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```STEEL AND TIMBER DESIGN
CE-506
SUBMITTED BY:
ESGUERRA, ALFA ROMEO C.
QUITO, FRANCES NICOLE
SUBMITTED TO:
ENGR. BRYAN DALE P. YU, MSCE
BEAM-COLUMN
1.
A steel column W 300 mm 203 kg/m
section is subjected to an axial load of
2670 kN and moments from framing
beams as follows:
Mx at the top=108 kN.m.
Mx at the bottom= -108 kN.m.
Find the critical buckling load (kN)
Prop. Of
the WF
Column
A= 2570 mm2
tr= 32 mm
d= 340 mm
Ix= 5.16x108 MPa
br= 315 mm
Iy= 1.65x108 MPa
tw= 20 mm
E= 200 GPa
Fy= 345.6 MPa
Unbraced column length= 3m
Effective length factor, K=1.0
Sidesway is prevented
P α=
π 2 EI
( KL)2
P α=
π 2 ( 200000 ) (1.65 x 108 )
(1 x 3000)2
Pα = 36188550 N
Pα = 36189 kN
2.
A W12 x 106 beam-column connection
shown is pin-connected at both ends. It is
subjected to an axial load of 1338 kN and
a mment of 122 kN.m. at the top of a
moment of 163 kN.m. at the bottom.
Bending is about the strong axis.
Sidesway is not prevented and K=1.18.
Use A36 steel with Fy= 248 MPa.
Unsupported height of the column is 4.27
m.
Determine the allowable bending stress
about the strong axis if only bending is
acting.
Prop. Of W12 x 106
A= 20129 mm2
tf= 25.15 mm
d= 327.41 mm
Ix= 388x106 mm4
bf= 310.39 mm
Iy= 125x106 mm4
tw= 15.49 mm
Sx= 2376x103 mm3
rx=138.94 mm
Sy= 808x103 mm4
ry=78.99 mm
L
b
=
4270 mm
L c=
200 b f
√ Fy
L c=
200(310.39)
√ 248
Lc= 3942
137900
L u=
Fyd
bftf
L u=
137900
248 (327.41)
( 310.39 ) ( 25.15)
Lu=13258 mm
Lb&gt;Lc
Lb&lt;Lu
Use Fbx=0.60Fy
Fbx= 0.60(248)
Fbx=148.8MPa
3.
A W310 x 79 column of A36 steel
supports a vertical load of 178 kN. The
length is 3.6 m. and K=1.0 with respect to
both axes. Sidesway is prevented by
bracing. Fy= 250 MPa, E= 200,000 MPa.
Determine the allowable compressive
stress if only axial load is acting
L/r 57.23
=
=0.455
Cc 125.66
L 2
( )
r
Fy
Fa= 1 −
2
2 Cc FS
[ ]
L
L 3
3( ) ( )
5
r
r
FS= +
−
3 8Cc 8 Cc3
3
5 3(0.455) (0.455)
FS= +
−
3
8
8
FS=1.83
Prop. Of W310 x 79
A= 10100 mm2
d= 306 mm
bf= 254 mm
tw= 8.8 mm
rx=132 mm
ry=62.9 mm
KL 1(3600)
=
=27.27
rx
132
KL 1(3600)
=
=57.23
ry
62.9
L
Use =57.23
r
Cc=
√
2 π 2( 200,000)
250
Cc=125.66&gt;57.23
tf= 14.6 mm
Ix= 177x106 mm4
Iy= 39.9x106 mm4
Sx= 1160x103 mm3
Sy= 314x103 mm3
(0.455)2 250
Fa= 1−
2
1.83
[
]
Fa=122.47 MPa allowable compressive
stress
4.
A W310 x 58 beam-column carries an
axial load of 1338 kN and a moment of
122 kN.m at the top and 163 kN.m. at the
bottom causing a reverse curvature.
Bending is about the strong axis.
Sidesway is not prevented by adequate
bracings. Use K=1.18 and A36 steel
Fy=250 MPa. Length of column is 4.2 m.
Determine the allowable compressive
strength if only axial load is acting.
