Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 7, Problem 7.7PFS
To determine
Design the right handed and left handed column using both the LRFD and the ASD method
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Q2) The members of the truss structure shown below is plain concrete. The compressive
strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by
the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use
material factors and do not consider slenderness) Comment on your results briefly.
P
A&
2m
SC
2 m
1380
2m
D
The rigid frame shown in Figure is unbraced in the plane of the frame. In the direction perpendicular to the frame, the frame is braced at the joints. The connections at these points of bracing are simple (moment-free) connections. Roof girders are W14 x 30, and floor girders are W16 x 36. Member BC is a W10 x 45. Use A992 steel and select a W-shape for AB. Assume that the controlling load combination causes no moment in AB. The service dead load is 25 kips and the service live load is 75 kips. a. Use LRFD. b. Use ASD
Answers:
AB =
kN
E
AH =
kN
Determine the force in each member of the loaded truss. All triangles are 3-4-5. Enter a
the member is in tension, negative if in compression.
BC -
kN
C
34kN
kN
BG
E
34 kN
BH -
kN
15 kN
BI =
kN
D
B
kN
kN
A
E
DE
kN
H
G
F
DF =
kN
- 4 panels at 8 m-
DG =
kN
DI
kN
EF =
kN
FG -
kN
GH =
kN
Chapter 7 Solutions
Structural Steel Design (6th Edition)
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Similar questions
- A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.arrow_forwardSelect all zero-force members in the truss shown below. Check the box for zero- force members 3 m 3 m 12 m, 8 @ 1.5 m DE O LK ЕР O HF O BC BM EF OM CD BN LO O DK FI O coarrow_forwardCurrent Attempt in Progress A pin-connected structure is supported and loaded as shown. Member ABCD is rigid and is horizontal before the load Pis applied. Bars (1) and (2) are both made from steel (E - 30600 ksi) and both have a cross-sectional area of 0.87 in.?. Assume L;-87 in, Lg-111 in., a-50 in., b-93 in, and c-39 in. If the normal stress in each steel bar must be limited to 19.4 ksi, determine the maximum load Pthat may be applied to the rigid bar. (2) (1) D Answer: P- kips.arrow_forward
- Select all zero-force members in the truss shown below. Check the box for zero- force members 3m 12 m, 8 @ 1.5 m O DE LK EP OHF O BC O BM O EF O OM O CD O BN LO O DK FI O co O O O O O O 0 0 O 0arrow_forwardW3D 25 Y= 155 %31 1oow= 2500 KN 27=310mm X= 475 Z= 60 %3D 102=600mmarrow_forward1. Determine the force in each member of the loaded truss. All triangles are 3-4-5. Indicate whether it is compression or tension. Use method of joint. C 40KN B XX H G -4 panels at 8 m- A 32 KN 22 KN F D Earrow_forward
- determine the member stresses of the truss shown. indicate if the stress in the member is in tension or compression. present complete and neat solutions. p1-30kN W-25kN/m P2=51kN AH Av Im P₂ Im P₂ Im P₂ Im Av W 1.6marrow_forward7) Determine the force in each member of the truss shown below. C F 100 kips 1S' 1S D - 20 ァー 20arrow_forwardН WWWN 160kN 140kN 120kN 100kN 130kN 150kN 8 panels @ 6m ea. = 48m 10.8m 8.1m D Determine the force in the members and identify each if (T) - tension or (C) - compression. HI= DG= DE= IJ = 419.444kN T HJ= НК - KJ = JM = 425kN T KL - EG GF EF FI = 419.444kN T FH = GH = || = 9.6m 10.5marrow_forward
- Where is/are the location(s) of the maximum compressive flexural stress? A simple I-beam is loaded as shown. 20 mm P KN PKN P KN B 20 mm- B с D L/4 m L/4 m L/4 m Pin support at the NA Midspan at point B Mid span at point D Midspan at the top fiber Roller support at top fiber Section D at the top fiber Section C at +170 mm from the NA L/4 m 20 mm C 250 mm 150 mm 150 mm D Aarrow_forwardA bracket of length 1.5m is rigidly attached to the simply-supported beam at its end. The beam is subjected to the combined effect of M,, M, and compressive force P caused by the factored loads as shown. There are lateral supports only at the ends. The member is made of HE300A section and steel grade is S275 (Fy-275MPA, Fu= 430 MPa). The loads acting are determined from LRFD combinations. Determine whether the member can safely carry the applied loads. Multiply first-order moment diagrams by a factor of 1.2 to account for second-order effects. No need to check shear limit state. You do not need to additionally consider the beam self-weight. E Use the following simple expression to calculate the value of L:: L, = T*rts 0.7*Fy y P = 400kN 1.5 m F = 150kN 2 т 4 т X: Lateral support НЕЗ00Аarrow_forwardA WT205 × 30 structural steel section (see Appendix B for cross-sectional properties) is used for a 7.1 m column. Assume pinned connections at each end of the column. Determine: (a) the slenderness ratio. (b) the Euler buckling load. Use E = 200 GPa for the steel. (c) the axial stress in the column when the Euler load is applied.arrow_forward
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