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# Design of Steel Structures CE 405

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This 59 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 405 at Michigan State University taught by Staff in Fall. Since its upload, it has received 130 views. For similar materials see /class/207372/ce-405-michigan-state-university in Civil Engineering at Michigan State University.

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CE 405 Design ofSteel Structures Prof Dr A Viier Tension Member Design Chapter 4 TENSION MEMBER DESIGN 41 INTRODUCTORY CONCEPTS Stress The stress in an axially loaded tension member is given by Equation 41 41 where P is the magnitude of load and A is the crosssectional area normal to the load The stress in a tension member is uniform throughout the crosssection except near the point of application of load and at the crosssection with holes for bolts or other discontinuities etc For example consider an 8 x 12 in bar connected to a gusset plate and loaded in tension as shown below in Figure 41 O O Gussetplate IEEI bl o o lb Section bb 78 in diameter hole al 1 a I Section 1111 AV 8x 12 in bar Figure 41 Example of tension member Area ofbar at section a 7 a 8 x 12 4 in2 Areaofbarat sectionbib872x78x312in2 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 0 Therefore by de nition Equation 41 the reduced area of section b 7 b will be subjected to higher stresses However the reduced area and therefore the higher stresses will be localized around section b 7 b o The unreduced area of the member is called its gross area Ag 0 The reduced area of the member is called its net area An 42 STEEL STRESS STRAIN BEHAVIOR o The stressstrain behavior of steel is shown below in Figure 42 A lt Stress f 7 Strain s Figure 42 Stressstrain behavior of steel 0 In Figure 42 E is the elastic modulus 29000 ksi Fy is the yield stress and Fu is the ultimate stress 8y is the yield strain and 8 is the ultimate strain CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 0 Deformations are caused by the strain 8 Figure 42 indicates that the structural de ections will be small as long as the material is elastic flt Fy o Deformations due to the strain 8 will be large after the steel reaches its yield stress Fy 43 DESIGN STRENGTH o A tension member can fail by reaching one of two limit states 1 excessive deformation or 2 fracture 0 Excessive deformation can occur due to the yielding of the gross section for example section aa from Figure 41 along the length of the member 0 Fracture of the net section can occur if the stress at the net section for example section bb in Figure 41 reaches the ultimate stress Fu 0 The objective of design is to prevent these failure before reaching the ultimate loads on the structure Obvious o This is also the load and resistance factor design approach recommended by AISC for designing steel structures 431 Load and Resistance Factor Design The load and resistance factor design approach is recommended by AISC for designing steel structures It can be understood as follows Step 1 Determine the ultimate loads acting on the structure The values of D L W etc given by ASCE 798 are nominal loads not maximum or ultimate During its design life a structure can be subjected to some maximum or ultimate loads caused by combinations of D L or W loading CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design The ultimate load on the structure can be calculated using factored load combinations which are given by ASCE and AISC see pages 210 and 211 of AISC manual The most relevant of these load combinations are given below 14D 4271 12D 16L05Lor S 4272 12D 16L or S05Lor 08 W 4273 12D 16W05L05LrorS 4274 09D16W 4275 Step II Conduct linear elastic structural analysis Determine the design forces Pu V and Mn for each structural member Step 111 Design the members The failure design strength of the designed member must be greater than the corresponding design forces calculated in Step 11 See Equation 43 below M gt Zvi Q1 43 Where Rn is the calculated failure strength of the member is the resistance factor used to account for the reliability of the material behavior and equations for Rn Q is the nominal load vi is the load factor used to account for the variability in loading and to estimate the ultimate loading condition 432 Design Strength of Tension Members 0 Yielding of the gross section will occur when the stress f reaches Fy P fFy Ag CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Therefore nominal yield strength Pn Ag Fy 44 Factored yield strength PH 45 where 09 for tension yielding limit state 0 See the AISC manual section on mifications Chapter D page 161 724 o Facture of the net section will occur after the stress on the net section area reaches the ultimate stress Fu F u P f AC Therefore nominal fracture strength Pn Ae Fu Where A6 is the effective net area which may be equal to the net area or smaller The topic of Ae will be addressed later Factored fracture strength 11 Ae Fu 46 Where 075 for tension fracture limit state See page 16124 of AISC manual 433 Important notes 0 M Why is fracture amp not yielding the relevant limit state at the net section Yielding will occur rst in the net section However the deformations induced by yielding will be localized around the net section These localized deformations will not cause excessive deformations in the complete tension member Hence yielding at the net section will not be a failure limit state 0 m Why is the resistance factor smaller for fracture than for yielding The smaller resistance factor for fracture 075 as compared to 090 for yielding re ects the more serious nature and consequences of reaching the fracture limit state 0 Note 3 What is the design strength of the tension member CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design The design strength of the tension member will be the lesser value of the strength for the two limit states gross section yielding and net section fracture m Where are the Fy and Fu values for different steel materials The yield and ultimate stress values for different steel materials are noted in Table 2 in the AISC manual on pages 16 17141 and 1617142 W What are the most common steels for structural members See Table 21 in the AISC manual on pages 2724 and 2 25 According to this Table the preferred material for W shapes is A992 Fy 50 ksi Fu 65 ksi the preferred material for C L M and S shapes is A36 