Loads. Moayyad Al Nasra, PhD, PE. (c) Al Nasra

Transcription

1 Loads Moayyad Al Nasra, PhD, PE (c) Al Nasra 1

2 Loads - Types Dead Load, D: Load of constant magnitude, specified direction, and well defined point of action. Ex. Self weight of beams, slabs, etc. Typical unit weight of structural materials Reinforced concrete 150 lb/cu ft Structural steel 490 lb/ cu ft Plain concrete 145 lb/cu ft Hardwood flooring(7/8) in 4 psf Suspended ceiling 2 psf (c) Al Nasra 2

3 Loads types cont d Live load, L: Load that may change in magnitude, direction, or position such as: Floor loads, specified by the design codes Apartments Public rooms Office building Storage, light Storage, heavy 40 psf 100 psf 100 psf 125 psf 250 psf (c) Al Nasra 3

4 Loads, types, live loads, cont d Traffic Loads: Loads of varying magnitude caused by groups of truck or train wheels Impact Loads: Loads due to vibration of moving or moving loads. Live load impact factor Elevator machinery 100% Motor driven machinery 20% Reciprocating machinery 50% Hangers for floors or balconies 33% (c) Al Nasra 4

5 Loads, types, live loads, cont d Other live loads Blast loads Thermal loads Longitudinal loads, stopping loads of trains, trucks, etc. Centrifugal loads, curved bridges Hydrostatic pressure, dams, reservoirs, water tanks, etc. Earth pressure, retaining walls, basements, etc. (c) Al Nasra 5

6 Loads, types, cont d Environmental loads Snow loads: 1 inch of snow is equivalent of 5 psf. Common values ranges between 10 psf to 40 psf. Rain loads: The best method of preventing ponding is to have an appreciable slope of the roof at least ¼ in per foot. Wind load, Hurricanes and tornados, etc. Earthquake loads, seismic forces (c) Al Nasra 6

7 LRFD load combinations U= 1.4 D U= 1.2 D L (Lr or s or R) U= 1.2 D (Lr or s or R) + (0.5L or 0.8W) U= 1.2 D W L +0.5( Lr or s or R) U= 1.2 D +/- 1.0 E L S U= 0.9 D +/- (1.6 W or 1.0 E) (c) Al Nasra 7

8 LRFD load combinations, cont d U= the design or ultimate load D = dead load L = live load due to occupancy Lr = roof live load S= snow load R = nominal load due to initial rainwater or ice, exclusive of the ponding contribution W= wind load E= earthquake load (c) Al Nasra 8

9 Example, Load combination» W12X120, 10 O.C Floor system has W12X120 beams spaced 10 ft on center and is supporting a floor dead load of 60 psf, and a live load of 100 psf. Determine the governing load in lb/ft that each beam must support. (c) Al Nasra 9

10 Solution Self dead load = 120 lb /ft Floor dead load = 10X 60 = 600 lb/ft Total dead load = 720 lb/ft Live load = 10 X 100 = 1000 lb/ft U= 1.4 D = 1.4 X 720 = 1008 lb/ft U= 1.2 D L + 0= 1.2 (720)+1.6(100)= 2464 lb/ft The governing factored load = 2464 lb/ft (c) Al Nasra 10

11 ASD Load Combination D D+L D+(Lr or S or R) D+0.75L+0.75(Lr or S or R) D+/- (W or 0.7E) D+0.75(W or 0.7E)+0.75L+0.75(Lr or S or R) 0.6D+/- (W or 0.7E) (c) Al Nasra 11

12 Exercise, Load combinations Determine the maximum combined loads using the recommended AISC expressions for both LRFD and ASD. D= 80 psf L= 60 psf R= 10 psf S= 20 psf (c) Al Nasra 12

13 Solution - LRFD Wu = 1.4 D = 1.4 (80) = 112 psf Wu = 1.2 D+1.6L+0.5(Lr or S or R)= 1.2(80)+1.6(60)+0.5(20) = 202 psf Wu= 1.2D+1.6(Lr or S or R) +0.5(L or 0.8W)= 1.2(80)+1.6(20)+0.5(60)=158psf Wu= 1.2D+1.6W+0.5L+0.5(Lr or s or R) = 1.2(80)+0.5(60)+0.5(20) = 136 psf Wu= 1.2D+/-1.0E+0.5L+0.2S = 1.2(80)+0.5(60)+0.2(20) = 130 psf Wu= 0.9D+/-(1.6W or 1.0E) = 0.9(80)=72 psf The governing factored load is 202 psf (c) Al Nasra 13

14 Solution, ASD Wa=D=80 psf Wa=D+L=80_60=140 psf Wa=D+(lr or S or R) = 80+20=100 psf Wa= D+0.75L+0.75(Lr or S or R) = (60)+0.75(20)= 140 psf Wa=D+(W or 0.7E)=80+0=80 psf Wa=D+0.75(Wor 0.7E)+0.75L+ 0.75(Lr or S or R)= (60)+0.75(20)=140 psf Wa=0.6D+/-(W or 0.7E) = 0.6(80)= 48 psf The governing load is 140 psf (c) Al Nasra 14

