3
PLASTIC SECTIONAL MODULUS (zx ), in
• Shape Property, Independent of material
• Higher Z x indicates more bending strength
zx   Ai Yi
y1  1"
y1  0.5"
A 1  12 in
A 1  12 in
2
2
PNA
y2  0.5"
zx  A1Y1  A2 Y2
 12x0.5  12x0.5
 66
Zx  12in
3
A 2  12 in
2
y 2  1"
A 2  12 in
zx  A1Y1  A2 Y2
 12x1  12x1
Zx  24 in3
2
A 1  12 in
A 1  12 in
2
2
y1  3"
y1  1.5"
PNA
y2  1.5"
A 2  12 in
y2  3"
2
A 2  12 in
2
zx  12x1.5  12x1.5
zx  12x3  12x3
Zx  36 in3
Zx  72 in3
 18  18
 36  36
A 1  8 in
2
A 2  4 in
2
y1 
y2 
PNA
y3 
A 3  4 in
A 4  8 in
y4 
2
zx   Ai Yi
2
zx  A1Y1  A2 Y2  A3 Y3  A4 Y4
 8x4.5  4x2  4x2  8x4.5
 36  8  8  36
Zx  88 in3
A 1  10 in
2
A 2  2 in
TOTAL A  A1  A 2  A3  A 4
2
 10  2  2  10
 24 in 2
y1 
y2 
PNA
(Plastic Neutral Axis)
y3 
y4 
A 4  10in
A 3  2 in
2
2
Zx  A1Y1  A2 Y2  A3 Y3  A 4 Y4
 10x4.5  2x2  2x2  10x4.5
 45  4  4  45
 98 in 3
A 1  10 in
TOTAL A  A1  A 2  A3  A 4
2
A2
 2 in 2
 10  2  2  10
 24 in 2
y1 
y2 
PNA
(Plastic Neutral Axis)
y3 
y4 
A 4  10in
A 3  2 in
2
2
Zx  A1Y1  A2 Y2  A3 Y3  A 4 Y4
 10x8.5  2x4  2x4  10x8.5
 85  8  8  85
 186 in 3
Zx  12 in3
Zx  88 in3
Zx  24 in3
Zx  36 in3
Zx  72 in3
Zx  98 in3
Zx  186 in3
fy
 
 f y  bd
2 2
 

1 d
3 2

2 d
3 2

2 d
3 2
 
 f y  bd
2 2
 

1 d
3 2
f y FULLY LOADED
M(INT ) 
FORCE X LEVERARM
  
2
 f y  bd
2d
bd


 
x
 fy
 6 
3
2 2
M(INT )  f y .S x
S x For rectangular sections=
bd 2
6
fy
fy
f  bd2 
f   bd2 
f  bd2 
f  bd2 
y
y
y
y
M(INT )  FORCE X LEVERARM
 fy 
 
bd x d  f  bd 2 
y
 4 
2
2
M(INT )  f y .Z x
zx   Ai Yi
zx   Ai Yi
 A1Y1  A 2 Y2
  bd  .  d    bd  .  d 
 2  4  2  4
  
2
2
2
bd
bd
bd



8
8
4
8Fy
8Fy
SECTION
4Fy
4Fy
4Fy
4Fy
8Fy
8Fy
ELEVATION
M(INT )  8Fy x9  4Fy x4
M  72Fy  16Fy
M  88Fy
M  ZxFy
Zx  88 in3
LATERAL
SUPPORT OF
STEEL BEAMS
47 kip-ft
CASE I
CASE II
CASE III
W10X12
28.5 kip-ft
CASE I
CASE II
CASE III
W10X12
BRACING DISTANCE
CASE I if lateral brace is spaced 0-2.75’
CASE II if lateral brace is spaced 2.75’-8’
CASE III if lateral brace is spaced more than 8’
CASE I
BRACES
W10X12
0L b 2.75'
Design Moment Strength =
M
N
= 47 kip-ft
CASE II
LATERAL BRACES
Lb
W10X12
Design Moment Strength reduces as L b increases
AT L b  2.75'
AT L b  8 '
Linear variation in L b &
 M  47 kip-ft
 M  28.5 kip-ft
N
N
M
N
2.75'L b 8'
CASE III
Lb
L b  8'
Should be avoided for load bearing floor beams.
Design Moment Strength reduces as L b increases
AT L b  8 '
AT L b  18 '
M
M
N
 28.5 kip-ft
N
 8.25 kip-ft
CASE I CONSTRUCTION DETAILS
x
CONCRETE
STEEL STUDS
W-SHAPE
x
X-X
Lb  0
Most Common Current Practice Using Metal Deck
and Shear Studs. Steel and Concrete Deck can
be designed as Composite or Non-Composite
Use Fy  50 ksi and select a shape for a typical floor beam AB. Assume that the
floor slab provides continuous lateral support. The maximum permissible live load
deflection is L/180. The service dead loads consist of a 5-inch-thick reinforcedconcrete floor slab (normal weight concrete), a partition load of 20 psf, and 10 psf
to account for a suspended ceiling and mechanical equipment. The service live
load is 60 psf.
• Fy = 50ksi
GIRDER
• Case 1
•
LL
not to exceed
L
180
• DEAD LOADS
1) 5” Slab
2) Partition = 20psf
3) Ceiling, HVAC = 10psf
• LIVE LOAD = 60psf
GIRDER
(PRIMARY BEAMS)
Figure 1
DESIGN LOAD (kips/ft) on AB = wu x TRIBUTORY AREA
LENGTH OF BEAM

[1.2x(62.5  20  10)  1.6(60)]x[6x30] x 1
30
1000
= 1.242 kips/ft
12” of Slab = 150 psf
6” of Slab = 75 psf
1” of Slab = 12.5 psf
5” of Slab = 62.5 psf
(12.5 psf for every inch of concrete
thickness)
*5 x 12.5 = 62.5
wu = 1.242 kips/ft
wuL2 1.242x302
Mu =
= 139.7 ft-kips
8 =
8
CASE 1
Page # 3-127
STRENGTH OF W14 X 26
= 150.7 ft-kips
STRENGTH OF W16 X 26
= 165.7 ft-kips
139.7
Page # 1-22
Page # 1-23
Page # 1-20
Page # 1-21
SHAPE
IX
AREA
W 14x26
7.69
245
W16x26
7.68
301
SELECT W16X26
2
1.2x0.026x30
EXTRA SELF WEIGHT MOMENT =
= 3.3 ft-kips
8
MOMENT STRENGTH  APPLIED MOMENT
165.7  139.7 + 3.3
W16X26 is OK
CODE
=
30x12
180
=
360
180
= 2”
ACTUAL
=
5 wl
4
384 EI
Page # 3-211
w=
(60)x(6x30)
30
= 360 lb/ft = 0.360 kips/ft
0.360 kips/in = 0.03 kips/in

12
=
5x0.03x(30x12)4
384x29,000x301
4
5 wl
;
=
384 EI
= 0.75”  2” OK
 kips 

 . (in)4
 in 
kips (in)4
.
2
in
in 2
=
in
= in
Note that duct holes have to be
strengthened by plates. Also, holes
are at third point where shear &
moment are not maximum.
Typical Copes for a shear connection of
a large girder to column web.
Cantilever construction
for projected balcony.
If shear studs are noticed on
beams and column then those
members have to be encased in
concrete for increasing fire
resistance of steel.
Details of web opening in steel girders for HVAC ducts.