1. An A992 steel W150 x 22 column is braced at the top against both x and y displacement but not against rotation. It is also braced at the midpoint against y displacement and is fixed at the bottom. Apply a F.S. of 2 against buckling. Determine the maximum allowable force that the column can support without buckling or exceeding the yield stress. 3 m P

1. An A992 steel W150 x 22 column is braced at the top against both x and y displacement but not against rotation. It is also braced at the midpoint against y displacement and is fixed at the bottom. Apply a F.S. of 2 against buckling. Determine the maximum allowable force that the column can support without buckling or exceeding the yield stress. 3 m P

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.14.2P

See similar textbooks

Similar questions

A D -0.5 m- 400 KN ?AD, ?CF, ?BE = B E -0.5 m- C 0.4 m CL F The assembly consists of two posts AD and CF made of A-36 steel and having a cross-sectional area of 1000 mm2, and a 2014-T6 aluminum post BE having a cross-sectional area of 1700 mm2. If a central load of 400 kN is applied to the rigid cap, determine the normal stress in each post. There is a small gap of 0.09 mm between the post BE and the rigid member ABC. (Figure 1)
An 80-foot long plate girder (see below) is fabricated from a ½-inch x 78-inch web and two 3inch x 22-inch flanges. Continuous lateral support is provided. The steel is A992. The loading consists of a uniform service dead load of 1.0 kip/ft (including the self-weight), a uniform service live load of 2.0 kips/ft, and a concentrated service live load of 500 kips at midspan. Stiffeners are placed at each end and at 4 feet, 16 feet, and 28 feet from each end. Once stiffener is placed at midspan. Determine whether the flexural strength is adequate using LRFD.
SOLVE FOR THE FORCE IN ALL MEMBERS OF THE TRUSS. SHOW SOLUTIONS CLEARLY USING METHOD OF JOINTS AND WRITE WELL. 0.8m 2.4m A BD BD 3 KN MEMBER AB AC BC BE BF EF GE FG 2m Av(REACTION @ hinge) Gv(REACTION @ roller) C 1m 10KN B 1m 8KN 3.5m SUMMARY OF RESULTS FORCE E 8KN 1m F 2.4m 6KN G NATURE(T or C)
1. Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends, and A992 steel (E345 MPa and Fu= 450 MPa). The only uniform service dead load is the weight of the beam. The 70 KN service loads are 30%DL and 70%LL. Use LRFD. IR₁ 0.90m 70,KN * W14 x 68 1.20m 70 KN * 0.90m R₂ W14x68 70 KN
S f A structure is to be built as shown supporting a uniform load of 18 kN/m. The location for the supports at A and B has been determined. The connection at C may be placed anywhere along the member AD 18 kN/m A C D 3 m B L -7 mi -5 m- If the allowable bending stress for member AD is 8 MPa, which among the choices gives the most economic section? Select the correct response: 275 mm x 550 mm 250 mm x 500 mm 300mm x 600 mm 225 mm x 450 mm rigid
A 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 НЕЗ00А
Where is/are the location(s) of the maximum transverese shear stress? A simple I-beam is loaded as shown. 20 mm P KN PKN PKN Į Į -C B с D L/4 m L/4 m Midspan at point C Roller at E at the NA Section B at the top fiber Pin A at point C Midspan at point D Roller E at point D L/4 m L/4 m OE 20 mm 20 mm- 250 mm 150 mm 150 mm A
5. The beam-column has a height of 7.5m and is a member of a braced frame. It is subjected to an axial load of DL-65KN and LL= 1OOKN. It is pinned at both ends. A W12X35 section was used and was made up of A-36 steel with Fy-250MPA. Compute the interaction value for beam- column based on NSCP 2015 code provisions. Ngelect the weight of the section. A = 6645 mm2 Zx- 839x10^3 mm3 d- 317.50mm Sx- 747x10^3 mm3 tw= 7.62 Rx= 133.35 bf= 166.62 ly= 10x10^6 mm4 tf- 13.21 Sy= 122x10^3 mm3 Ix= 119x10^6 mm4 Ry3 39.12mm 3.5 Note: Use C=D1.00%--0.2
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.
4/47 Compute the force in member HN of the loaded truss. BE 2 m A 4/40 n L L D R Q L L F 6 m L P O N -8 panels at 3 m Problem 4/47 age to L BODJEN H M L I L L 2- J [2 m EM K
i n. In. . • U d d Y:0E O HW4 modified03... Use the method f sections to find the force the member s BD,CD, and EC You can 3m use the methad of Joints to check your 3 m/ 3m 3m 3 m C 3m A 120 kN results II Mach 2l2o Due Tuesday March 25 at 12:0op
Grath fas A cantlever AB o erath lm carries a from klev. uniformly distributed load, wNm fe fired end A to the free end B. Te the cantilever Consista a horizontal thicknees lomm and with 120mm as shown in fhe below, plars plate4 figure calculate the total permiseiblé diafribte load,wif the meximum bending Stress is not exceeding 10OMH/mt. Ignore the grore te weight a the tlefe Im 120mm