Loads

Moayyad Al Nasra, PhD, PE

1 (c) Al Nasra

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

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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 40 psf

• Public rooms 100 psf

• Office building 100 psf

• Storage, light 125 psf

• Storage, heavy 250 psf

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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%

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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.

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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

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LRFD load combinations

• U= 1.4 D

• U= 1.2 D + 1.6 L + 0.5 (Lr or s or R)

• U= 1.2 D + 1.6 (Lr or s or R) + (0.5L or 0.8W)

• U= 1.2 D + 1.6 W + 0.5 L +0.5( Lr or s or R)

• U= 1.2 D +/- 1.0 E + 0.5 L + 0.2 S

• U= 0.9 D +/- (1.6 W or 1.0 E)

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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

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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.

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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 + 1.6 L + 0= 1.2 (720)+1.6(100)= 2464 lb/ft

The governing factored load = 2464 lb/ft

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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)

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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

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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

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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) = 80+0.75(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)= 80+0.75(60)+0.75(20)=140 psf

• Wa=0.6D+/-(W or 0.7E) = 0.6(80)= 48 psf

• The governing load is 140 psf

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Analysis of Tension Members

Moayyad Al Nasra, Ph.D, PE

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Analysis of Tension Members

• Types of tension members:

– L- section,

– round bars,

– flat bars,

– double-angle,

– T-section,

– I-section,

– built-up

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Strength of tension members AISC 14th edition page 16.1-26

a.) Gross-section

Pn= Fy.Ag AISC D2-1

Pu=Φt.Fy.Ag

Φt=0.90

b.) Net-section

Pn=Fu.Ae AISC D2-2

Pu=Φt.Pn=Φt.Fu.Ae

Φt=0.75

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The design strength of a tension member

Φt.Pn, is to be the smaller of the above equations

Where:

Pn = nominal tensile force

Pu = ultimate tensile strength

Fy = yield stress

Fu = ultimate stress

Ae = effective area

Ag = gross area

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Net Area

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• 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=An=(3/8)(6)-2[3/4+1/8](3/8)=1.59 in2

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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

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• (concentration of stresses around the holes). Therefore AISC introduced a reduction

• factor, U, so that – Ae=AnU AISC D3-1 (AISC 14th pp.

16.1-27)

• Where U = shear lag factor determined from table D3.1 AISC 14th pp.16.1-28 – U=1-x/L for tension members see case 2 table

D3.1

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• (table D3.1 AISC 14th edition steel design

manual shows that for W, M, S section use

U=0.9 for bf≥2/3 d, and use U=0.85 for bf<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.

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• 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.

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• From AISC manual W10x45 (Ag=13.3 in2, d=10.10 in., bf=8.02 in., tf=0.620 in.), Also

• WT5x22.5, x=0.907 in. ( half of W10x45)

• Pu= Φt. Fy.Ag=(0.90)(50)(13.3)=598.5 k

• An= 13.3-4(3/4+1/8)(0.620)=11.13 in2

– But, bf=8.02”>(2/3)d=(2/3)10.1=6.73, from case 7 AISC 14th manual table 3.1, U=0.9 pp. 16.1-28

• Use U=0.9

• Ae=U.An=((0.90)(11.13)=10.02 in2

• Pu=Φt.Fu.Ae=(0.75)(65)(10.02)=488.5 K

• Therefore the LRFD tensile strength of the section is 488.5 K

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Bolted splice plates

• For bolted splice plates

• Ae=An≤0.85 Ag

• Example 2-2

• Same as example 2-1

• Ag=(3/8)(6)=2.25 in2

• 0.85Ag=0.85(2.25)=1.91 in2

• Ae=An=1.59 in2 < 0.85Ag=1.91 in2

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Effect of Staggered Holes

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• 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 + S2/(4g)

•

• Example Determine the critical net area of

¼ “ thick plate. The holes are punched for

¾” bolts

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• 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 = 6.125 in.

– Abde: 7-2(7/8)+42/(4x3)=6.58 in.

• Therefore the section abc controls, net width=6.125

• Net area=An=6.125(1/4)=1.53 in2

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• 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.

• 6.125=7-2(7/8)+S2/(4x3) then solve for

S=3.24 inch

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Net Area • In General

– bn=b – Σ dh + Σ (s2/4g)

b dh

s

g

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Net Area

• Where

– bn = net width

– b = gross width

– dh = width of hole ( diameter)

– s= pitch spacing

– g = gage spacing

For members of uniform thickness, t, then

An = bn.t

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Net Area

• For members of non-uniform thickness, the net

area can be calculated as follows:

• An= Ag – Σ dh.t + [Σ (s2/4g)] . t

• The critical net area is the net area having the

least value. It is obtained by checking all

possible failure paths

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Net Area, Example

• Compute the net area of the 7/8 X 12 plate

shown. The holes are for 3/4 –in bolts

3

in

3

in

3

in 3

in

1.5-in

¾ - in.

bolts

7/8 X 12 –in

plate

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Net Area, solution

3

in

3

in

3

in 3

in

1.5-in

A

B

C

D

E

=

s (c) Al Nasra 38

Net Area, Solution

• Net width

– ABCD = 12 – (2)(3/4+1/8)=10.25 in.

– ABECD = 12 – (3)(3/4+1/8) + (2)[1.52/(4x3)]=9.75

in

– The 9.75 is the lowest and it governs

Net area = (9.75)(0.875)= 8.53 in2

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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

Block Shear

• The formulae used by LRFD (Φt.Pn) and

ASD (Pn/Ωt) 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

out. (c) Al Nasra 41

Block Shear

The cross-hatched parts may tear out

Bolted angle

Bolted W

Section

Shear

plane

Tension plane

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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

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Block Shear

• The AISC specification (J4.3) states that the available strength, Rn, for the block shear rupture design strength is as follows:

• Rn=0.6FuAnv+UbsFuAnt ≤ 0.6FyAgv+UbsFuAnt (AISC 14th ed. Eq. J4-5,pp. 16.1-129 )

• Φ =0.75 (LRFD), Ω = 2.00 (ASD)

• In which

– Anv =net area subjected to shear, in2 (mm2)

– Ant = net area subjected to tension, in2 (mm2)

– Agv = gross area subjected to shear, in2 (mm2)

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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, Ubs. Should the tensile stress distribution be uniform, Ubs. = 1.0 according to AISC specification (J4.3) (i.e. gusset plates, single-row beam connection,…). For non-uniform stress tensile stresses, Ubs. =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.

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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

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Block shear, Solution

Using a W12X45 (tf=0.575-in, bf=8.05-in)

Agv=(4)(8)(0.575)=18.4 in2 (2 each flange)

Anv=(4)[8-2.5(7/8+1/8)](0.575)=12.65 in2 (2.5 bolts each of the 4

sides)

Ant=(4)[1.275-(.5)(7/8+1/8)](0.575)=1.78 in2 (1/2 bolt in tension)

8-in

1.275-

in

5.5-in

1.275-

in

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• Ubs = 1.0 uniforn tensile stress

• Rn=0.6FuAnv+UbsFuAnt ≤ 0.6FyAgv+UbsFuAnt

• Rn= 0.6(70)(12.65)+1.0(70)(1.78) = 655.9 k

• <0.6(50)(18.4)+1.0(70)(1.78)=676.6 k

• Therefore Rn =655.9 k

• LRFD

– Φ. Rn = (0.75)(655.9)=491.9k

ADS

Rn/Ω=655.9/2=327.9k

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