bgscmc04exprob

7/30/2019 BGSCMC04ExProb

1/32

A Beginner's Guide to the Steel Construction Manual last modified: 11 AugExample Problem 4.1 by: TBQ

Given: A lap splice is connected with (11) A325-X bolts as shown. Assume that the applied load consists of 40% dead load and 60% live load.

db = 0.75 in% DL 40%% LL 60%

bolts:3/4" diaA325-Xpretensionedin standard holes

Class A faying surfaces

Wanted: Determine the capacity of the connection based on bolt strength as specified below.Consider both LRFD and ASD.Express your results in terms of service load levels.

a. Determine the slip capacity of the connection.b. Determine the bearing capacity of the connection.

Solution:

Results Summary: Computations follow

Table of CapacitiesLRFD ASD

Limit State f R n P s,eq R n/W P s,eq(k) (k) (k) (k)

Slip 122 84.6 81.2 81.2


2/32

a. Slip Capacity - See specification J3.8

m 0.35 (Class A Surfaces)Du 1.13h sc 1 (Standard Holes)T

b 28 kips/boltNs 1r n 11.1 kips/boltNb 11 boltsR n 122 kips/connection

LRFD ASDf = 1 W = 1.5f R n = 122 kips R n / W = 81 kips

b. Bearing Capacity - See specification J3.6

F nv 60 ksi (from Table J3.2) Ab 0.4418 in

r n 26.5 kips/shear planeNs 1 shear planes/boltNb 11 boltsR n 292 kips/connection

LRFD ASDf = 0.75 W = 2f R n = 219 kips R n / W = 146 kips


3/32

007

ures no slipures no bolt shear failure


4/32


Given: A splice plate connection is used to connect two WTs together as shown.The connection is subjected to axial tension, P.

Assume that the load transferred consists of Dead Load (30%) and Seismic Load (70%).

db = 0.75 in% DL 30%% E 70%

bolts:3/4" diaA325-Xpretensionedin standard holes

Class B faying surfaces



Solution:


Table of CapacitiesLRFD ASD

Limit State f R n P s,eq R n/W P s,eq(k) (k) (k)

Slip 435 410 290 367


5/32

a. Slip Capacity - See specification J3.8

m 0.5 (Class B Surfaces)Du 1.13h sc 1 (Standard Holes)T

b 35 kips/boltNs 2 shear planes/boltr n 39.6 kips/boltNb 11 boltsR n 435 kips/connection

LRFD ASDf = 1 W = 1.5f R n = 435 kips R n / W = 290 kips




LRFD ASDf = 0.75 W = 2f R n = 437 kips R n / W = 292 kips


6/32

007

ures no slipures no bolt shear failure

C5

P s = 0.790 P s,equiv


7/32


Given: A WT is bolted to a W section as part of a beam/brace connection as shown.The connection is subjected to axial tension, P. Bolt size, type, location, etc. areall given on the drawing. Assume that the load transfer consists of only Wind Load (100%).

db = 0.75 in% W 100%

bolts:3/4" diaA325-Npretensionedin standard holes


Wanted: Determine the capacity of the connection based on bolt strength as specified below.Consider both LRFD and ASD.Express your results in terms of comparable service level loads.


Solution:


Allowable Brace Force (at comparable service load levels)

Part (a) Slip CriticalLRFD ASD

(k) (k)Slip 41.1 43.9


8/32

P t = 0.6 P braceP v = 0.8 P brace

ASCE 7 Load Combinations:

The controlling load case is LRFD LC4 The controlling load case is ASD

Composite load factor for LRFD: P u = 1.60 P s,equiv Composite load factor for ASD:

a. Slip Critical Connection

Tensile Rupture Limit State - Section J3.6

F nt = 90 ksi (Table J3.2) Ab = 0.4418 in

r n = 39.8 k/boltNb = 6 boltsR n = 239 k/connection

LRFD ASDf = 0.75 W = 2f R n = 179 kips R n / W = 119 kipsP vs,eq < 112 kips P vs,eq < 119 kipsP brace < 186 kips P brace < 199 kips

Slip Limit State - Section J3.8 and J3.9

m 0.35 (Class B Surfaces)

Du 1.13h sc 1 (Standard Holes)Tb 28 kips/boltNs 1 shear planes/boltr n 11.1 kips/boltNb 6 boltsR n 66.4 kips/connection

LRFD ASD

f = 1 W = 1.5

W

W

WW

bbu

n

n

brace

u

n s

nbracevs

N T D R

R P

D Rk R P P

90.080.0

.05.1180.0

bbu

n

nbrace

bbu

bracen snbracevu

N T D R

R P

N T D P Rk R P P

f

f

f f

96.028.1

96.0128.1


9/32

f R n = 66 kips R n / W = 44 kipsP brace = 41.1 kips P brace = 43.9 kips

b. Bearing Connection

Shear Rupture Limit State - Section J3.6



LRFD ASDf = 0.75 W = 2

f R n = 95.4 kips R n / W = 63.6 kipsPv,seq = 59.6 kips Pv,seq = 63.6 kipsP brace = 74.6 kips P brace = 79.5 kips

