TOPIC5a - DESIGN OF COMPACT BEAMScee.pdf

ce sub

1. CE497
DESIGN OF COMPACT BEAMS
ENGR. DJ LEX YNOT
STEEL AND TIMBER DESIGN

2. BEAMS
If a beam can be counted on to remain stable up to the fully plastic condition,
the nominal moment strength can be taken as the plastic moment capacity
�
??????=�
??????
Otherwise, �
?????? will be less than �
??????

3. BEAMS
As with a compression member, instability can be in an overall sense or it can
be local. When a beam bends, the compression region (above the neutral axis)
will buckle if the member is slender enough. Unlike a column, however, the
compression portion of the cross section is restrained by the tension portion,
and the outward deflection (flexural buckling( is accompanied by twisting
(torsion). This form of instability is called lateral-torsional buckling
Lateral-torsional buckling can be prevented by bracing the beam against
twisting at sufficiently close intervals

4. BEAMS
Classification of Shapes
AISC classifies cross-sectional shapes as compact, noncompact, or slender,
depending on the values of the width-thickness ratios
Let?????? = width-thickness ratio
??????
?????? = upper limit for compact category
??????
?????? = lower limit for noncompact category
Then
If ??????≤??????
?????? the shape is compact
If ??????>??????
?????? the shape is slender
If ??????
??????<??????≤??????
?????? the shape is noncompact

5. BEAMS
Width-thickness parameters
Element ?????? ??????
?????? ??????
??????
Flange
??????
??????
&#3627409360;&#3627408533;
??????
&#3627409358;.&#3627409361;&#3627409366;
&#3627408492;
&#3627408493;
??????
&#3627409359;.&#3627409358;
&#3627408492;
&#3627408493;
??????
Web
??????
&#3627408533;
??????
&#3627409361;.&#3627409365;&#3627409364;
&#3627408492;
&#3627408493;
??????
&#3627409363;.&#3627409365;&#3627409358;
&#3627408492;
&#3627408493;
??????

6. BEAMS
Compact Section
For compact section, &#3627408448;
?????? = &#3627408448;
??????
&#3627408448;
??????≤??????
??????&#3627408448;
??????
&#3627408448;
??????≤??????
??????&#3627408441;
&#3627408486;??????
where
&#3627408448;
?????? = plastic moment
?????? = plastic section modulus

7. SAMPLE PROBLEM 1
The beam shown is a W16 x 31 of A992 steel. It supports a reinforced concrete
floor slab that provides continuous lateral support of the compression flange.
What is the design strength of this beam? Given: ??????
&#3627408467;/2??????
&#3627408467; = 6.28
??????
????????????
?????? = &#3627409360;&#3627409358;&#3627409360;.&#3627409363; ??????????????????&#3627408428;.??????&#3627408429;
If the service dead load is 450 lb/ft (excluding the weight of the beam) and the
service live load is 550 lb/ft, does this beam have adequate moment strength?

8. SAMPLE PROBLEM 1
Solution:
Check for compactness:
??????
??????=0.38
&#3627408440;
&#3627408441;
&#3627408486;
=0.38
29000
50
=9.15
??????=
??????
&#3627408467;
2??????
&#3627408467;
=6.28
??????
??????=1.0
&#3627408440;
&#3627408441;
&#3627408486;
=1.0
29000
50
=24.08
Since, ??????<??????
?????? the member is compact
&#3627408448;
?????? = &#3627408448;
??????
&#3627408448;
?????? = &#3627408441;
&#3627408486;??????
&#3627408485;
&#3627408448;
?????? = 5054=2700 kips.in
&#3627408448;
?????? = 225 kips.ft
Design strength:
??????
??????&#3627408448;
?????? = 0.90(225) kips.ft
??????
??????&#3627408448;
?????? = &#3627409360;&#3627409358;&#3627409360;.&#3627409363; ??????????????????&#3627408428;.??????&#3627408429;

9. SAMPLE PROBLEM 1
Solution:
Total service dead load:
??????
????????????=450+31=481 lb/ft
Total service live load:
??????
????????????=550 lb/ft
Maximum dead load moment: Maximum live load moment:
&#3627408448;
????????????=
1
8
??????
????????????&#3627408447;
2
&#3627408448;
????????????=
1
8
??????
????????????&#3627408447;
2
&#3627408448;
????????????=
1
8
48130
2
&#3627408448;
????????????=54112.5 lb.ft
&#3627408448;
????????????=54.11 kip.ft
&#3627408448;
????????????=
1
8
55030
2
&#3627408448;
????????????=61875 lb.ft
&#3627408448;
????????????=61.88 kip.ft