Prop. Of W250x58
A= 20100 mm2
d= 327 mm
bf= 310 mm
tw= 15.5 mm
rx=139 mm
ry=78.9 mm
tf= 25.1 mm
Ix= 386x106 mm4
Iy= 125x106 mm4
Sx= 2360x103 mm3
Sy= 805x103 mm3
Cc=
√
2 π 2( 200,000)
250
Cc=125.66
L/r 62.81
=
=0.50
Cc 125.66
L 2
( )
r
Fy
Fa= 1 −
2
2 Cc FS
[ ]
L
L 3
3( ) ( )
5
r
r
FS= +
−
3 8Cc 8 Cc3
3
5 3(0.50) (0.50)
FS= +
−
3
8
8
FS=1.84
(0.50)2 250
Fa= 1−
2
1.84
[
]
Fa=118.89 MPa allowable compressive
stress
KL 1.18(4200)
=
=36.65
rx
139
KL 1.18(4200)
=
=62.81
ry
78.9
L
Use =62.81
r
5.
A W14x90 column in a building frame for
a 7m. story height supports a 1160 kN
axial compression load and a moment at
the top Mx1=+136 kN.m. and a moment of
Mx2=+163 kN.m. at the bottom ends as
shown. The column is braced against
sidesway in the weak bending plane of
the column (x-z plane) but with possible
sidesway in the strong bending plane (yz
plane). Use A36 steel.
Calculate the allowable bending stress
L c=
200(368.81)
√ 250
Lc= 4665
137900
L u=
Fyd
bftf
L u=
137900
250(356.11)
( 368.81 ) (18.03)
Lu=10300 mm
Lb&gt;Lc
Lb&lt;Lu
Use Fbx=0.60Fy
Fbx= 0.60(250)
Fbx=150 MPa
Prop.
W14x90
A= 17097 mm2
d= 356.11 mm
tw= 11.18 mm
rx=155.96 mm
ry=93.98 mm
Kx=1.20
Lb= 7000
L c=
200 b f
√ Fy
Of
tf= 18.03 mm
bf= 368.81 mm
Sx= 2343x103 mm3
Sy= 818x103 mm3
Ky=1.0
Fy=250 MPa
6.
A W360x134 steel is used to support
an axial load of 1248 kN and a weak
axis moment of 80 kN.m. The member
is a part of a frame in which sidesways
is not prevented. The length is 4.2 m.
and K is estimated to be 1.3.
Fy=350MPa.
KL 1.3(4200)
=
=58.09
ry
156
Use
KL
=58.09
r
2 π 2( 200,000)
Cc=
350
√
Cc=106.21&gt;58.09
KL/r 58.09
=
=0.547
Cc
106.21
KL 2
)
r
Fy
Fa= 1 −
2
2Cc FS
[ ]
(
KL
KL 3
3(
) (
)
5
r
r
FS= +
−
3 8 Cc
8 Cc3
Prop. Of W360x134
d=369 kk
bf=369 mm
tf=28 mm
tw=11.2 mm
ry= 94 mm
3
2
A=17100 mm
Sx=2330x103 mm3
Sy=817x103 mm3
rx=156 mm
Determine the allowable bending stress if
only bending moment alone existed.
Fb=0.75Fy (along weak axis)
Fb=0.75(350)
Fb=263.50 MPa
Determine the allowable bending stress if
only axial load is acting.
KL 1.3(4200)
=
=35
rx
156
5 3(0.547) (0.547)
FS= +
−
3
8
8
FS=1.85
[
Fa= 1−
(0.547)2 350
2
1.85
]
Fa=160.89 MPa allowable compressive
stress
1.
BOLTED CONNECTIONS
An eccentrically loaded bolted connection
is shown. Diameter of A325 bolts in
standard holes is 20 mm. The applied load
is 100 kN. Determine the polar moment of
inertia of the group of bolts.
2. A lapped, bolted tension member is
shown. Diameter of bolts are 18 mm &oslash;
and the plate material is A36 steel
Fy=250 MPa., Fu=400 MPa. Assume the
fasteners are adequate and do not
control the tensile capacity. Diameter of
hole is 3mm bigger than the diameter of
the bolt.
a. Determine the tensile capacity of the
lapped joint based on gross area.
P=0.60FyAg
Ag=Width x thickness
P=0.60 (250) (300) (12)
P=540kN
b. Determine the tensile capacity of the
lapped joint based on net area.