Fy 36 ksi Fu 58 ksi All these shapes are also available in A572 Gr 50 Fy 50 ksi Fu 65 ksi Note 6 What is the amount of area to be deducted from the gross area to account for the presence of boltholes The nominal diameter of the hole dh is equal to the bolt diameter db 116 in However the bolthole fabrication process damages additional material around the hole diameter Assume that the material damage extends 1 16 in around the hole diameter Therefore for calculating the net section area assume that the gross area is reduced by a hole diameter equal to the nominal holediameter 116 in CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design EXample 31 A 5 X 12 bar of A572 Gr 50 steel is used as atension member It is connected to a gusset plate with siX 78 in diameter bolts as shown in below Assume that the effective net area Ae equals the actual net area An and compute the tensile design strength of the member Gussetplate bl coxlb 78 in diameter bolt 1 l A 5 x 12 in bar A572 Gr 50 Solution Gross section area Ag 5 X 12 25 in2 Net section area An Bolt diameter db 7 8 in Nominal hole diameter dh 78 116 in 1516 in Hole diameter for calculating net area 1516 116 in 1 in Net section areaAn 572X1X12151n2 0 Gross yielding design strength Pn Fy Ag Gross yielding design strength 09 X 50 ksi X 25 in2 1125 kips o Fracture design strength Pn I Fu Ae Assume Ae An only for this problem Fracture design strength 075 X 65 ksi X 15 in2 73125 kips 0 Design strength of the member in tension smaller of 73125 kips and 1125 kips CE 405 Design ofSteel Structures Prof Dr A Viier Tension Member Design Therefore design strength 73125 kips net section fracture controls EXample 32 A single angle tension member L 4 X 4 X 38 in made from A36 steel is connected to a gusset plate with 58 in diameter bolts as shown in Figure below The service loads are 35 kips dead load and 15 kips live load Determine the adequacy of this member using AISC speci cation Assume that the effective net area is 85 of the computed net area Calculating the effective net area will be taught in the next section 0 Gross area of angle Ag 286 in2 from Table 1 7 on page 1 36 ofAISC V L4x4x38 d17 58 in Section aa Gussetplate 0 Net section area An Bolt diameter 58 in Nominal hole diameter 58 116 1116 in Hole diameter for calculating net area 1116 116 34 in Net section area Ag 7 34 X 38 286 7 34 X 38 2579 in2 0 Effective net area Ae 085 X 2579 in2 2192 in2 0 Gross yielding design strength 1AAg Fy 09 X 286 in2 X 36 ksi 92664 kips 0 Net section fracture Ae Fu 075 X 2192 in2 X 58 ksi 95352 kips CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 0 Design strength 92664 kips gross yielding governs 0 Ultimate design load acting for the tension member Pu The ultimate design load can be calculated using factored load combinations given on page 2 11 ofthe AISC manual or Equations 42 1 to 42 5 ofnotes see pg 4 According to these equations two loading combinations are important for this problem These are 1 14D and 2 12D 16L The corresponding ultimate design loads are 14 X PD 14 35 49 kips 12 PD 16 PL 66 kips controls The ultimate design load for the member is 66 kips where the factored dead live loading condition controls 0 Compare the design strength with the ultimate design load The design strength of the member 92664 kips is greater than the ultimate design load 66 kips Pn 92664 kips gt PH 66 kips o The L 4 X 4 X 38 in made from A36 steel is adequate for carrying the factored loads 44 EFFECTIVE NET AREA 0 The connection has a significant in uence on the performance of a tension member A connection almost always weakens the member and a measure of its in uence is called joint efficiency 0 Joint efficiency is a function of a material ductility b fastener spacing c stress concentration at holes d fabrication procedure and e shear lag CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design All factors contribute to reducing the effectiveness but shear lag is the most important Shear lag occurs when the tension force is not transferred simultaneously to all elements of the crosssection This will occur when some elements of the crosssection are not connected For example see Figure 43 below where only one leg ofan angle is bolted to the gusset plate OOO Figure 43 Single angle with bolted connection to only one leg A consequence of this partial connection is that the connected element becomes overloaded and the unconnected part is not fully stressed Lengthening the connection region will reduce this effect Research indicates that shear lag can be accounted for by using a reduced or effective net area Ae Shear lag affects both bolted and welded connections Therefore the effective net area concept applied to both types of connections For bolted connection the effective net area is Ae U An For welded connection the effective net area is Ae U Ag Where the reduction factor U is given by 47 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Where i is the distance from the centroid of the connected area to the plane of the connection and L is the length of the connection If the member has two symmetrically located planes of connection E is measured from the centroid of the nearest one 7 half of the area Additional approaches for calculating E for different connection types are shown in the AISC manual on page 1611 78 The distance L is de ned as the length of the connection in the direction of load For bolted connections L is measured from the center of the bolt at one end to the center of the bolt at the other end For welded connections it is measured from one end of the connection to other If there are weld segments of different length in the direction of load L is the length of the longest segment Example pictures for calculating L are given on page 1611 79 of AISC The AISC manual also gives values of U that can be used instead of calculating E L They are based on average values of E L for various bolted connections For W M and S shapes with widthtodepth ratio of at least 23 and for Tee shapes cut from them if the connection is through the anges with at least three fasteners per line in the direction of applied load U 090 For all other shapes with at least three fasteners per line U 085 For all members with only two fasteners per line U 075 For better idea see Figure 38 on page 41 ofthe Segui textbook These values are acceptable but not the best estimate of U If used in the exam or homeworks full points for calculating U will