Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
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Question
Chapter 6, Problem 6.6.12P
To determine
(a)
If the given member satisfies the AISC specification using Load and Resistance Factor Design (LRFD) method.
To determine
(b)
If the given section satisfies the AISC specification using Allowable Strength Design (ASD) method.
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4.3-4
Determine the available strength of the compression member shown in Figure P4.3-4.
in each of the following ways:
a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and
the allowable strength for ASD.
15
HSS 10x6x
ASTM A500, Grade B steel
(Fy=46 ksi)
2/3
The beam shown in Figure is a two-span beam with a pin (hinge) in the center of the left span,making the beam statically determinate. There is continuous lateral support. The concentratedloads are service live loads. Determine whether a W12 × 79 of A992 steel is adequate.a. Use LRFD.b. Use ASD.
A W14X120 is used as a tension member in atruss. The flanges
of the member are connected to a gusset plate by 3/4 inch
boltas shown below. Use A36 steel with Fy-36 ksi and Fu=58
ksi
Determine the Yielding Capacity of the section based on
LRFD (kips)
Determine the Tensile Rupture capacity of the section
based on LRFD
Determine the Demand to Governing Capacity Ratio (based
on yielding and rupture only) if the Demand load carried by
the section are DL=200 kips LL=400 kips use LRFD
Properties and Dimension
Ag=35.30 in^2
x = 6.24 in
ry= 3.74 in
d=14.5 in
tf=0.94 in
bf=14.7 in
tw=0.59 in
k=1.54
d=14.5
Y
k1=1.5
bf=14.7
tf-0.94
X
-tw=0.59
H
Chapter 6 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 6 - Prob. 6.2.1PCh. 6 - Prob. 6.2.2PCh. 6 - Prob. 6.6.1PCh. 6 - Prob. 6.6.2PCh. 6 - Prob. 6.6.3PCh. 6 - The member shown in Figure P6.6-4 is part of a...Ch. 6 - Prob. 6.6.5PCh. 6 - Prob. 6.6.6PCh. 6 - Prob. 6.6.7PCh. 6 - Prob. 6.6.8P
Ch. 6 - Prob. 6.6.9PCh. 6 - Prob. 6.6.10PCh. 6 - Prob. 6.6.11PCh. 6 - Prob. 6.6.12PCh. 6 - Prob. 6.6.13PCh. 6 - Prob. 6.7.1PCh. 6 - Prob. 6.7.2PCh. 6 - Prob. 6.8.1PCh. 6 - Prob. 6.8.2PCh. 6 - Prob. 6.8.3PCh. 6 - Prob. 6.8.4PCh. 6 - Prob. 6.8.5PCh. 6 - Prob. 6.8.6PCh. 6 - Prob. 6.8.7PCh. 6 - Prob. 6.8.8PCh. 6 - Prob. 6.8.9PCh. 6 - Prob. 6.8.10PCh. 6 - Prob. 6.9.1PCh. 6 - Prob. 6.9.2P
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Similar questions
- A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by ¾ inch boltas shown below. Use A36 steel with Fy=36 ksi and Fu=58 ksi Determine the Yielding Capacity of the section based on LRFD (kips) Determine the Tensile Rupture capacity of the section based on LRFD Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFDarrow_forward2) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNarrow_forwardDetermine the maximum axial compressive service load that can be supported if the live load is twice as large as the dead load. Use AISC Equation E3-2 or E3-3. a. Use LRFD. b. Use ASDarrow_forward
- A plate is used as a tension member, to carry a deadload = 300 KN and live load of 260 KN. Steel used is A36, Fy= 248 MPa, Fu = 400 MPa. If the width of the tension plate is 210 mm, determine: a) The thickness of the plate based on NSCP 2015 ASD, bult used is M20.------ b) The thickness of plate based on NSCP 2015 LRFD. Bolt used is M20.-------- I M20 bolts- 210 mm -Tension Platearrow_forwardA PL 38 x 6 tension member is welded to a gusset plate as shown in figure. The steel is A36. PL ½ x 6 The design strength based on yielding is nearest to: The design strength based on rupture is nearest to: The design strength for LRFD is nearest to: The allowable strength based on yielding is nearest to: The allowable strenath based on rupture is nearest to: The allowable strength for ASD.arrow_forwardA built-up section was made using PL414x12mm thk plates as shown in the figure below. It is pinned at both ends with additional support against weak axis at middle point. Assume A50 steel. PL414x12 DO Section W16x67 L x-axis a) Calculate moment of inertia at both axes in mm*. b) Determine the design compressive strength in kN if L-3m. c) Find the design compressive strength in kN if L=18m. Elevation y-axisarrow_forward
- The given beam is laterally supported at the ends and at the 1 3 points (points 1, 2, 3, and 4). The concentrated load is a service live load. Use Fy=50 ksi and select a W-shape. Do not check deflections. a. Use LRFD. b. Use ASD.arrow_forwardIf the beam in Problem 5.5-9 i5 braced at A, B, and C, compute for the unbr Cb aced length AC (same as Cb for unbraced length CB). Do not include the beam weight in the loading. a. Use the unfactored service loads. b. Use factored loads.arrow_forwardThe member shown in Figure P6.6-4 is part of a braced frame. The load and moments are computed from service loads, and bending is about the x axis (the end shears are not shown). The frame analysis was performed consistent with the effective length method, so the flexural rigidity. EI, was unreduced. Use Kx=0.9. The load and moments are 30 dead load and 70 live load. Determine whether this member satisfies the appropriate AISC interaction equation. a. Use LRFD. b. Use ASD.arrow_forward
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