P=0.50FuAn
An= 300(12)-3(21)(12)
An= 2844 mm2
P=0.50(400)(2844)
P=568.8 kN
∑(x2+y2) =6(75)2+4(100)2
∑(x2+y2) =73750 mm2
3.
A double shear butt connection is
connected by 9-22mm dia. Bolts
Thickness of the outer plate is 12.5 mm
while the main plate is 22 mm. Spacing of
bolts is 75 mm with an edge distance of
37.5 mm. A36 steel is used. Fy=248 MPa,
Fu= 400 MPa.
Allowable shear stress of bolts= 117 MPa
Allowable bearing stress of bolts= 1.2 Fu
Dia. Of hole= 3mm greater than bolt
diameter.
a.
4. An eccentrically loaded connection is
subjected to an axial vertical load P acting
at an eccentricity “e”. The bracket is a
structural tee having a web thickness of 6
mm. A325 high strength bolts is connected
in two vertical rows and has a diameter of
20 mm. The bolts are in standard holes
having a diameter of 22 mm &oslash; and are
spaced at 90 mm vertically. Assume that
the column and brackets are adequate.
Allowable shear stress of bolts= 145 MPa
Determine the capacity of the connection
based on the strength of the bolts
SHEAR OF BOLTS
a.
Compute the value of P based on the
allowable shear stress.
fv= P/A
T=AvFv
π
T = ( 22 )2( 2)&iquest;
4
T= 800559 N
BEARING OF BOLTS
T= AbFp
Fp=1.2Fu
T= 22(22)(9)(1.2)(400)
T=2090880 N
Use T= 800.56 kN
fv=
P
π
( 20 )2 (8)
4
P=328.89 kN
5.
Two plates each with thickness t=18mm
are bolted together with 6 – 22 mm &oslash;
bolts forming a lap connection. Bolts
spacing are as follows: S1= 40 mm, S2= 80
mm, S3=100 mm. Bolt hole diameter= 25
mm.
Allowable stress:
Shear stress of the bolt: Fv=100 MPa
Bearing stress of the bolt= Fp=1.2Fu
a.
Calculate permissible tensile load P based
on shear capacity of bolts
π
P= ( 22 )2 (2)(6)(100)
4
P= 228.08 kN
b.
Calculate permissible tensile load P based
on bearing capacity of bolts
P= AbFp
P= 18(22)(6)(1.2)(400)
P=1140.48 kN
6. The bracket shown is a structural tee. It is
subjected to an eccentric load of 112 kN
at an eccentricity of 200 mm. Diameter
of A325 N high strength bolts is 19 mm.
The bolts are placed in two vertical rows
and are spaced 75 mm vertically.
Assume the column and brackets are
adequate. Allowable shear stress is 145
MPa.
Determine the max. tensile stress of the
bolt.
I=Ad2
π
2
2
2
I&iquest; ( 19 ) [ 4(112.5) + 4(37.5) ]
4
I=15.95x106
M=112000(200)
M=22.4 x106 N.mm
Mc
I
22.4 x 106 (112.5)
f t=
15.95 x 10 6
ft =158 MPa
WELDED CONNECTIONS
Two plates 300 mm x 16mm is to be
connected using butt weld connections.
Use A36 steel. Fy=250 MPa, Fu= 400 MPa
and electrodes E70 with Fu=485 MPa.
f t=
1.
a. Determine the capacity of the plate
connection assuming the butt weld is
adequate. Use a reduction factor of
0.85 for the Net Area.
Based on gross area:
T=0.60FyAg
T=0.60(250)(300)(16)
T=720 000 N
Based on net area:
T= 0.50FuAe
T=0.50(400)(0.85)(300)(16)
T=816 000 N
T=720 kN
2.
A welded bracket shown is required to
assist a load of 90 kN. The steel is A36
(Fy= 250 MPa) and the welding is to be
R=√ (470.55)2 +(352.91)2
R=588.19 N/mm
performed by using E70 electrodes
(SMAW) Fu= 485 MPa. J=10.4 x 106 mm4
Determine the force on the weld due to
torsional moment
3.
A plate used as a tension member is
connected to a gusset plate, as shown in
Figure 7.40. The welds are 3/16 – inch
fillet welds made with E70XX electrodes.