not be given CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design EXample 33 Determine the effective net area and the corresponding design strength for the single angle tension member ofExample 32 The tension member is an L 4 X 4 X 38 in made from A36 steel It is connected to a gusset plate with 58 in diameter bolts as shown in Figure below The spacing between the bolts is 3 in centertocenter Compare your results with those obtained for Example 32 L4 X4 X38 L4 X4 X38 Cmsset pl ate Gross area of angle Ag 286 in2 from Table 1 7 on page 1 36ofAISC Net section area An Bolt diameter 58 in Hole diameter for calculating net area 1116 116 34 in Net section area Ag 7 34 X 38 286 7 34 X 38 2579 in2 X is the distance from the centroid of the area connected to the plane of connection For this case X is equal to the distance of centroid of the angle from the edge This value is given in the Table 1 7 on page 1 3 6 of the AISC manual X 113 in L is the length ofthe connection which for this case will be equal to 2 X 30 in CE 405 Design ofSteel Structures Prof Dr A Viier Tension Member Design 1 08116 in 60 o Effective net area Ae 08116 x 2579 in2 2093 in2 0 Gross yielding design strength Ag Fy 09 X 286 in2 X 36 ksi 92664 kips 0 Net section fracture Ae Fu 075 X 2093 in2 X 58 ksi 91045 kips 0 Design strength 91045 kips net section fracture governs o In EXample 32 Factored load Pu 660 kips Design strength Pn 9266 kips gross section yielding governs Net section fracture strength Pn 95352 kips assuming Ae 085 0 Comparing EXamples 32 and 33 Calculated value of U 08166 is less than the assumed value 085 The assumed value was unconservative It is preferred that the U value be specifically calculated for the section After including the calculated value of U net section fracture governs the design strength but the member is still adequate from a design standpoint CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design EXample 34 Determine the design strength of an ASTM A992 W8 X 24 with four lines if 3A in diameter bolts in standard holes two per ange as shown in the Figure below Assume the holes are located at the member end and the connection length is 90 in Also calculate at what length this tension member would cease to satisfy the slendemess limitation in LRFD speci cation B7 34 in diameter balm OOOO W8x24 we we 0 P l 3 1n 3 m 3 m Holes inbeam ange Solution 0 For ASTM A992 material Fy 50 ksi and Fu 65 ksi o For the W8 X 24 section Ag 708 in2 d 793 in tw 0285 in bf 65 in tf04 in ry 161 in Gross yielding design strength Pn Ag Fy 090 X 708 in2 X 50 ksi 319 kips Net section fracture strength Pn Ae Fu 075 X Ae X 65 ksi Ae U An for bolted connection An Ag 7 no of holes X diameter of hole X thickness of ange An 708 7 4 X diameter ofbolt l8 in X 04 in Ag 568 in CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design What is X for this situation X is the distance from the edge of the ange to the centroid of the half 7 section b tf d 2tf d2tf gtltt gtlt gtltt X f f 2 2 W 4 65 X 04gtlt 023565 gtlt 0285 X 21825 65 X 043565 gtlt 0285 f 076 d b gtltt gtltt f f 2 w Xcan be obtained from the dimension tables for Tee section WT 4 x 12 See page 150 and 151 ofthe AISC manual 3 0695 in The calculated value is not accurate due to the deviations in the geometry 31 0923 L 90 But US 090 Therefore assume U 090 Net section fracture strength AAe Fu 075 X 09 X 568 X 65 2492 kips The design strength of the member is controlled by net section fracture 2492 kips According to LRFD speci cation B7 the maXimum unsupp01ted length of the member is limited to 300 ry 300 X 161 in 543 in 403 ft CE 405 Deszgn of Steel Structures 7 Prof Dr A mea Tension Member Design 441 Special cases for welded connections I If some elements of the crosssection are not connected then Ae Will be less than An For a rectangular bar or plate Ae Will be equal to An However if the connection is by longitudinal welds at the ends as shown in the figure below then Ae UAg Where U 10 forL 2 W U087 for15WSLlt2W U075 forW Lltl5W L length of the pair of welds 2 W W distance between the welds or Width of platebar 1h I4 I AISC Specification B3 gives another special case for welded connections For any member connected by transverse welds alone Ae area of the connected element of the crosssection Transverse c ms Dcslgvl ofstsal stmtms erf m A Vanna TensmnMEmberDesly steel Examgle n to 1 shown below Caleulate the tenslon deslgn strength 4 168quot LGXGX A 5V1quot 4 gt Solutlon Ag 5 00 tn2 An 5 00 tn2 abecause lt ls awelded eonneetlon i AUAn alumni17 e i 168m forthlsweldedconnecnon e L 6 0 m forthls welded connection Gross yleldlng deslgn strenth 435A 0 9 n 36 n 5 00 162 klps 75X58X072X5001566kps Net seetaon fracture strength a F A Deslgn strength 156 s klps netsemanfmcmre gnvms CE 415 Design ositcci Stmcmws 7 P70 D7 A Vinmz TensmnMemberDesign 45 STAGGERED BOLTS e eia ea uu 1 To get more capacity by increasing the effective net area 2 To achieve a smaller connectionlength 3 To t the geometry ofthe tension connection itself the tel atinn hin the inclined portion has CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 0 Net section fracture can occur along any zigzag or straight line For example fracture can occur along the inclined path a b c d in the gure above However all possibilities must be examined 0 Empirical methods have been developed to calculate the net section fracture strength According to AISC Specification B2 2 net width gross width Z d Z g where d is the diameter of hole to be deducted dh ll6 or db 18 s24g is added for each gage space in the chain being considered s is the longitudinal spacing pitch of the bolt holes in the direction of loading g is the transverse spacing gage of the bolt holes perpendicular to loading dir net area An net width X plate thickness effective net area As U An where U l EL net fracture design strength Ae Fu 075 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design EXAMPLE 36 Compute the smallest net area for the plate shown below The holes are for 1 in diameter bolts 1 a o o jO b0 0 0 oc gt 311311 3m3in3in 3m 0 The effective hole diameter is 1 18 1125 in o For line a b d e wn 160 7 2 1125 1375 in o For line a b c d e wn 160 7 3 1125 2 X 32 4 X 5 1352 in o The line a b c d e governs o Antwn07513521014in2 w 0 Each fastener resists an equal share of the load 0 Therefore different potential failure lines may be subjected to different loads 0 For example line a b c d e must resist the