10. SAMPLE PROBLEM 1
Solution:
Total factored moment:
1.4&#3627408448;
????????????=1.454.11=75.75 kip.ft
1.2&#3627408448;
????????????+1.6&#3627408448;
????????????=1.254.11+1.6(61.88)=163.94 kip.ft
Use ??????
&#3627408534;=&#3627409359;&#3627409364;&#3627409361;.&#3627409367;&#3627409362; kip.ft
Since the design moment (202.5 kip.ft) is greater than the factored moment (163.94), the
section W16 x 31 is satisfactory

11. SAMPLE PROBLEM 2
The beam shown below is a W10 x 77 and has continuous lateral support. The
load P is a service live load. If &#3627408441;
&#3627408486; = 50 ksi, what is the maximum permissible
value of ??????? Use LRFD and ASD
Properties of W10 x 77
??????
&#3627408468;=22.6 in
2
ℎ=10.6 in
??????
&#3627408484;=0.53 in
??????
&#3627408467;=10.19 in
??????
&#3627408467;=0.870 in
??????
&#3627408485;=97.6 in
3

12. SAMPLE PROBLEM 2
Solution:
Check for compactness:
??????
&#3627408467;
2??????
&#3627408467;
=
10.19
20.87
=5.86 0.38
&#3627408440;
&#3627408441;
&#3627408486;
=0.38
29000
50
=9.15>5.86(flange is compact)
ℎ
??????
&#3627408484;
=
10.6
0.53
=20
3.76
&#3627408440;
&#3627408441;
&#3627408486;
=3.76
29000
50
=90.55>20(web is compact)
Because the beam is compact and laterally supported, the nominal flexural strength is:
&#3627408448;
??????=&#3627408448;
??????
&#3627408448;
??????=&#3627408441;
&#3627408486;??????
&#3627408485;
&#3627408448;
??????=5097.6=4880 kips.in×
1ft
12in
=406.67 kip.ft

13. SAMPLE PROBLEM 2
Solution:
(a) LRFD
&#3627408448;
??????=∅&#3627408448;
??????
&#3627408448;
??????=0.90406.67
Total service dead load:
??????
????????????=77 lb/ft
Total service live load:
??????
???????????? (concentrated at midspan)
Maximum dead load moment: Maximum live load moment:
&#3627408448;
????????????=
1
8
??????
????????????&#3627408447;
2
&#3627408448;
????????????=
??????
????????????&#3627408447;
4
&#3627408448;
????????????=
1
8
7730
2
&#3627408448;
????????????=8662.5 lb.ft
&#3627408448;
????????????=8.66 kips.ft
&#3627408448;
????????????=
??????
????????????(30)
4
&#3627408448;
????????????=7.5??????
???????????? lb.ft
&#3627408448;
????????????=0.0075??????
???????????? kips.ft
&#3627408448;
??????=366 kips.ft
&#3627408448;
??????=1.2&#3627408448;
????????????+1.6&#3627408448;
????????????
366=1.2(8.66)+1.60.0075??????
????????????
??????
????????????=29634 lb=&#3627409360;&#3627409367;.&#3627409364;&#3627409361; ??????????????????&#3627408428;

14. SAMPLE PROBLEM 2
Solution:
(b) ASD
&#3627408448;
??????=&#3627408448;
??????/Ω
&#3627408448;
??????=406.67/1.67
Total service dead load:
??????
????????????=77 lb/ft
Total service live load:
??????
???????????? (concentrated at midspan)
Maximum dead load moment: Maximum live load moment:
&#3627408448;
????????????=
1
8
??????
????????????&#3627408447;
2
&#3627408448;
????????????=
??????
????????????&#3627408447;
4
&#3627408448;
????????????=
1
8
7730
2
&#3627408448;
????????????=8662.5 lb.ft
&#3627408448;
????????????=8.66 kips.ft
&#3627408448;
????????????=
??????
????????????(30)
4
&#3627408448;
????????????=7.5??????
???????????? lb.ft
&#3627408448;
????????????=0.0075??????
???????????? kips.ft
&#3627408448;
??????=243.51 kips.ft
&#3627408448;
??????=&#3627408448;
????????????+&#3627408448;
????????????
243.51=8.66+0.0075??????
????????????
??????
????????????=31313.33 lb=&#3627409361;&#3627409359;.&#3627409361;&#3627409359; ??????????????????&#3627408428;