, R / Effect of load, Q

ASD

Since the specifications sets limit on stresses, it became Allowable Stress Design (ASD)

Mostly reasonable where stresses are distributed un iformly (such as determinate trusses, arches, cables etc)

Drawback of ASD

1. ASD does not give reasonable measures of strengt h

2. Factor Of Safety (FOS) is applied only to stress not in load

LRFD

Considers variability not only in resistance but al so in the effects of load

Quality of material should maintain highest level

Now different factored load can be assign to differ ent types of loads (dead, live, wind etc)

Trusses

In architecture and structural engineering, a truss is a structure comprisingone or more triangular units constructed with straight members whose ends

are connected at joints referred to as nodes (hinge or pin).

Assumptions

1. Members are connected at their end by frictionless pins

2. Loads and reactions applied only at joints

3. Two force members

Types of Trusses

Two broad categories:

Roof truss

Bridge truss

Roof Trusses

Flat: The most economical flat truss for a roof is provided when the depth of the truss in inchesis approximately equal to the span in inches.

Mono: Used where the roof is required to slope in only one direction. Also in pairs with their high ends abuttingon extremely long span with a support underneath the high end.

Common: Truss configurations for the most widely designed roof shapes.

Scissors: Provides a cathedral or vaulted ceiling. Most economical when the difference in slope between the top and bottom chords is at least 3/12or the bottom chord pitch is no more than half the

top chord pitch.

Roof Trusses

Bridge Trusses

Pratt

Warren

Bridge Trusses

Howe

K-truss

Stability and Determinacy of Truss

Minimum members required to make a rigid framework is Three

3 members3 joints

5 members4 joints

7 members5 joints

Thus to form a rigid and stable truss of ‘n’ joints number of members (j) required is:

j = 3+2*(n-3)

For first 3 joints For remaining joints

j= 2n-3

Stability and Determinacy of Truss

If j < 2n-3 Unstable truss

If j > 2n-3 Indeterminate truss

Bottom chord

DiagonalTop chordVertical

Industrial roof truss system

Span

Bay

Parlin

Top chord bracing

Sagrod

Column

Beam


Vertical bracing

Bottom chord strutR

ise


When Span > 40 ft Truss system

Trusses are normally spaced 15-25 ft apart

For longer span trusses must be simply supported

Hinge support Roller support

Why this is called simply supported ??

Loads on structures

Load

DynamicStatice.g. Earthquake

Live loadDead load

e.g. Self wt. or fixed object e.g. Wind, snow, wheel of vehicle or any moving objects

Wind load analysis for roof truss

Basic wind pressure at 30 ft, q (psf) = 0.00256 V 2

where, V is wind speed (mph)

Pressure on a pitched roof

α

P

Windward side

Wind

Leeward side

P


For windward surface

≤≤≤−≤≤−

≤≤−

=

ααααα

α

0

00

00

0

609.0

6030)9.003.0(

3020)1.207.0(

2007.0

q

q

q

q

P

For leeward surface

qP 7.0−=

where, ‘P’ is the pressure normal to the roof surfac e

If P is +ve

If P is -ve

Pushing the roof surface

Pulling or suction or uplift

Leeward side always experience upliftMaximum P will be taken

Analysis and design of an Industrial roof truss sys tem

Steps

1. Selection of truss type2. Estimation of loads

3. Analysis and design of purlins4. Analysis and design of sagrods

5. Dead load and wind load analysis

6. Combination of D.L and W.L to determine the design bar forces7. Design of members8. Design of bracing system

9. Design of connections (welded)10. Detailing

Your Truss !!!

6@ A ft

B ft

L0 L1 L2 L3 L4 L5

L6

U1

U2

U3

U4

U5

Bay = 20 ft; F y=36 ksi; E = 29 ksi

11.511109.5B

76.55.55A

4321Group


1. Selection of truss type

6@6 ft = 36 ft

10 [email protected] ft

L0 L1 L2 L3 L4 L5

L6

U1

U2

U3

U4

U5

Bay = 20 ft; F y=36 ksi; E = 29 ksi

29.050


2. Estimation of loadsDead loads

1. CGI (Corrugated Galvanized Iron ) 2.0 psf2. Purlins 1.5 psf3. Sagrods, bracings 1.0 psf

Sub total 4.5 psf

4. Self weight 60 lb/ft of span

Live loads

Design wind speed 100 mph

Given loads should be transformed into Equivalent j oint loads


3. Analysis and design of purlins

Roofing

Purlin

SagrodTop chord

Purlins are nothing but beams. They span between the adjacent trusses, i.e the spacing of the trusses (Bay) is the span of purlins . Normally channel section is used to design purlins.

Design criteria for purlin is “bending stress”

First we will check for dead load and then for the wind load



Since the principal axes of the purlins section are inclined, the dead load causes biaxial bending in the purlins.