full load whereas i j f h will be subjected to 811 of the applied load The reason is that 311 of the load is transferred from the member before i j f h received any load 20 c ms Desan ofsml Stmntuns erf m A Wm TensmnMEmberDesly Thls pl ate ls than analyzedllke shown above sum ome leg lengths mlnus the angle thckness T n m u quot reduced by an amount equal to the angle thlckness See Flgure below Forthls slmauonlhe dlstance g Wlll be 3 2 e 2ln 3 1 o o o E 4239 We 0 o o 0 L5 x 5 x 1 c ms Dcszganstzdstmnmnsrl rqf m A mm b mu 2 Tensmn MEmber Deng mu manly m cmnmh um mm mm In manna A16 M n um and halal m or Ainclldilm u um sounlax Camp m nnwmm wl 639A135m I mm m 339 39 3 u x a x mmmum v A m Fm mm wquot as 1unsin Pnrllnzuucz 135309 mum quot N15 mm W um mu m rm mm mmemm a m 1mm ly warm 12mm m m man a Aanbhmznuundmmmbcwmpudmm 10129 u nmmunnakdqy 05 u 571 159 1 g a 3 Wquot 3935 0 ms was 40 u A Thummsncnnlmk A mun Manon 501m m Imm angku mm m Itldmg x5 mums 219 ms ANSWER memnmummmnzm 218kquot CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 46 BLOCK SHEAR o For some connection con gurations the tension member can fail due to tearout of material at the connected end This is called block shear o For example the single angle tension member connected as shown in the Figure below is susceptible to the phenomenon of block shear a I i o o o ltgt T b Shearfailure C Tensionfailure Figure 44 Block shear failure of single angle tension member For the case shown above shear failure will occur along the longitudinal section ab and tension failure will occur along the transverse section bc AISC Specification SPEC Chapter D on tension members does not cover block shear failure explicitly But it directs the engineer to the Specification Section J43 23 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Block shear strength is determined as the m of the shear strength on a failure path and the tensile strength on a perpendicular segment Block shear strength net section fracture strength on shear path gross yielding strength on the tension path Block shear strength gross yielding strength of the shear path net section fracture strength of the tension path Which of the two calculations above governs See page 161 67 Section J43 ofthe AISC manual When Fu Am 2 06FuAnv an 4 06 Fy Agv Fu Am 5 gt 06 FHA v Fu Am When Fu Am lt 06Fu Anv tRn I 06 Fu Am Fy Ag 5 gt 06 FuAm Fu Am Where 075 Agv gross area subject to shear Agt gross area subject to tension Am net area subject to shear Am net area subject to tension 24 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design EXAMPLE 38 Calculate the block shear strength of the single angle tension member considered in Examples 32 and 33 The single angle L 4 x 4 x 38 made from A36 steel is connected to the gusset plate with 58 in diameter bolts as shown below The bolt spacing is 3 in centertocenter and the edge distances are 15 in and 20 in as shown in the Figure below Compare your results with those obtained in Example 32 and 33 L4x4 x38 db 58 in metplate L 4 X 4 X 38 0 Step 1 Assume a block shear path and calculate the required areas Gusset plate Agt gross tension area 20 X 38 075 in2 Am net tension area 075 7 05 X 58 18 X 38 0609 in2 Agv gross shear area 30 30 15 X 38 2813 in2 Am net tension area 2813 725 X 58 18 X 38 2109 in2 25 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 0 Step II Calculate which equation governs 06 Fu AW 06 X 58 X 2109 73393 kips Fu Am 58 X 0609 35322 kips 06 Fu Am gt Fu Am Therefore equation with fracture of shear path governs 0 Step III Calculate block shear strength Rn 075 06 Fu Am Fy Agt 1Rn 075 73393 36 X 075 75294 kips 0 Compare with results from previous eXamples EXample 32 Ultimate factored load Pu 66 kips Gross yielding design strength Pn 92664 kips Assume Ae 085 An Net section fracture strength 95352 kips Design strength 92664 kips gross yielding governs EXample 33 Calculate Ae 08166 An Net section fracture strength 91045 kips Design strength 91045 kips net section fracture governs Member is still adequate to carry the factored load Pu 66 kips EXample 38 Block shear fracture strength 75294 kips Design strength 75294 kips block shear fracture governs Member is still adequate to carry the factored load Pu 66 kips 26 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design 0 Bottom line Any of the three limit states gross yielding net section fracture or block shear failure can govern The design strength for all three limit states has to be calculated The member design strength will be the smallest of the three calculated values The member design strength must be greater than the ultimate factored design load in tension Practice Example Determine the design tension strength for a single channel C15 X 50 connected to a 05 in thick gusset plate as shown in Figure Assume that the holes are for 34 in diameter bolts and that the plate is made from structural steel with yield stress Fy equal to 50 ksi and ultimate stress Fu equal to 65 ksi gusset plate 3 3 9 T centertocenter ltgt lgt C15 x50 Limit state of yielding due to tension Tn 09 50 147 662kz39ps Limit state of fracture due to tension Aquot Ag nd2t147 4Z07161219m2 8 A UA 1 An 1 03976981219 10571712 Check U 0867 S 09 OK 27 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Note The connection eccentricity X for a C15X50 can be found on page 151 LRFD Tn 075 65 1057 515kz39ps Limit state of block shear rupture 06FMAWV 06 65 2 75 25 0716 2966925 FuAm 65 9 0716 2966925 FMA gt 06FuAW nt 464kz39ps Rn 06FyAgV FuAm 07506 50 15 0716 65 M 5 Block shear rupture is the critical limit state and the design tension strength is 464kips 28 CE 405 Design ofSteel Structures Prof Dr A Viier Tension Member Design 47 Design of tension members The design of a tension member involves nding the ligl1test steel section angle wide ange or channel section with design strength Pn greater than or equal to the maximum factored design tension load Pu acting on it P11 2 Pu Pu is determined by structural analysis for factored load combinations Pn is the design strength based on the gross section yielding net section fracture and block shear rupture limit states For gross yielding limit state Pn 09 X Ag X Fy Therefore 09 X Ag X Fy 2 PH P Therefore Ag 2 u 09 X Fy For net section fracture limit state Pn 075 X Ae X Fu Therefore 075 X Ae X Fu 2 PH Therefore Ac 2 quot 075 X Fu But Ae U An Where U and An depend on the end connection Thus designing the tension member goes handinhand with