Cee 312(6)(structural analysis)

CEE-312Structural Analysis and Design Sessional-I

(1.0 credit)Lecture: 6

Bijit Kumar Banik

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

[email protected]

Department of Civil and Environmental Engineering

Analysis and design of an Industrial roof truss system

R L

(- 8.88) (-8.88) (-6.35) (-4.19) (-4.43) (-4.43)(7.14) (5.61) (4.07) (4.85) (5.00) (5.14)

6@6 ft = 36 ft

L0 L1 L2 L3 L4 L5L6

U1

U2

U3

U4

U5

2.60 k4.54 k

(0)

(-0.

13)

(-1.

41)

(0)(-3.

07)(3.78)(2.90) (0

.35)

(0.27)

3.24k

L→R

6@6 ft = 36 ft

L0 L1 L2 L3 L4 L5L6

U1

U2

U3

U4

U5

4.54 k2.60 k

(- 1.19) (-1.19) (-0.95) (-3.11) (-5.64) (-5.64)(5.14) (5.00) (4.85) (4.07) (5.61) (7.14)(0

)

(-0.

13)

(-1.

41)

(0)(-3.

07)(0.35)(0.27) (3

.78)

(2.90)

3.24k

(4.41) (4.41) (3.53) (3.53) (4.41) (4.41)(- 5.05) (- 4.04) (- 3.03) (- 3.03) (- 4.04) (- 5.05)

6@6 ft = 36 ft

L0 L1 L2 L3 L4 L5L6

U1

U2

U3

U4

U5

2.94 k2.94 k

(.18

)

(0.6

7)

(0.6

7)

(.18

)

(2.1

4)(- 1.32)(- 1.01) (-

1.32

)

(- 1.01)

Dead Load


Load combinations

Condition 1: F1 = DL [No wind]

Condition 2: F2 = DL + WL (L→R)

Dead Load →DL

Wind Load →WL

Condition 3: F3 = DL + WL (R→L)

For design compressive force → Minimum of F1, F2 & F3 (If positive leave it !)

For design tensile force → Maximum of F1, F2 & F3 (If negative leave it !)


Compr.TensionWL(R L)WL(L R)DL

-1.23+4.41-4.43-5.64+4.41L5L6

-1.23+4.41-4.43-5.64+4.41L4L5Member

-0.66+3.53-4.19-3.11+3.53L3L4Chord

-2.82+3.53-6.35-0.95+3.536.00L2L3Bottom

-4.47+4.41-8.88-1.19+4.41L1L2

-4.47+4.41-8.88-1.19+4.41L0L1

-5.05+2.09+5.14+7.14-5.05L6U5

-4.04+1.57+5.00+5.61-4.04U4U5Member

-3.03+1.82+4.85+4.07-3.03U3U4Chord

-3.03+1.82+4.07+4.85-3.036.86U2U3Top

-4.04+1.57+5.61+5.00-4.04U1U2

-5.05+2.09+7.14+5.14-5.05L0U1

Design Member Force (k)Member Force (k)Length

(ft)

MemberRemarks


-1.32+2.46+0.35+3.78-1.328.97L3U4

-0.93+2.14-3.07-3.07+2.1410.0L3U3Member

-1.32+2.46+3.78+0.35-1.328.97L3U2Web

-0.74+0.67-1.41-0.13+0.676.67L2U2

-1.01+1.89+2.90+0.27-1.016.86L2U1

-+0.1800+0.183.33L1U1

Compr.TensionWL(R L)WL(L R)DL

Length

(ft)

MemberRemarks

-+0.1800+0.183.33L5U5

-1.01+1.89+0.27+2.90-1.016.86L4U5

-0.74+0.67-0.13-1.41+0.676.67L4U4


During analysis we assumed as if truss members are pin ended

Actually they are continuous or welded/riveted to other part ofthe truss

So, ends are somewhat rigid

We assume the effective length factor k = 0.6

Whereas k = 0.5 for fixed endk = 1.0 for pin end

Le = 1.0 L Le = 0.5 L

L/4

L/4Pin Fixed


2L0.25One end fixed other free

L1Both end hinged

0.7L2One end fixed other hinged

0.5L4Fixed

Le = Effective length

N = Number of times strength of hinged columns

End connection

Design of chord members

yc F

EC

2π=Cc = slenderness ratioE = 29,000 ksiFy = 36 ksi

Fa = allowable stressP = maximum bar force

Choose angle (based on Areq)Take A, rmin

rmin = min. radius of gyration out of rx, ry & rz

3

2

/

8

1/

8

3

3

5

/5.01

−

+

−

=

cc

cy

a

C

rKL

C

rKL

C

rKLF

F( )2/

000,149

rKLFa =

If KL/r < CcIf KL/r < C c

Compare Fa with f Take another member with greater ‘L’(if any); check for that L & P