Tension Rupture Limit State - Sections J3.6 & J3.7

F nt = 90 ksi (Table J3.2) Ab = 0.4418 in

r n = 39.8 k/boltNb = 6 bolts

R n = 239 k/connection

LRFD ASDf = 0.75 W = 2

Without J3.9 reduction: Without J3.9 reduction:f R n = 179 kips R n / W = 119 kipsP ts,eq < 112 kips P ts,eq < 119 kipsP brace < 186 kips P brace < 199 kips

With req'd J3.7 reduction: With req'd J3.7 reduction:

nv

bnt nb

bb

b

nv

nt nt bbb

bbvnv

nt nt bbnt btu

F P F

R P

N A P

F F

F N A P

N A f F F F N A F P P

28.13.196.0

80.06.13.196.0

3.160.06.1

f

f f

f f f

bnv

nt nb

bnv

nt nt b

bbnt

bts

P F F R

P

A F F

F P

N A F P P

80.03.160.0

80.03.160.0

.160.0

W

W

W


10/32

P brace < 69.2 kips P brace < 73.8 kips

nv

nt

nb

F F

R P

28.196.0

3.1 f

nv

nt

n

b

F F

R

P 80.060.0

3.1W


11/32

007


12/32

C4

P s = 1.00 P s,equiv

bb

brace

N P 6


13/32

bb

b

b

bbv

nv

nt nt

N A P

N A f F F F 3

W

WW


14/32


15/32

A Beginner's Guide to the Steel Construction Manual last modified: 24 SeptExample Problem 4.4a by: TBQ

Given: A bracket is connected to a column flange as shown. Assume that the applied load is 50% D and 50% L.The bolt information is given on the drawing.

db = 0.75 in% D 50%% L 50%




a. Determine the bearing capacity of the connection using the elastic method.b. Determine the bearing capacity of the connection using the IC method.

Solution (a):


Bearing Capacity - Elastic Method

LRFD ASD(k) (k)

P s,eq < 32.9 30.7



Composite load factor for LRFD: P u = 1.40 P s,eq Composite load factor for ASD:




16/32

r n 26.5 kips/shear planeNs 1 shear planesr n 26.5 kips/bolt

LRFD ASDf = 0.75 W = 2f R n = 19.9 kips R n / W = 13.3 kips

Compute the coefficient that relates applied load to maximum bolt force:

Let P = 10 k P x = 5.000 kd = 30 deg P y = -8.660 ke = 7.75 in M = Pe = 77.5 in-k

Compute bolt geometry quantities:

Bolt cx cy c Ix Iy

(in) (in) (in) (in4) (in4)1 -1.75 4.5 4.83 8.95 1.352 1.75 4.5 4.83 8.95 1.353 -1.75 1.5 2.30 0.99 1.354 1.75 1.5 2.30 0.99 1.355 -1.75 -1.5 2.30 0.99 1.356 1.75 -1.5 2.30 0.99 1.357 -1.75 -4.5 4.83 8.95 1.358 1.75 -4.5 4.83 8.95 1.35

39.76 10.82

Ip = 50.58 in

Compute the bolt forces:

Bolt r px r py r mx r my r x r y r (k) (k) (k) (k) (k) (k) (k)

1 -0.625 1.083 -3.046 -1.184 -3.671 -0.102 3.6722 -0.625 1.083 -3.046 1.184 -3.671 2.267 4.3143 -0.625 1.083 -1.015 -1.184 -1.640 -0.102 1.6434 -0.625 1.083 -1.015 1.184 -1.640 2.267 2.7985 -0.625 1.083 1.015 -1.184 0.390 -0.102 0.4036 -0.625 1.083 1.015 1.184 0.390 2.267 2.3007 -0.625 1.083 3.046 -1.184 2.421 -0.102 2.423

8 -0.625 1.083 3.046 1.184 2.421 2.267 3.317-5.000 8.660 r max = 4.314

C = P/r max = 2.318

Using the coefficent, we can now compute the capacities using P = C*r max :

LRFD ASD


17/32

r max = f R n 19.9 k/bolt r max = R n/W = 13.3P u < 46.1 kips P a < 30.7P s,eq < 32.9 kips P s,eq < 30.7


18/32

2007

C2

P s = 1.00 P s,eq


19/32


20/32

k/bolt

kips

kips


21/32

A Beginner's Guide to the Steel Construction Manual last modified: 24 SeptExample Problem 4.4b by: TBQ

Given: A bracket is connected to a column flange as shown. Assume that the applied load is 50% D and 50% L.The bolt information is given on the drawing.

db = 0.75 in% D 50%% L 50%




a. Determine the bearing capacity of the connection using the elastic method.b. Determine the bearing capacity of the connection using the IC method.