designing the end connection which we have not covered so far Therefore for this chapter of the course the end connection details will be given in the eXamples and problems The AISC manual tabulates the tension design strength of standard steel sections Include wide ange shapes angles tee sections and double angle sections 29 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design The gross yielding design strength and the net section fracture strength of each section is tabulated This provides a great starting point for selecting a section 0 There is one serious limitation The net section fracture strength is tabulated for an assumed value of U 075 obviously because the precise connection details are not known For all W Tee angle and doubleangle sections A6 is assumed to be 075 Ag The engineer can rst select the tension member based on the tabulated gross yielding and net section fracture strengths and then check the net section fracture strength and the block shear strength using the actual quot details 0 Additionally for each shape the manual tells the value of Ae below which net section fracture will control Thus for W shapes net section fracture will control if Ae lt 0923 Ag For single angles net section fracture will control if Ae lt 0745 Ag For Tee shapes net section fracture will control if Ae lt 0923 For double angle shapes net section fracture will control if Ae lt 0745 Ag 0 Slenderness limits Tension member slenderness lr must preferably be limited to 300 as per LRFD specification B7 30 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Example 310 Design a member to carry a factored maximum tension load of 100 kips a Assume that the member is a wide ange connected through the anges using eight 3 in diameter bolts in two rows of four each as shown in the gure below The centertocenter distance of the bolts in the direction of loading is 4 in The edge distances are 15 in and 20 in as shown in the figure below Steel material is A992 34 in diameter bolts 2 15 in Holes in beam ange SOLUTION 0 Step 1 Select a section from the Tables Go to the TEN section ofthe AISC manual See Table 31 on pages 317 to 319 From this table select W8X10 with Ag 296 inz Ae 222 inz Gross yielding strength l33 kips and net section fracture strength108 kips This is the lightest section in the table Assumed U 075 And net section fracture will govern if Ae lt 0923 Ag 0 Step II Calculate the net section fracture strength for the actual connection According to the Figure above An Ag 4 db 18 X tf An296 434 l8X0205 224in2 The connection is only through the anges Therefore the shear lag factor U will be the distance from the top of the ange to the centroid of a WT 4 X 5 3l CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design See DIM section ofthe AISC manual See Table l8 on pages l50 l5l X 0953 U l XL l 09534076 Ae 076 An 076 X 224 170 in2 APn 075 X Fu X Ae 075 X 65 X 170 829 kips Unacceptable because Pu 100 kips REDESIGN required Step III Redesign Many ways to redesign One way is shown here Assume Pn gt 100 kips Therefore 075 X 65 X Ae gt 100 kips Therefore Ae gt 2051 in2 Assume Ae 076 An based on previous calculations step 11 Therefore An gt 27 in2 But Ag An 4 db l 8 X tf based on previous calculations step 11 Therefore Ag gt 27 35 X tf Go to the section dimension table ll on page l22 ofthe AISC manual Select neXt highest section I For W 8 X13 tf 0255 in Therefore Ag gt 27 35 x 0255 359 in2 From Table 11 ws x 13 has Ag 384 in2 gt 359 in2 I Therefore W8 X 13 is acceptable and is chosen 32 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Step IV Check selected section for net section fracture Ag 384 in2 An38435X0255295in2 From dimensions of WT4 X 65 X 103 in Therefore U1 XL 11034 074 Therefore Ae U An 074 x 295 219 in2 Therefore net section fracture strength 075 X 65 X 219 1067 kips Which is greater than 100 kips design load Therefore W 8 X 13 is acceptable Step V Check the block shear rupture strength 0 Identify the block shear path V 15m F gt 2111I 4111 1 The block shear path is show above Four blocks will separate from the tension member two from each ange as shown in the gure above 6 X 0255 X 4 612 in2 for four tabs Agv 42 x tf x 4 Anv 42 15 x db18 x tf x 4 478 in2 Agt 15 thX 4 153in2 Am 15 05 x db18thX 4 1084 in2 0 Identify the governing equation 33 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design FuAm 65 X 1084 704 kips 06FuAnV 06 X 65 X 478 18642 kips which is gt FuAm 0 Calculate block shear strength Rn 075 06FuAm FyAgt 075 18642 50 x 153 1972 kips Which is greater than Pu 100 kips Therefore W8 X 13 is still acceptable 0 Summary of solution Mem Design Ag An U Ae Yield Fracture Block shear load strength strength strength W8X13 100 kips 384 295 074 219 173 kips 1067 kips 1972 kips Design strength 1067 kips net section fracture governs W8 X 13 is adequate for Pu 100 kips and the given connection 34 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design EXAMPLE 311 Design a member to carry a factored maximum tension load of 100 kips b The member is a single angle section connected through one leg using four 1 in diameter bolts The centertocenter distance of the bolts is 3 in The edge distances are 2 in Steel material is A36 20151 0 O O Hlt gtlt gtH gt 20 i 30 in 30 in 30 in 0 Step 1 Select a section from the Tables Go to the TEN section ofthe AISC manual See Table 32 on pages 320 to 321 From this table select L4X3Xl2 with Ag 325 inz Ae 244 inz Gross yielding strength 105 kips and net section fracture strength106 kips This is the lightest section in the table Assumed U 075 And net section fracture will govern if Ae lt 0745 Ag 0 Step II Calculate the net section fracture strength for the actual connection According to the Figure above An Ag 1 db 18 X t An 325 11 18 X 05 26875 in2 The connection is only through the m Therefore the shear lag factor U will be the distance from the back of the long leg to the centroid of the angle See DIM section ofthe AISC manual See Table l7 on pages l36 l37 i 0822 in U l EL l 0822 9 0908 But Umust be S 090 Therefore let U 090 35 CE 405 Design ofSteel Structures Prof Dr A Varma Tension Member Design Ae 090 An 090 X 26875 241 in2 APn 075 X Fu X Ae 075 X 58 X 241 1048 kips Acceptable because Pu 100 kips 0 Step V Check the block shear rupture strength 0 Identify the block shear path W DH DH H 20 i 30 in 30 in 30 in Agv92x05 55 in2 Anv1135X118X05353in2 Agt20X05l0in2 Am 20 05 x 1 18 x 05 072 in2 0 Identify the