The connected parts are A36 steel/
Assume that the tensile strength of the
member is adequate, and determine the
availability strength of the welded
connection.
37.5+e=150+250
e=362.50
ɵ=0&deg;
M=90000(362.5)
Fnw = 0.60FEXX
M=32.625x106N.mm
My
J
32.625 x 106 (112.50)
f x=
10.4 x 106
fx=470.55 N/mm
f x=
Rn= 0.707wFnw = 0.707(3/16)(0.6 x 70)
= 5.568 kips/in.
The allowable strength of the weld is
Rn/Ω = 5.568 / 2.00 = 2.784 kips/in.
The shear yield strength is
Rn/Ω = 0.6Fyt / 1.50 = 0.6(36)(1/4) /
1.50 = 3.6 kips/in.
The shear rupture strength is
Rn/Ω = 0.6Fut / 2.00 = 0.6(58)(1/4) /
2.00 = 4.35 kips/in.
The weld shear strength controls. For the
connection,
Rn/Ω = 2.784 kips/in. x (4 + 4)in =
22.3kips
Allowable strength of the weld is 22.3
kips
4.
A double fillet lap joint is shown in the
figure. The steel is A36 steel Fy=250 MPa
and the electrode used was E70 (Fu=480
MPa) The fillet weld is 8 mm and the
shielded metal arc welding (SMAW) as
used.
x=47.5 mm
5. The 150mm x 18 mm plate is welded to a
300 mm x 18 mm plate by a 10 mm fillet
weld. The steel is A36 and the electrode
use was E70 (SMAW) shielded metal arc
welding.
For welding:
Fu=483.8 MPa
For steel plate:
Fy= 248 MPa, Fu= 400 MPa
a. Determine the allowable tensile load
that may be applied to the connection
based on shear capacity of the fillet weld.
T=0.707tL0.30Fu
T=0.707(8)(200+200)(0.3)(480)
T=325785.6 N
T=325.8 kN
b. Determine the allowable tensile load
that may be applied to the connection
based on gross area of the plate
T= 0.60FyAg
T=0.60(250)(200)(9.5)
T=285 000 N
T= 285 kN
c. Determine the minimum length of the
lap
x= 5(9.5)
(5x the thinner part joined but not less
than 25 mm)
a.
Determine the tensile load of the welded
lap joint based on shear capacity of fillet
T=0.707tL0.30Fu
T=0.707(10)(125+125)(0.3)(483.8)
T=410 456 N
T=410.46 kN
a. Determine the tensile load of the
welded lap joint based on gross area of
plate.
T= 0.60FyAg
T=0.60(248)(150)(18 )
T=401 760 N
T= 401.8 kN
6.
The bracket carrying a load P is 14 mm
thick. It is welded. Allowable weld stress,
Fv= 145 MPa, weld thickness=9.5mm
a.
If P =200 kN and it acs at the centroidal
axis of the weld group, calculate the force
per unit length (kN/m) on the weld.
F=P/L
F=200/ (0.125(2)+(0.25))
F= 400 kN/m
SHEAR IN BEAMS
1.
A W 420 x 85 steel is fully restrained with
a uniformly distributed superimposed
load of 25 kN/m. The beam has a span of
10 m.
Properties of W420x85
A = 10839 mm2
d = 420 mm
tw = 11 mm
Ix = 310 x 10⁶ mm⁴
tf = 18 mm
bf = 180 mm
a. Compute the bending stress in MPa.
Bending stress in MPa.
85(9.81)
1000
W = 25.834 KN/m
W = 25 +
W L2
12
25.834(10)2
Ma =
12
Ma = 215.28 KN.m
Ma =
fb =
MC
I
215.28 x 106 ( 420)
2(310)106
fb = 145.83 MPa
fb =
b. Compute the maximum web shear
stress in MPa.
V
dtw
2R = 25.834(6)
R = 77.50 KN
V = 77.50 KN
V
fv =
dtw
77500
fv =
420(11)
fv = 16.77 MPa
fv =
2.