15 (c) Al Nasra 15

16 Analysis of Tension Members Moayyad Al Nasra, Ph.D, PE (c) Al Nasra 16

17 Analysis of Tension Members Types of tension members: L- section, round bars, flat bars, double-angle, T-section, I-section, built-up (c) Al Nasra 17

18 Strength of tension members AISC 14 th edition page a.) Gross-section P n = F y.a g AISC D2-1 P u =Φ t.f y.a g Φ t =0.90 b.) Net-section P n =F u.a e AISC D2-2 P u =Φ t.p n =Φ t.f u.a e Φ t =0.75 (c) Al Nasra 18

19 The design strength of a tension member Φ t.p n, is to be the smaller of the above equations Where: P n = nominal tensile force P u = ultimate tensile strength F y = yield stress F u = ultimate stress A e = effective area A g = gross area (c) Al Nasra 19

20 Net Area (c) Al Nasra 20

21 Net area = gross area areas of holes Holes are usually punched 1/16 inch larger than the diameter of the bolt. Also the punching of the hole is assumed to damage or even destroy 1/16 inch more of the surrounding metal. Therefore the area of the hole is 1/8 inch in diameter larger than the bolt diameter. Example 2-1: Net Area=A n =(3/8)(6)-2[3/4+1/8](3/8)=1.59 in 2 (c) Al Nasra 21

22 Effective area To account for the non-uniform distribution of stresses at the connection sections, a reduction factor will be introduced. The further the section from the connection the more uniform the stress becomes (c) Al Nasra 22

23 (concentration of stresses around the holes). Therefore AISC introduced a reduction factor, U, so that A e =A n U AISC D3-1 (AISC 14 th pp ) Where U = shear lag factor determined from table D3.1 AISC 14 th pp U=1-x/L for tension members see case 2 table D3.1 (c) Al Nasra 23

24 (table D3.1 AISC 14 th edition steel design manual shows that for W, M, S section use U=0.9 for b f 2/3 d, and use U=0.85 for b f <2/3 d) Where: L= length of a connection X= distance measured from the plane of the connection to the centroid of the area of the whole section. Can be obtained from the AISC manual. (c) Al Nasra 24

25 Example: Determine the LRFD tensile strength of a W10X45 with two lines of ¾ inch bolts in each flange using A572 grade 50 steel, with Fy=50 ksi, and Fu = 65 ksi, and the AISC specification. There are assumed to be at least three bolts in each line 4 inches on center, and the bolts are not staggered with respect to each other. (c) Al Nasra 25

26 (c) Al Nasra 26

27 From AISC manual W10x45 (A g =13.3 in 2, d=10.10 in., b f =8.02 in., t f =0.620 in.), Also WT5x22.5, x=0.907 in. ( half of W10x45) P u = Φ t. F y.a g =(0.90)(50)(13.3)=598.5 k A n = (3/4+1/8)(0.620)=11.13 in 2 But, b f =8.02 >(2/3)d=(2/3)10.1=6.73, from case 7 AISC 14 th manual table 3.1, U=0.9 pp Use U=0.9 A e =U.A n =((0.90)(11.13)=10.02 in 2 P u =Φ t.f u.a e =(0.75)(65)(10.02)=488.5 K Therefore the LRFD tensile strength of the section is K (c) Al Nasra 27

28 Bolted splice plates For bolted splice plates A e =A n 0.85 A g Example 2-2 Same as example 2-1 A g =(3/8)(6)=2.25 in A g =0.85(2.25)=1.91 in 2 A e =A n =1.59 in 2 < 0.85A g =1.91 in 2 (c) Al Nasra 28

29 Effect of Staggered Holes (c) Al Nasra 29

30 The joint will fail at the weakest section To compute the net width of a tension member along a zig-zag section: Net width=gross width-diameter of holes along the zig-zag section + S 2 /(4g) Example Determine the critical net area of ¼ thick plate. The holes are punched for ¾ bolts (c) Al Nasra 30

31 (c) Al Nasra 31

32 Solution: Possible sections are abc, and abde. Hole diameters to be subtracted =3/4+1/8=7/8 in. Net width: Abc: 7-7/8 = in. Abde: 7-2(7/8)+4 2 /(4x3)=6.58 in. Therefore the section abc controls, net width=6.125 (c) Al Nasra Net area=a =6.125(1/4)=1.53 in 2 32

33 Example: Design bolts configuration by optimizing the failure mechanism, using the previous example. Change the value of, S, to make the net width of abc, the same as the net width of abde =7-2(7/8)+S 2 /(4x3) then solve for S=3.24 inch (c) Al Nasra 33

