Cee 312(7 & 8)(structural analysis)

CEE-312Structural Analysis and Design Sessional-I

(1.0 credit)Lecture: 7&8

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

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

Analysis and design of an Industrial roof truss system

-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

Design bracing system

Bracing system consists of

a) Top chord bracing

b) Vertical bracing

c) Bottom chord strut

Exact analysis is seldom done in practice

Following guideline will be used

Tension member : KL/r < 400

Compression member: KL/r < 300 K = 0.7

Vertical & top chord bracing will be designed as tension member

Bottom chord strut will be designed as compression member

Industrial roof truss system

Vertical bracing

Bottom chord strutR

ise

Vertical bracing

Member of vertical bracing will be tied to each other at their crossing point. So, half of length will be considered

Length of each member = 22 2010 + = 22.36 ft = 269 inch

400)2/269(*7.0 <

r

r<400

15.94

inchr 235.0>

From AISC chart, select 16

1

4

11

4

11 XXL ; for which rmin = 0.244

Design of Web chord members

1/8

0.291

0.984 in2

0.244


Span

Bay

Parlin

Top chord bracing

Sagrod

Column

Beam

Top chord bracing

Member of top chord bracing also tied to each other at their crossing point. So, half of length will be considered

Length of each member = inchft 29125.2420)86.6*2( 22 ==+

400)2/291(*7.0 <

r

r<400

85.101

inchr 255.0>


1

2

11

2


Design of Web chord members

1/8

0.291

0.984 in2


Bottom chord strutR

ise

Tie

Bottom chord strut

To economize our design we will use lateral tie at the midspan of the strut, verysimilar to sagrod. So, half of the bay length will be considered.

Length of strut ,

300120*7.0 <

r

r<300

0.84

inchr 28.0>


1

2

11

2


inchL 12012*2

20 ==

Tie: ½ inch dia steel bar, connected with ½ inch nuts.

Design of Welded Connections

Here, All joints of the truss will be welded connections

Gusset plates will join the members at joints

Thickness of gusset plate will be assumed as the thickness of the members

Two types of joints here-

a) Joints where all members end (L0, U3, L6)

b) Joints where there is one continuous member (L1, L2,…. L2 etc)

For all the joints we select the thickness of the gusset plate as 3/8 inch

Maximum forces on members

L0 L1 L2 L3 L4 L5

L6

U1

U2

U3

U4

U5

(- 4.47) (-4.47) (3.53) (3.53) (4.41) (4.41)

(-5.05)

(-4.04)

(-3.03) (-3.03)

(-4.04)(-5.05)

(0.1

8)

(-0.

74)

-0.7

4)

(0.1

8)

(2.1

4)(2.46)(1.89) (2.4

6)

(1.89)

Type ‘a’ joints

Weld design of joint U3

16

322 XXL

8

3

2

11

2

11 XXL

16

322 XXL

U3

U 2U 3 U

3 U4

L 3U

3

P = 3.03 k P = 3.03 k

P =

2.1

4 k

Gusset plate (3/8 in)assumed

Gusset Plate


Allowable weld shear, FV = 0.3 Fy = 0.3 * 36 = 10.8 ksi

Weld design for member U2U3 or U3U4

P = 3.03 k

Weld size, s = 1/8 inch (assumed)= 0.128 inch

Total weld length required, 71.3125.0*707.0*8.10

03.3

707.0*′′===

sF

PL

v

So, L1+L2 = 3.17 inch


x

y

16

322 XXL

L1

L2

3.03 k


1/8

0.2910.84 in2

3/16


0.569

1.431

16

322 XXL

L1

L2

3.03 k

515.2569.0

431.1

2

1 ==L

L

L1 = 2.515 L2 L1+L2 = 3.17 inch

2.515 L2 + L2 = 3.17

L2 = 0.902 ≈ 1 inch

L1 = 0.902* 2.515 = 2.27 ≈ 2.5 inch

Same way calculate U3L3


16

322 XXL

8

3

2

11

2

11 XXL

16

322 XXL

U3

U 2U 3 U

3 U4

L 3U

3

P = 3.03 k P = 3.03 k

P =

2.1

4 k

Gusset plate (3/8 in)assumed

1/8 11/82.5

Weld design of joint L3

For L2L3L4 continuous bottom chord member

2.16(+3.53-3.11)

= -0.42

(+3.53-0.95)

= -2.58

DL+WL

(L→R)

2.16(+3.53-4.19)

= -0.66

(+3.53-6.35)

= -2.82

DL+WL

(R→L)

0+3.53+3.53Dead load

Resultant

[A~B]

L3L4(member force)

B

L2L3(member force)

A

Condition

P = 2.16 k Now, same as previous !!

Cee 312(7 & 8)(structural analysis)

Engineering

Transcript of Cee 312(7 & 8)(structural analysis)