Check the strength of each type of member in the one story steel frame building below.

Transcription

1 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 1 / 9 Check the strength of each type of member in the one story steel frame building below. 8 ft B 20 f ft Side Elevation 3 4 Plan View 32 ft F y = 50 ksi all members F u = 65 ksi 20 ft Shape Strength Purlins W12x40 M n = 214 k ft Girders W21x44 M n = 298 k ft Columns W16x36 P n = 302 k X Brace 3x2x3/16 T n = 30 k Front Elevation oads: 3.5 thick light weight concrete slab (unit weight = 120 pcf) = 40 p W = 30 p oad Combinations: 1.2D D + 1.6W

2 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 2 / 9 Identify purlins (or joists) and girders. The roof deck shown below is supported by the Z shaped purlins, which run transverse to the deck corrugations (see exploded building in figure below). For the example on Page 1, the owner wants to add another story at a later date, so the roof is a 3.5 inch thick concrete floor slab. Floors are supported by joists (in this case W12x40 steel wide flange beams). The joists are in turn supported by the girders, which run transverse to the joists. The girders are supported at their ends by the columns. 1 8 ft B The H shaped symbols represent the columns. 25 ft 2 3 wtrib In the Plan View at left, we see that some of the vertical members in the sketch are attached directly to columns, but that some of the vertical members are attached at their ends to other beams. The ends of the horizontal members, on the other hand, all attach to columns. Therefore, the horizontal members in the sketch are girders, and the vertical members are joists. 4 Plan View

3 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 3 / 9 Joist max moment due to factored loads (M u ) The joists are three span continuous beams. oading all of the spans of a continuous beam may not cause the maximum bending moment. lthough the position of the dead load is given, the position of live load is variable and the structural engineer must determine the loading causing the maximum bending moment. One way of determining the loading causing the maximum bending moment is to apply all possible load configurations, one at a time, and select the loading causing the maximum effect. In this class we will take a short cut that provides the same answer most of the time: we will assume that the location of the maximum bending moment due to dead plus live loads is the location with the maximum bending moment due to dead loads. ocation of max M D+ = ocation of max M D The statement above is true for continuous beams with equal span lengths. Our procedure for calculating the maximum moment due to factored loads will be: 1. pply the dead load to all spans and calculate the moment (M D ) using the chart from the ISC manual for moments in 3 and 4 span continuous beams. 2a. ssume that the location of the max M D+ = the location of the max M D. Draw the influence diagram for moment for this location. 2b. pply the live load to the spans indicated by the influence diagram and calculate the moment (M ) using the ISC charts. 3. Calculate M u from 1.2 M D M. 1. Dead oads: weight of slab = 3.5 /12 / x 120pcf = 35 p self weight of W12x40 joist = 40 plf wt in plf 25 ft Trib. Width* oad on Joist slab 35 p 8 ft = (35 p)(8 ft) = klf w D Joists 40 plf = klf w D = = klf *see sketch on bottom of Pg. 2 M D, k 0.080w w 2 Max M D = w 2 from ISC charts M D = (0.320 klf )(25 ft ) w 2 = 20.0 k ft M D = 20.0 k ft

4 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 4 / 9 2. ive oads: ssume T = area supported by one span of the joist (conservative) reduc_ factor, reduction (pg143fereference) k T k 2 ( beams) T tributaryareaof joist (8 reduc_ factor 0.25 ft 15 (2)(200 )(25 ft ) ) Therefore w = ( reduc_factor )()(tributary width) = (1.0)(40.0 p )(8 ft )/(1000 lb/k ) = klf 2a. ssume max M D+ occurs at location of max M D Influence Diagram for M at Support 2: 2b. Span loading to cause max. M at Support w max M = (0.320 klf )(25 ft ) 2 = 23.3 k ft M w 2 = 23.3 k ft 3. M u = moment due to factored loads M = 23.3 k ft Use oad Combination for gravity loads (dead and live loads) from page 1: 1.2 D M u = 1.2( 20.0 k ft ) + 1.6( 23.3 k ft ) M u = 61.3 k ft M n = 214 k ft > 61.3 k ft = M u, OK

5 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 5 / 9 Girder max. moment due to factored load (M u ) Since all of the girders are the same size, the girder with the largest unity check will be the girder with the largest loads. Therefore, analyze a girder from an interior frame. The loads on this girder are indicated in the sketch below, where P ext = the loads on the girder from exterior joists and P int = the loads on the girder from interior joists 8 ft P ext P int P int P int P ext 8 trib for load on girder from exterior joist 25 The tributary areas used to calculate these loads are shown in the sketch at right. Since only the interior loads will cause bending of the girder for this example, we only need to calculate P int. Dead oads: weight of slab = 35 p self weight of joist = 40 plf self weight of W21x44 girder = 44 plf wt in plf trib for load on girder from interior joist Trib. rea or Trib. ength oad on Girder slab 35 p = (8 ft)(25 ft) = 200 = (35 p)(200 ) = 7.00 k Joists 40 plf 25 ft = (40 plf)(25 ft) = 1.00 k Girder 44 plf 8 ft = (44 plf)(8ft) = k P D = = 8.35 k