governing equation FuAm 58 X 072 4176 kips 06FuAnV 06 X 58 X 353 122844 inz which is gt FuAm 0 Calculate block shear strength ARn 075 06FuAm FyAgt 075 12284 36 X 10 119133 kips Which is greater than Pu 100 kips Therefore L4X3X12 is still acceptable 36 c ms Dcszganstzdstmnmnsrl rqf m A Wm CHAPTERS BOLTED CONNECTION 51 INTRODUCTORY CONCEPTS loadmg the shank Hanger connecnon The hanger Connection puts the bolts m tension 0 o lt 070 gt O O c ms Dcszanfstzdstmnmnsrl rqf m A Vanna d e The bolts are subjectedto shear ortenslon loadmg 7 1n rnostbolted eonneetaon the bolts are subjectedto shear e Bolts ean fad tn shear orm tenston 7 You ean ea1eu1ate the shear strength or the tenst1e strength ofabolt I tmnV nu h v V L the eenter gravtty of the ofthe load lndeldual bolts tnthe eonneetton We will rst eoneentrate on bolted shear eonneenons CE 405 Design ofSteel Structures Prof Dr A Varma 52 BOLTED SHEAR CONNECTIONS 0 We want to design the bolted shear connections so that the factored design strength R is greater than or equal to the factored load 0 So we need to examine the various possible failure modes and calculate the corresponding design strengths 0 Possible failure modes are Shear failure of the bolts Failure of member being connected due to fracture or block shear or Edge tearing or fracture of the connected plate Tearing or fracture of the connected plate between two bolt holes Excessive bearing deformation at the bolt hole 0 Shear failure of bolts Average shearing stress in the bolt fV PA Pn dbl4 P is the load acting on an individual bolt A is the area of the bolt and db is its diameter Strength of the bolt P fV X 11 dbl4 where fV shear yield stress 06Fy Bolts can be in single shear or double shear as shown below When the bolt is in double shear two crosssections are effective in resisting the load The bolt in double shear will have the twice the shear strength of a bolt in single shear ca 455 Desxgn mm Stmctm 7 Pm D7 A mea P gtP2 w p m f p P12 P j HE P P ne P PZ gt 1I39 m P P I lt I Pang PZw 122 gt 2 Single Shear 1 Double Slleur Failure ofconnected membex r We have covexed this In detail In Ch 2 on tmslon membexs r Membex can fall due to tension fracture 0 block Shem cucumfexmce ofthe fastenex e As such the stxess well be highest at the mdml contactpomt A Howevex the avenge r e Mano bolt spacing and edge descmce well have an effect on the beanng stx the connected plate not the bolt c ms Dcszanfstzdstmnmnsrl rqf m A Wm occur between two holes m the dArecnon othe beanng load Rn2x06FLct12Fquc CE 405 Design UfSteel Stmrmres erf Dr A Vanna To prevent excessive deformation of the hole an upper limit is placed on the bearing load This upper limit is proportional to the fmcture stress times the projected bearing area RnC xFuxbearingareaCFudbt If deformation is not a concern then C 3 If deformation is a concern then C24 C 24 corresponds to a deformation of025 in Finally the equation for the bearing strength of a single bolts is Rn Where 1 075 and RH 12 LE t F lt 24 dthu LC is the clear distance in the load direction from the edge of the bolt hole to the edge of the adjacent hole or to the edge of the material This relationship can be simpli ed as follows The upper limit Will become effective When 12 LE t F 24 d tFu ie the upper limit Will become effective WhenLE 2 db IfLElt2 dtJ Rn l2LEtFu lfLEgt2dtJ Rnl4dthu Failure surface Failure surface 1 LL i Ian2 1 R RHIZ b CE 405 Design ofSteel Structures Prof Dr A Varma 53 DESIGN PROVISIONS FOR BOLTED SHEAR CONNECTIONS o In a simple connection all bolts share the load equally e gt 6 Tn TVquot 0 e gt Tn TVquot 6 6 TVquot TVquot o In a bolted shear connection the bolts are subjected to shear and the connecting connected plates are subjected to bearing stresses Bearing Stresses in plate 0 The shear strength of all bolts shear strength of one bolt X number of bolts 0 The bearing strength of the connecting connected plates can be calculated using equations given by AISC speci cations 0 The tension strength of the connecting connected plates can be calculated as discussed earlier in Chapter 2 531 AISC Design Provisions 0 Chapter J of the AISC Speci cations focuses on connections 0 Section J3 focuses on bolts and threaded parts CE 405 Design ofSteel Structures Prof Dr A Varma AISC Specification J33 indicates that the minimum distance s between the centers of bolt holes is 2db A distance of 3db is preferred AISC Specification J34 indicates that the minimum edge distance Le from the center of the bolt to the edge ofthe connected part is given in Table J34 on page 16161 Table J34 specif1es minimum edge distances for sheared edges edges of rolled shapes and gas cut edges AISC Specification J35 indicates that the maximum edge distance for bolt holes is 12 times the thickness of the connected part but not more than 6 in The maximum spacing for bolt holes is 24 times the thickness of the thinner part but not more than 12 in Specification J36 indicates that the design tension or shear strength of bolts is FnAb Table J32 gives the values of and FH Ab is the unthreaded area of bolt In Table J32 there are different types of bolts A325 and A490 The shear strength of the bolts depends on whether threads are included or excluded from the shear planes If threads are included in the shear planes then the strength is lower We will always assume that threads are included in the shear plane therefore less strength to be conservative We will look at specifications J37 7 J39 later AISC Specification J310 indicates the bearing strength ofplates at bolt holes The design bearing strength at bolt holes is Rn Rn 12 LC t Fu S 24 db t Fu deformation at the bolt holes is a design consideration Where Fu specified tensile strength of the connected material CE 405 Design ofSteel Structures Prof Dr A Varma LC clear distance in the direction of the force between the edge of the hole and the edge of the adjacent hole or edge of the material in t thickness of connected material 532 AISC Design Tables Table 710 on page 733 of the AISC Manual gives the design shear of one bolt Different bolt types A325 A490 thread condition included or excluded loading type single shear or double shear and bolt diameters 58 in to ll 2 in are included in the Table Table 711 on page 733 ofthe AISC Manual is an extension of Table 710 with the exception that it gives the shear strength of n bolts Table 712 on page 734 ofthe AISC manual gives the design bearing strength at bolt holes for