Y

X

W

Check for Dead Load



So, different support condition of purlin for X and Y direction

For Y-direction Y

XW

20 ft

For X-direction

10 ft 10 ft

Additional support for sagrod



Dead loads coming on purlinsRoofing

Purlin

Sagrod

Roofing 2.0 psf

Self wt. of purlin 1.5 psf

Total 3.5 psf (neglecting sagrod)


3. Analysis and design of purlins Parlin

6.86 ft

3.5 psf

UDL on purlin, w DL= 3.5 psf X 6.86 ft = 24.01 lb/ft



wDLx= wDLsin θ = 24.01 X sin29.01 0 = 11.66 lb/ft

X

Y

wDL= 24.01 lb/ft

29.050

29.050

wDLy= wDLcos θ = 24.01 X cos29.01 0 = 20.99 lb/ft



For the detail computation of BMD please go through : Strength of materials-by Singer; pp-285, Prob.-828

For Y-direction

L = 20 ft

For X-direction

L/2 =10 ft L/2 = 10 ft

BMD

wDLy wDLx

8

2LwMxx DLy=

32

2LwMyy DLx=



Y

X

wDLy

wDLx

L

Mxx = (wDLy *L) * L/2

Myy = (wDLx *L) * L/2



8

2LwMxx DLy=

32

2LwMyy DLx=

ftkip −== 05.18

20*99.20 2

ftkip −== 15.032

20*66.11 2



Select American standard channel: C 3 X 4.1

Sxx = 1.10 in3 Syy=0.202 in3



Maximum bending stress:

= 0.66*36 = 23.76 ksi

yy

yy

xx

xxb S

M

S

Mf +=

ksi37.20202.0

12*15.0

10.1

12*05.1 =+=

For bending; Allowable stress ,

> fb(20.37 ksi)

Fb= 0.66Fy

So, section C 3 X 4.1 is ok for dead load



Check for Wind Load

3@6 ft = 18 ft

10 ft

L0 L1 L2 L3 L4 L5

L6

U1

U2

U3

U4

U5 29.050

Pitch angle, 01 05.29)18

10(tan == −α V = 100 mph

psfVq 6.25100*00256.000256.0 22 ===Basic wind pressure,


For windward surface

≤≤≤−≤≤−

≤≤−

=

ααααα

α

0

00

00

0

609.0

6030)9.003.0(

3020)1.207.0(

2007.0

q

q

q

q

P

For leeward surfaceqP 7.0−=

P = (0.07*29.05-2.1)*25.6 = -1.7 psf

UDL on the windward side = -1.7*6.86 = -11.66 lb/ft

Purlin spacing

UDL on the Leeward side = - 17.92 *6.86 = -122.93 lb/ftP = -0.7*25.6 = -17.92 psf

Here Leeward side load will governWhat does the negative sign mean? Suction



Maximum bending stress:

yy

yy

xx

xxb S

M

S

Mf +=

202.0

12*15.0

10.1

12*10.5 +=

Why pound is expressed as ‘lb’ ?

libra→ lbpound weight which was libra pondo in Latin.

Resultant load in y-direction = -122.93+20.99 = -10 1.94 lb/ftDead load in y-direction = 20.99 lb/ft ( from previous DL calculation )

→101.94 lb/ft

8

2LwMxx y= ftkip −== 10.5

8

20*94.101 2

ftkipsMyy −= 15.0

> Fb(23.76 ksi)ksi57.63= Not ok



Next trail channel: C 4 X 7.25

Sxx = 2.29 in3 Syy=0.343 in3



yy

yy

xx

xxb S

M

S

Mf +=

343.0

12*15.0

29.2

12*10.5 +=

ksi97.31= > Fb(23.76 ksi)

Not ok



Next trail channel: C 5 X 6.7

Sxx = 3.00 in3 Syy=0.378 in3



yy

yy

xx

xxb S

M

S

Mf +=

378.0

12*15.0

00.3

12*10.5 +=

ksi16.25= > Fb(23.76 ksi)

Not ok

�� !!!



Next trail channel: C 5 X 9

Sxx = 3.56 in3 Syy=0.450 in3



i.e. 9/6.86 = 1.31 psf

yy

yy

xx

xxb S

M

S

Mf +=

450.0

12*15.0

56.3

12*10.5 +=

So, C 5 X 9 is the final Purlin section

ksi19.21= < Fb(23.76 ksi)

য��য��য��য�� !!!!!!!! okCheck

Self weight of purlin = 9 lb/ft

< 1.5 psf ( assumed self wt .)


4. Analysis and design of sagrods

For the detail computation of reaction please go th rough: Strength of materials-by Singer; pp-285, Prob.-828

For X-direction

L/2 =10 ft L/2 = 10 ft

wDLx = 11.66 lb/ft

LwDLx8

3 LwDLx8

3LwDLx8

5

Tensile Force on sagrod = Midspan reaction

LwF DLx8

5=

A round bar of dia (3/8) inch will be adequate.

≈== lbF 75.14520*66.11*8

5kips15.0


4. Analysis and design of sagrods

Assuming that the bolt threads will reduce the effe ctivediameter by (1/16) inch.

Net X-sectional area = 22

077.016

1

8

3*

4in=

−π

Allowable stress in tension, F t = 0.6 Fy = 21.6 ksi

The rod is able to carry a load of 21.6*0.077 = 1.66 kips >> F

So, #3 or (3/8) inch round rod will be used as sagr od

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Cee 312(2)

Engineering

Transcript of Cee 312(2)