6 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 6 / 9 ive oads: Since the girder is a single span, there is no need to consider span load patterns for live load. For live load reduction, tributary area ( t ) equals tributary width (25 ) times the span length (32 ). t = 25 ft x 32 ft = k 18 k 18 k reduction (between 0.4 and (2)(800 ) 1.0, OK) 27k 27 k P = (40 p )(0.625)(25 ft )(8 ft ) = 5.0 k 27 k 9 k Mu: P = 1.2(8.35 k ) + 1.6(5.0 k ) = 18.0 k u V k ft fter drawing the shear and moment diagrams (at right), M u = 288 k ft M M n = 298 k ft > 288 k ft = M u, OK Column max. axial force due to factored loads (P u ) The critical column is one of the edge columns (as opposed to a corner column). Dead oads: weight of slab = 35 p wt joists = 40 plf wt girder = 44 plf Trib. rea or Trib. ength n oad on Column slab 35 p = (16 ft)(25 ft) = 400 = (35 p)(400 ) = k Joists 40 plf 25 ft 2.5 = (40 plf)(25 ft)(2.5 joists) = 2.50 k Girder 44 plf 16 ft = (44 plf)(16 ft) = k P D = = k

7 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 7 / 9 ive oads: reduc k T _ factor ( columns) tributaryarea of joist (16 reduc _ factor k T, 15 (4)( ft )(25 ft reduction ) ) 1.0 (pg143 FE Reference) P = (40 p )(0.625)(400 ) = 10.0 k P u = 1.2(17.2 k ) + 1.6(10.0 k ) P u = 36.6 k P n = 302 k > 36.6 k = P u, OK (but over designed) End Wall Cross Bracing max tension due to factored sidesway loads (T u ) ets follow the wind loads applied to a long wall of the building (see sketch below). ssume that the wall acts like a simply supported beam spanning between the foundations on the bottom and the roof diaphragm on the top, then half of the wind load goes to the foundations along the bottom of the wall, and the other half is distributed to the roof or floor diaphragm. Roof Diaphragm Wind oad Wind oad Foundation End Wall The diaphragm acts like a beam in the horizontal plane: it supports the distributed load from the top of the wall, and is in turn supported at its ends by the end walls. ssume that only the end walls are braced against sidesway (x bracing on the internal frames would restrict the use of the building). Then half of the wind load to the roof diaphragm is applied to the top of each end wall. The end wall acts like a vertical cantilever beam: the horizontal force applied to its free end results in a horizontal reaction at its base and a couple (called the overturning moment due to wind load). Rather than

8 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 8 / 9 specify one diagonal brace that can carry tension or compression (depending on the wind direction), structural engineers usually specify x bracing and assume that the diagonal brace in compression buckles elastically. Therefore the horizontal wind load at the top of the end wall is resisted by the horizontal component of the tension diagonal brace. The portion of the long wall area that collects the wind load distributed to the top of an end wall is indicated in the sketch below. wall area collecting wind load applied to top of left end wall h/2 Isometric View bldg /2 For the current example, the force at the top of the end wall due to factored wind load (see load combination on Page 1) is: W hcol ength of Bldg 20' 75' P U end wall ( 1.6)( W)( )( ) (1.6)(30p )( )( ) k C k 21.2 k T End Wall Elevation 20 ft 32 ft 37.7 ft The tension force in the diagonal brace due to factored wind loads is therefore: T U end wall brace = (18 k )(37.7 ft / 32 ft ) T U end wall brace = 21.2 k For buildings having interior frames with sidesway resistance (either moment frames of braced frames), the distribution of sidesway forces to the frames depends on the type of diaphragm. Flexible diaphragms (for example wood roofs) behave like horizontal beams that are simply supported between the frames. The sidesway load to each frame is in proportion to its tributary width.

9 CE 331, Summer 2015 nalysis of Steel Braced Frame Bldg 9 / 9 Tributary width for Frame B /2 /2 Flexible Diaphragm Flexible Diaphragm Flexible Diaphragm B C D Sidesway loads to frames for flexible diaphragms Rigid diaphragms behave like a rigid horizontal beam. The horizontal deflections at the top of each frame () are equal. The sidesway load to each frame is proportional to its lateral (sidesway) stiffness. Rigid Diaphragm D B C If all frames resisted sidesway in the steel building of this example, the sidesway loads to the top of the frames would be: a) with a flexible diaphragm: P Frame = P Frame D = (1.3) (30 p) (10 ft) ( 25 ft / 2 ) = 4.88 k P Frame B = P Frame C = (1.3) (30 p) (10 ft) ( 25 ft ) = 9.75 k Sidesway loads to frames for rigid diaphragms b) with a rigid diaphragm and frames with identical sidesway stiffness: P Frame = P Frame B = P Frame C = P Frame D = (1.3) (30 p) (10 ft) ( 75 ft ) (1 / 4 ) = 7.31 k