various bolt spacings These design bearing strengths are in kipsin thickness The tabulated numbers must be multiplied by the plate thickness to calculate the design bearing strength of the plate The design bearing strengths are given for different bolt spacings 267db and 3db different FH 58 and 65 ksi and different bolt diameters 58 7 1 12 in Table 712 also includes the spacing smn required to develop the full bearing strength for different Fu and bolt diameters Table 712 also includes the bearing strength when s gt Sfull Table 712 also includes the minimum spacing 223 db values Table 713 in the AISC manual on page 735 is similar to Table 712 It gives the design bearing strength at bolt holes for various edge distances CE 405 Design ofSteel Structures Prof Dr A Varma These design bearing strengths are in lapsin thickness The tabulated numbers must be multiplied by the plate thickness to calculate the design bearing strength of the plate The design bearing strengths are given for different edge distances 125 in and 2 in different FH 58 and 65 ksi and different bolt diameters 58 7 1 12 in Table 713 also includes the edge distance Le mu required to develop the full bearing strength for different Fu and bolt diameters Table 713 also includes the bearing strength when Le gt Le full CE 405 Design ofSteel Structures Prof Dr A Varma Example 51 Calculate and check the design 38 in N 5x 2 strength of the connection shown below Is H i 5 o 0 A36 2 50 the connection adequate for carrying the 125 0 0K imbth H o l 125 250125 factored load of 65 kips Solution Step I Shear strength of bolts 0 The design shear strength of one bolt in shear Fn Ab 075 X 48 X 11 X 07524 FH Ab 159 kips per bolt See Table 132 and Table 710 Shear strength of connection 4 X 159 636 kips See Table 711 Step 11 Minimum edge distance and spacing requirements 0 See Table 134 minimum edge distance 1 in for rolled edges ofplates The given edge distances 125 in gt 1 in Therefore minimum edge distance requirements are satis ed 0 Minimum spacing 267 db 267 X 075 20 in Preferred spacing 30 db 30 X 075 225 in The given spacing 25 in gt 225 in Therefore spacing requirements are satisfied Step III Bearing strength at bolt holes 0 Bearing strength at bolt holes in connected part 5 X 12 in plate At edges LC 125 7 hole diameter2 125 7 34 1162 0844 in Rn 075 X 12 Let F 075 X 12 X 0844 X 05 X 58 2202 kips But Rn S 075 24 db t F 075 X 24 X 075 X 05 X 58 3915 kips Therefore Rn 2202 kips at edge holes 0 Compare with value in Table 7 13 Rn 440x 05 220 kips CE 405 Design ofSteel Structures Prof Dr A Varma At other holes s 25 in LC 25 7 34 116 1688 in Rn 075 X 12 Let F 075 X 12 X 1688 X 05 X 58 4405 kips But Rn S 075 24 db t Fu 3915 kips Therefore Rn 3915 kips Therefore Rn 3915 kips at other holes 0 Compare with value in Table 7 12 Rn 783 x 05 3915 kips Therefore bearing strength at holes 2 X 2202 2 X 3915 12234 kips o Bearing strength at bolt holes in gusset plate 38 in plate At edges LC 125 7 hole diameter2 125 7 34 1162 0844 in glRn 075 X 12 LctFu 075 X 12 X 0844 X m X 58 1652 k But Rn S 075 24 db t F 075 X 24 X 075 X m X 58 2936 kips Therefore Rn 1652 kips at edge holes 0 Compare with value in Table 7 13 Rn 440x 38 165 kips At other holes s 25 in LC 25 7 34 116 1688 in Rn 075 X 12 LctFu 075 X 12 X 1688 X M X 58 3304 kips But Rn S 075 24 dbt Fu 2936 kips Therefore Rn 2936 kips at other holes 0 Compare with value in Table 7 12 Rn 783 x 0375 2936 kips Therefore bearing strength at holes 2 X 1652 2 X 2936 9176 kips o Bearing strength of the connection is the smaller of the bearing strengths 9176 kips CE 405 Design ofSteel Structures Prof Dr A Varma Connection Strength Shear strength 633 kips Bearing strength plate 12234 kips Bearing strength gusset 9176 kips Connection strength MRquot gt applied factored loads 7Q Therefore ok CE 405 Design ofSteel Structures Prof Dr A Varma Example 52 Design a double angle tension member and a gusset plated bolted connection system to carry a factored load of 100 kips Assume A36 36 ksi yield stress material for the double angles and the gusset plate Assume A325 bolts Note that you have to design the double angle member sizes the gusset plate thickness the bolt diameter numbers and spacing Solution Step I Design and select a trial tension member 0 See Table 3 7 on page 333 ofthe AISC manual Select 2L 3 X 2 X 38 with Pn 113 kips yielding and 114 kips fracture While selecting a trial tension member check the fracture strength with the load Step 11 Select size and number of bolts The bolts are in double shear for this design may not be so for other designs 0 See Table 7 11 on page 733 in the AISC manual Use four 34 in A325 bolts in double shear Rn 127 kips shear strength of bolts from Table 711 Step 111 Design edge distance and bolt spacing 0 See Table J34 The minimum edge distance 1 in for 3 4 in diameter bolts in rolled edges Select edge distance 125 in 0 See speci cation J35 Minimum spacing 267 db 20 in Preferred spacing 30 db 225 in Select spacing 30 in which is greater than preferred or minimum spacing CE 405 Design ofSteel Structures Prof Dr A Viier Step IV Check the bearing strength at bolt holes in angles 0 Bearing strength at bolt holes in angles Angle thickness 38 in See Table 7 13 for the bearing strength per in thickness at the edge holes Bearing strength at the edge holes Le 125 in Rn 440 X 38 165 k See Table 7 12 for the bearing strength per in thickness at nonedge holes Bearing strength at nonedge holes s 3 in Rn 783 X 38 294 k Bearing strength at bolt holes in each angle 165 3 x 294 104 7 kips Bearing strength of double angles 2 x 104 7 kips 2094 kips Step V Check the fracture and block shear strength of the tension member 0 This has been covered in the chapter on tension members anal is left to the students Step VI Design the gusset plate 0 See speci cation J52 for designing gusset plates These plates must be designed for the limit states of yielding and rupture Limit state of yielding 0 RH 09 Ag Fy gt 100 kips 0 Therefore Ag L X t gt 309 in2 0 Assume t 12 in Therefore L gt 618 in 0 Design gusset plate 65 X 12 in 0 Yield strength Rn 09 X 65 X 05 X 36 1053 kips Limit state for fracture CE 405 Design ofSteel Structures Prof Dr A Varma 0 An Ag7db18 Xt 0 An 65 x 05 7 34 18X 05 281 in2 0 But An 5 085 Ag 085 x 325 276 m2 o Rn 075 x An x Fu 075 x 276 x 58 120 kips Design gusset plate 65 X 05 in 0 Step VII Bearing strength at bolt holes in gusset plates Assume Le 125 in same as double