Choose another angle of greater secn; Take A, rmin

If Fa < f If Fa > f

Final secn

of anglestop If all members have been checked

areq

ya

F

PA

FFAssume

=

= 5.0

A

Pf

stressecompressivActual

=

Compare KL/r with Cc

Design of Top chord members

Maximum compressive force found on L0U1 , having L = 6.86 ft ; P = 5.05 k

yc F

EC

2π=36

000,29*2π= 1.126=

Fa = 0.5Fy = 0.5*36 = 18 ksi

areq F

PA =

18

05.5= = 0.281 in2

From AISC chart, select 16

3

4

11

4

11 XXL


0.2440.434

A = 0.434 in2

rmin = 0.244


ksif 64.11434.0

05.5 ==

4.202244.0

)12*86.6(*6.0 ==r

KL> Cc

From AISC chart, now select 16

322 XXL

( ) fksiFa <== 64.34.202

000,1492 Not ok


0.2440.434

A = 0.715 in2

rmin = 0.394


ksif 06.7715.0

05.5 ==

36.125394.0

)12*86.6(*6.0 ==r

KL< Cc

So, Design Top chord is16

322 XXL

ok

3

2

/

8

1/

8

3

3

5

/5.01

−

+

−

=

cc

cy

a

C

rKL

C

rKL

C

rKLF

F3

2

1.12636.125

81

1.12636.125

83

35

1.12636.125

5.0136

−

+

−=

= 9.50 ksi > f

Design of Bottom chord members

Maximum compressive force found on L0L1 , having L = 6 ft ; P = 4.47 k

yc F

EC

2π=36

000,29*2π= 1.126=

Fa = 0.5Fy = 0.5*36 = 18 ksi

areq F

PA =

18

47.4= = 0.248 in2


3

4

11

4

11 XXL


0.2440.434

A = 0.434 in2

rmin = 0.244


ksif 3.10434.0

47.4 ==

05.177244.0

)12*6(*6.0 ==r

KL> Cc


122 XXL

( ) fksiFa <== 75.405.177

000,1492 Not ok


A = 0.484 in2

rmin = 0.398

1/8


ksif 24.9484.0

47.4 ==

54.108398.0

)12*6(*6.0 ==r

KL< Cc

So, Design Bottom chord is8

122 XXL

ok

3

2

/81/

83

35

/5.01

−

+

−

=

cc

cy

a

CrKL

CrKL

C

rKLF

F3

2

1.12654.108

81

1.12654.108

83

35

1.12654.108

5.0136

−

+

−=

= 11.88 ksi > f

Design of Web members

Maximum tensile force found on U2L3 , having L = 8.97 ft ; P = 2.46 k

yc F

EC

2π=36

000,29*2π= 1.126=

Fa = 0.5Fy = 0.5*36 = 18 ksi

areq F

PA =

18

46.2= = 0.137 in2


3

4

11

4

11 XXL

Design of Web chord members

0.2440.434

A = 0.434 in2

rmin = 0.244


ksif 67.5434.0

46.2 ==

69.264244.0

)12*97.8(*6.0 ==r

KL> Cc


5

2

11

2

11 XXL

( ) fksiFa <== 13.269.264

000,1492 Not ok


A = 0.84 in2

rmin = 0.291

1/8

0.2910.84 in2


ksif 93.284.0

46.2 ==

94.221291.0

)12*97.8(*6.0 ==r

KL> Cc

ok( ) fksiFa >== 02.3

94.221

000,1492

But we have a member U3L3 of length 10 ft with P = 2.14 k

Now check the section for this member

ksif 55.284.0

14.2 ==

42.247291.0

)12*10(*6.0 ==r

KL> Cc

( ) fksiFa <== 43.242.247

000,1492 Not ok


A = 0.984 in2

rmin = 0.289

1/8

0.291

0.984 in2


ksif 17.2984.0

14.2 ==

13.249289.0

)12*10(*6.0 ==r

KL> Cc

( ) fksiFa >== 40.213.249

000,1492 ok

So, Design web member is8

3

2

11

2

11 XXL

Design summery of members

Design web member8

3

2

11

2

11 XXL

Design Bottom chord member8

122 XXL

Design Top chord member16

322 XXL

Date post:	09-Aug-2015
Category:	Engineering
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