A W 310 x 74 of A36 steel Fy=250 MPa is
on a simple span of 9m. Assume full
lateral support. The beam supports a
uniformly distributed load of 3.6 kN/m
Determine the maximum concentrated
load that can be placed at the mid-span so
that it will not exceed the allowable
shearing stress
fv= 0.40Fy
fv =
V
dtw
0.40Fy =
V
dtw
0.40(250)=
V
305(17.4)
V=530.7 kN
P WL
V= +
2 2
P 3.6(9)
530.7= +
2
2
P=1029 kN
3. A fixed ended WF beam has a span of 6m
it carries a load of 135 kN acting at 2m
from B. I=91 x106 mm4, Vmax= 100 kN
Prop. Of WF section
d = 240 mm
bf = 160 mm
tw = 20 mm
tf = 20 mm
a. Determine the max. vertical shearing
stress of the beam
Q=160(20)(110)+100(20)(50)
Q=452000 mm3
VQ
Max. τ =
Ib
Max.
τ
=
100000(452000)
91 x 10 2( 20)
Max. τ= 24.84 MPa
b. Ave. shearing stress in the web
Q=160(20)(110)
Q=352000
VQ
τ=
Ib
100000 ( 352000 )
τ=
91 x 10 2 (20 )
τ=19.34 MPa
τave=19.34 + 2/3 (5.5)
1457 x 106
10958 x 103
f b=134.60 MPa
τave= 23 MPa
f b=
4. For the section shown, the shear force
acting is 670 kN and the bending moment
is 1475 kN-m
b. Compute the total shear carried by the
web and the flanges.
F v=
a.
Compute the max. bending stress of the
beam shown.
I=
+
+
[
[
10 ( 1050 )3
12
3
300 ( 16 )
+ 300 ( 16 )( 533 )2 2
12
]
]
250 ( 16 )3
+ 250 ( 16 )( 549 )2 2
12
I =6103.5 x 106
I
S=
C
6103.5 x 106
S=
557
S=10958 x 103
f b=
M
S
VQ
Ib
For the flanges:
Q=250 ( 16 )( 549 ) +300 ( 16 ) ( 533 )
Q=4754400
VQ
Ib
4754400 ( 670000 )
F v=
6103.5 x 10 6 ( 10 )
F v =52.19 MPa
F v=
Average stress:
V
d tw
670000
f v=
1114 ( 10 )
f v =60.14 MPa
f v=
Shear force carried by the two flanges:
1.74
( 2 ) [ 250 ( 16 ) +300 ( 16 ) ]
2
F f =15312 N
Ff=
5.
A W 24 x 94 beam is subjected to a shear
force of 900 KN.
900000(1.52 x 106 )
fv =
1119.7 x 106 (13.11)
Properties of W sections:
A = 17870 mm2
tf = 22.15 mm
bf=230.15mm
tw = 13.11 mm
d = 616.97 mm
Ix=1119.7 x 10⁶ mm⁴
fv = 93.19 MPa
b. Compute the shear stress at neutral
axis.
Q = 230.15(22.15)(297.41) +
286.335(13.11)(143.1675)
Q = 2.05 x 10⁶ mm3
VQ
fv =
Ib
900000(2.05 x 106 )
fv =
1119.7 x 106 (13.11)
fv = 125.69 MPa
a.
Compute the
shear stress at junctions of flange and
web.
VQ
Ib
Q = 230.15(22.15)(297.41)
Q = 1.52 x 10⁶ mm3
fv =
c. Compute the shear force carried by the
flange.
( 2 ) 5.31
(230.15)(22.15)
2
F = 27069 N
F=
d. Compute the shear force carried by the
web.
F = 900000 – 27069
F = 872931 N
e. Compute the average shear stress on
web.
b = 230.15
VQ
Ib
b = 13.11 mm
fv =
fv =
V
dtw
fv =
900000
616.97(13.11)
fv = 111.27 MPa
6.
A wide flange section is subject to a
b. Determine the max shearing stress at
vertical of 356 kN. The moment of inertia
the neutral axis.
at the neutral axis 619 x 106. Fy= 250 MPa.
a.
Determine
the shearing stress in the flange.
Q = 265(25)(203) + 190.5(14.9)(95.25)
Q = 1615237.4mm3
F v=
VQ
IB
F v=
356 000(1615237.4)
619 x 10 6 (14.9)
Fv= 62.35 MPa
F v=
VQ
IB
Q = 265(25)(203)
Q = 1344875 mm3
F v=
VQ
IB
F v=
356000(1344875)
619 x 106 (265)
Fv= 2.92MPa
```