angles Plate thickness 38 in Bearing strength at the edge holes Rn 440 X 12 220 k Bearing strength at nonedge holes Rn 783 X 12 3915 k Bearing strength at bolt holes in gussetplate 220 3 x 3915 1395 kips Summary of Member and Connection Strength Connection Member Gusset Plate Shear strength 127 kips Yielding 113 kips Yielding 1053 kips Bearing strength 2094 kips angles Fracture Fracture 120 kips Bearing Strength 1395 gusset Block Shear Overall Strength is the smallest of all these numbers 1053 kips Gusset plate yielding controls Resistance gt Factored Load 100 kips Design is acceptable CE 405 Design ofSteel Structures Prof Dr A Varma 54 SLIP CRITICAL BOLTED CONNECTIONS High strength A325 and A490 bolts can be installed with such a degree of tightness that they are subject to large tensile forces These large tensile forces in the bolt clamp the connected plates together The shear force applied to such a tightened connection will be resisted by friction as shown in the Figure below Tightened NTb H iv NTb P mm rmm FuN NTbllJTlJU mun Tb H N Tb l Thus slip critical bolted connections can be designed to resist the applied shear forces using friction If the applied shear force is less than the friction that develops between the two surfaces then no slip will occur between them However slip will occur when the friction force is less than the applied shear force After slip occurs the connection will behave similar to the bearingtype bolted connections designed earlier CE 405 Design ofSteel Structures Prof Dr A Varma Table J3l summarizes the minimum bolt tension that must be applied to develop a slip critical connection The shear resistance of fully tensioned bolts to slip at factored loads is given by AISC Speci cation J38 a Shear resistance at factored load Rn 113 u Tb NS where I 10 for standard holes u 033 Class A surface with unpainted clean mill scale surface CE 405 Tb minimum bolt tension given in Table J3l NS number of slip planes See Table 715 on page 736 of the AISC manual This Table gives the shear resistance of fully tensioned bolts to slip at factored loads on class A surfaces For example the shear resistance of ll8 in bolt fully tensioned to 56 kips Table J3l is equal to 209 kips Class A faying surface When the applied shear force exceeds the Rn value stated above slip will occur in the connection The shear resistance of fully tensioned bolts to slip at service loads is given by AISC Specification J38 b Shear resistance at service load Rn FV Ab Where 10 for standard holes FV slipcritical resistance to shear at service loads see Table A J32 on page 161116 ofthe AISC manual See Table 716 on page 737 ofthe AISC manual This Table gives the shear resistance of fully tensioned bolts to slip at service loads on class A surfaces CE 405 Design ofSteel Structures Prof Dr A Varma For example the shear resistance of ll8 in bolt fully tensioned to 56 kips Table J3l is equal to 169 kips Class A faying surface When the applied shear force exceeds the Rn value stated above slip will occur in the connection 0 The nal strength of the connection will depend on the shear strength of the bolts calculated using the values in Table 711 and on the bearing strength of the bolts calculated using the values in Table 712 713 This is the same strength as that of a bearing type connection CE 405 Design ofSteel Structures Prof Dr A Varma Example 53 Design a slipcritical splice for a tension member subjected to 300 kips of tension loading The tension member is a W8 X 28 section made from A992 50 ksi material The unfactored dead load is equal to 50 kips and the unfactored live load is equal to 150 kips Use A325 bolts The splice should be slipcritical at service loads Solution Step I Service and factored loads Service Load D L 200 kips Factored design load 12 D 16 L 300 kips Tension member is W8 X 28 section made from A992 50 ksi steel The tension splice must be slip critical ie it must not slip at service loads Step II Slipcritical splice connection 0 Rn of one fullytensioned slipcritical bolt I FV Ab See Spec A J38 b page 161 11 7 ofAISC o Ifdb 34 in 111n ofone bolt 10 x 17 x 11 x 07524 751 kips Note FV 17 ksi from Table AJ32 From Table 7 16 on page 7 37 Rn 751 kips Rn ofn bolts 751 X n gt 200 kips splice must be slipcritical at service Therefore n gt 2663 0 Ifdb 78 in Rn of one bolt 102 kips from Table 7 16 Rn ofn bolts 102 X n gt 200 kips splice must be slipcritical at service Therefore n gt 196 bolts 20 Example 53 Step 1 Service andFactoredLoads D 50 Kips L 150 Kips I Service Loads PS D L PS 200 Kips Factored Loads Pu 12D 16L Pu 300 Kips Step II Slip Critical connection I In Service loads consideration Rn ofone fully tenstioned slipcritical bolt FV Ab As given in Spec A138b page 161 1 17 10 FV 17 Ksi A325 Table A132 3 1t 2 I If diameter ofthe bolt db m Ab db 4 4 for one bolt Rn FvAb Rn 751 Kips Ps Number of bolts required n n 2663 min reqd Rn 7 7 1t 2 I If diameter ofthe bolt 7 db 1n Ab db 8 4 MO bolt Rn FvAb Rn 1022 Kips Ps Number of bolts required n n 19565 min reqd Rn say we provide 24 bolts on either side of the center line 6 oneitlier side of the anges top bottom Step 111 Connection Details and spacings for 24bolts on each W8 X 28 I Note that there are 24 bolts on either side ofthe centerline In allthere are 48 number 78 in dia bolts used in the connection I Minimum pretension applied to the bolts 390 Kips from Table 131 I Minimum Edge distance from Table 134 Lgmin 1 125 in I Provide Edge Distance Le 125 in I Minimum spacing Spec 133 s 267db s 2336 in mama Wang 7 Ma anubunz mam m Faxdeagqumndespacmg s z m ElEVRTiDN 7 49 m nan sag w c mman smgm at named lauds mam mu knps Themummghumebamammm l kxpybmtxlAbnlts5184Knps EemngslxznghufWXmbnltnzedgzhnlev E 457 mpmmkms mm Emmgmgmmmmmmmm ED mz mpmmkms mm Tambearingmenthanebalthnlesxnwxde argesechm a mun an a zzzzxm3 mp5 sugv Beam m ghee gm 55 Ks P mu mp5 Pu 2 Teman lxel n9 FgtPn mmA man m mg Agy g Buy z Pu 2 TemananureEl7S angtPn mmAn mmAn6lSA m nun WWWWmmmmmmwmmm 51m MW cm W mmme Aquot z m 7 m 2 A 4 Hm 454 Mm m Ag 8 8 Ag Sumnng W1 m z 2 42 m mm mm mm am W w W mmm Assume unh pm um memans a a Lhzxefaxe mm plate themes 7 ChzckfmA andAg Ag1635 m2 gt m2 525 m2 gt m2 2 lt nstfuxax m Skenghuf39hesphcephlem yiel ng WASP 7257 mp5 gt a mu mp5 mum n75 Any mm mp5 m t va quot Chzckfmbemngmenghuf hesphceplaus Chzckfmblacksheazmpmre sugvx Chzckmzmber 1mg fucmreandblack sh g A 352 vas 1L 4

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