34 In General b n =b Σ d h + Σ (s 2 /4g) Net Area g d h b s (c) Al Nasra 34

35 Where Net Area b n = net width b = gross width d h = width of hole ( diameter) s= pitch spacing g = gage spacing For members of uniform thickness, t, then A n = b n.t (c) Al Nasra 35

36 Net Area For members of non-uniform thickness, the net area can be calculated as follows: A n = A g Σ d h.t + [Σ (s 2 /4g)]. t The critical net area is the net area having the least value. It is obtained by checking all possible failure paths (c) Al Nasra 36

37 Net Area, Example Compute the net area of the 7/8 X 12 plate shown. The holes are for 3/4 in bolts ¾ - in. bolts 7/8 X 12 in plate 3 in 3 in 3 in 3 in 1.5-in (c) Al Nasra 37

38 Net Area, solution A B C D E 3 in 3 in 3 in 3 in 1.5-in = s (c) Al Nasra 38

39 Net width Net Area, Solution ABCD = 12 (2)(3/4+1/8)=10.25 in. ABECD = 12 (3)(3/4+1/8) + (2)[1.5 2 /(4x3)]=9.75 in The 9.75 is the lowest and it governs Net area = (9.75)(0.875)= 8.53 in 2 (c) Al Nasra 39

40 Slenderness Ratio The AISC steel design manual specification D1 lists a preferred (but not required) maximum slenderness ration (SR) of 300. Rods and wires are excluded from this recommendation, SR=l/r l= un-braced length r= radius of gyration = sqrt(i/a) I= moment of inertia A= cross-sectional area (c) Al Nasra 40

41 Block Shear The formulae used by LRFD (Φ t.p n ) and ASD (P n /Ω t ) to calculate the allowable strengths of tension members are not always the controlling criteria. The allowable strength in tension may be controlled by block shear strength, where the failure may occur along a path involving tension in one plane and shear on a perpendicular plane. So it is possible for a block of steel to tear (c) Al Nasra 41 out.

42 Block Shear The cross-hatched parts may tear out Bolted angle Shear plane Tension plane Bolted W Section (c) Al Nasra 42

43 Block Shear The AISC specifications (J4.3) states that the block shear design strength of a particular member is to be determined by Computing the tensile fracture strength on the net section in one direction and adding to that value the shear yield strength on the gross area on the perpendicular segment Computing the shear fracture strength on the gross area subject to tension and adding it to the tensile yield strength on the net area subject to shear on the perpendicular segment. The expression to apply is the one with the larger rupture term (c) Al Nasra 43

44 Block Shear The AISC specification (J4.3) states that the available strength, Rn, for the block shear rupture design strength is as follows: R n =0.6F u A nv +U bs F u A nt 0.6F y A gv +U bs F u A nt (AISC 14 th ed. Eq. J4-5,pp ) Φ =0.75 (LRFD), Ω = 2.00 (ASD) In which A nv =net area subjected to shear, in 2 (mm 2 ) A nt = net area subjected to tension, in 2 (mm 2 ) A gv = gross area subjected to shear, in 2 (mm 2 ) (c) Al Nasra 44

45 Shear Block To account for the fact that stress distribution may not be uniform on the tensile plane for some connections, AISC introduced a reduction factor, U bs. Should the tensile stress distribution be uniform, U bs. = 1.0 according to AISC specification (J4.3) (i.e. gusset plates, single-row beam connection, ). For non-uniform stress tensile stresses, U bs. =0.5 (i.e. multiple-row beam end connection, ) Should the block shear strength of a connection be insufficient, it may be increased by increasing the edge distance and/or the bolt spacing. (c) Al Nasra 45

46 Block shear, Example Determine the block shearing strength (LRFD, ASD), W12X45 (A242 Grade 50 steel and 7/8-in bolts 2- in 3- in 3- in 5.5- in (c) Al Nasra 46

47 Block shear, Solution Using a W12X45 (t f =0.575-in, b f =8.05-in) A gv =(4)(8)(0.575)=18.4 in2 (2 each flange) A nv =(4)[8-2.5(7/8+1/8)](0.575)=12.65 in2 (2.5 bolts each of the 4 sides) A nt =(4)[1.275-(.5)(7/8+1/8)](0.575)=1.78 in2 (1/2 bolt in tension) 8-in in 5.5-in in (c) Al Nasra 47

48 U bs = 1.0 uniforn tensile stress R n =0.6F u A nv +U bs F u A nt 0.6F y A gv +U bs F u A nt R n= 0.6(70)(12.65)+1.0(70)(1.78) = k <0.6(50)(18.4)+1.0(70)(1.78)=676.6 k Therefore R n =655.9 k LRFD Φ. R n = (0.75)(655.9)=491.9k ADS R n /Ω=655.9/2=327.9k (c) Al Nasra 48