Chapter 6-std
Chapter 6-std
Chapter 6-std
Steel Design
Chapter 6:
Compression Members under Combined
Axial Load and Bending Loads
Chapter 6-Content:
6.1 Introduction
We need to calculate:
-ve +ve
Moment Magnification
Factor
Case 2: For beam–columns
with transverse loads
between the supports
B C
W8x15
D
A
Example 1: (Solution)
Calculate the non-sway moment magnifier (B1) for the
column AB.
B C
D
A
Moment Magnification
Factor
Example 1: (Solution)
From Normal Force Diagram:
-ve +ve
Case 1: For beam–columns with
no transverse loads between the
supports
Single Curvature Double Curvature
M2 = 520
M1 = 0
-ve +ve
Cm = 0.6 – 0.4 * 0/520 = 0.6
Example 1: (Solution)
Calculate the non-sway moment magnifier (B1) for the
column AB.
B C
0.6 D
B1 = = 1.034 > 1.0
98
1 – 233.5
A
B1 = 1.034
Moment Magnification
Factor
Example 2:
Calculate the sway moment magnifier (B2) for the
columns in the first story (i.e., the groundfloor
of the typical moment frame in Figure
columns) the service gravity and lateral wind loads
considering
shown.
Example 2: (Solution)
Example 2: (Solution)
1.2 D:
1.2*(30*16 + 20*16 + 20*16) = 1344 kN
L:
1.0*(15*16 + 15*16) = 480 kN
Lr:
0.5*(10*16) = 80 kN
1.6 W:
1.6*(16 + 32 + 32) = 128 kN
H = 128 kN
Total Factored Moment
The total factored second-order moment or
required moment strength is:
Total Factored Moment
b = 0.90
Beam-Column Design
Pu: Ultimate factored axial compressive Force.
Mu = B1 Mnt + B2 Mlt
6.3 Beam-Column
Analysis
Beam-Column Design
Ultimate Factored Loads
Mu, Pu
E
Limit = 0.56 = 0.56 200,000
250 = 15.84
Fy
E
Limit = 1.49 = 1.49 200,000
250 = 42.1
Fy
Given:
-Lx = Ly = 5 m
-Column is hinged (==pinned) at both ends.
Beam-Column Design
Example 3: (Solution)
Column is pinned at both
ends.
K = 1.0
Beam-Column Design
Example 3: (Solution)
E
KL/r = 64.9 < 4.71 = 4.71 200,000
250 = 133.2
Fy
Fy 250
Fcr = 0.658Fe *Fy = 0.658468.6 *250 = 199.97 MPa
n2E n2×200,000
= 468.6 MPa
Fe = (KL/r)2 = (64.9)2
Beam-Column Design
Example 3: (Solution)
Calculate the design axial compressive capacity.
Zx = 1590x103 mm3
Beam-Column Design
Example 3: (Solution)
Calculate nominal moment (Mn) based on
unbraced length (Lb)
The beam has continuous lateral support Lb = 0
Mn = Fy.Zx
Pu 1500
= 2213.7 = 0.68 > 0.2
Øc Pn
Interaction Equation:
Pu 8 150
+ ( Mux ) = 0.68 + 9 (357.75 ) = 1.053 > 1.0
Øb Mnx
8
Øc Pn
Unsafe and section need to be increased.
9
6.4 Beam-Column
Design
Beam-Column Design
Example 3: (Solution)
Check on Local Buckling of Flange:
b = bf/2
= 306 / 2 = 153 mm
b / tf = 153 / 17 = 9.0
𝐸
Limit = 0.56 = 0.56 200,000
250 = 15.84
𝐹𝑦
𝐸
Limit = 1.49 = 1.49 200,000
250 = 42.1
𝐹𝑦
Given:
-Lx = Ly = 5 m
-Column is hinged (==pinned) at both ends.
Beam-Column Design
Example 3: (Solution)
Column is pinned at both
ends.
K = 1.0
Beam-Column Design
Example 3: (Solution)
Calculate slenderness ratio about both principle
axes: KxLx/rx and KyLy/ry
1.0*5*1000/77.3 = 64.68
𝐸
KL/r = 64.68 < 4.71 = 4.71 200,000
250 = 133.2
𝐹𝑦
𝐹𝑦 250
Fcr = 0.658𝐹𝑒 *Fy = 0.658471.83 *250 = 200.3 MPa
𝜋2𝐸 𝜋2∗200,000
Fe = = = 471.83 MPa
(𝐾𝐿/𝑟)2 (64.68)2
Beam-Column Design
Example 3: (Solution)
Calculate the design axial compressive capacity.
Zx = 1770x103 mm3
Beam-Column Design
Example 3: (Solution)
Calculate nominal moment (Mn) based on
unbraced length (Lb)
The beam has continuous lateral support Lb = 0
Mn = Fy.Zx
𝑃𝑢
1500
∅ 𝑐 𝑃𝑛 = = 0.6118 > 0.2
2451.6
Interaction Equation:
Safe
W12 x 72 is an adequate section.