Solution (b):

Results Summary: Computations follow LRFD(k)

Bearing Capacity - IC Method Elastic P s,eq < 32.9IC P s,eq < 36.3

LRFD ASD(k) (k)

P s,eq < 36.3 33.8



Composite load factor for LRFD: P u = 1.40 P s,eq Composite load factor for ASD:



r n 26.5 kips/shear plane


22/32

Ns 1 shear planesr n 26.5 kips/bolt

LRFD ASDf = 0.75 W = 2f R n = 19.9 kips R n / W = 13.3 kips

Compute the coefficient that relates applied load to maximum bolt force:

d = 30 deg Dmax = 0.340 inR ult = 74.0 k

Locate the IC relative to the bolt CGx y

(in) (in) Computations for roIC -1.41 -1.14 atan( d) = 0.4823

CG 0.00 0.00 r ox = 10.22 inLoad 7.75 6.00 r oy = 4.93 in

These values found by computingr o = 11.35 in the intersection of two line whose

slope we know and a point that eaCompute bolt geometry and deformation quantities: passes through.

Bolt c xcg cycg cxic cyic c ic D(in) (in) (in) (in) (in) (in)

1 -1.75 4.5 -0.341 5.636 5.65 0.302 1.75 4.5 3.159 5.636 6.46 0.343 -1.75 1.5 -0.341 2.636 2.66 0.144 1.75 1.5 3.159 2.636 4.11 0.225 -1.75 -1.5 -0.341 -0.364 0.50 0.036 1.75 -1.5 3.159 -0.364 3.18 0.177 -1.75 -4.5 -0.341 -3.364 3.38 0.188 1.75 -4.5 3.159 -3.364 4.62 0.24

max = 6.46 0.34

Compute the bolt forces:

Bolt r r x r y r*cic(k) (k) (k) (in-k)

1 71.9 -71.8 -4.3 406

2 72.6 -63.4 35.5 4693 63.3 -62.8 -8.1 1684 69.2 -44.3 53.1 2855 33.0 24.1 -22.6 166 66.0 7.6 65.6 2107 66.8 66.5 -6.7 2268 70.3 51.3 48.1 325

sums = -92.7 160.6 2105P = 185.48 185.48 185.48


23/32

Differences 0.00 0.00 0.00 Good

r max = 72.6

C = P/r max = 2.554

Using the coefficent, we can now compute the capacities using P = C*r max :

LRFD ASDr max = f R n 19.9 k/bolt r max = R n/W = 13.3P u < 50.8 kips P a < 33.8P s,eq < 36.3 kips P s,eq < 33.8


24/32

2007

ASD(k)

30.733.8

C2

P s = 1.00 P s,eq


25/32

-1.41 -1.140.00 0.007.75 6.00

-1.75 4.501.75 4.50

-1.75 1.50

ch1.75 1.50

-1.75 -1.501.75 -1.50

-1.75 -4.501.75 -4.50

-6.00

-4.00

-2.00

0.00

2.00

4.00

6.00

8.00

-4.00 -2.00 0.00 2.00 4.00 6.00 8.00 10.00

Series1


26/32

k/bolt

kips

kips


27/32

A Beginner's Guide to the Steel Construction Manual last modified: 18 JanExample Problem 4.5 by: TBQ

Given: The following data:

Eccentric Load

DL 50% P s,equivLL 50% P s,equiv

Eccentricit 7 in

Fy = 50 ksiFu = 65 ksidb = 0.875 in

Ab = 0.6013 in^2Tb 39 kips

Fnv = 48 ksi

Fnt = 90 ksi

Number of bolts, n b 8 bolts:number of rows 4 7/8" diaRow spacing 3 in A325-N

pretensionedbf tf in standard holes

Column: W14x61 10 0.645 inBracket: WT6x13 6.49 0.38 in Class A faying surfaces

Wanted: Determine the capacity of the connection based on bolt strength as specified below. ConsidExpress your results in terms of comparable service level loads.

part (a) use the SCM Case I methodpart (b) use the SCM Case II Method

Solution:

Solution Summary LRFD ASD(kips) (kips)

a Case IShear limit 123.70 115.45Tension limit 77.35 48.75

Controlling Value 77.35 48.75


28/32

part (a)

Using SCM Case I (SCM page 7-10):

Shear Strength of Bolts (J3.6)

F nv 48 ksi Ab 0.6013 in^2r n = F nv Ab = R n / n b 28.9 k/shear planeShear Planes 1r n 28.9 k/bolt

LRFD ASDr uv < r nuv = f r n r av < r nav = r n / Wvf v = 0.75 Wv =

f v r n = 21.6 kips/bolt r n / Wv =

P u,max < f v r n nb = 173.18 kips/connection P a,max = n b r n / Wv =CLF 1.40 CLFP s,eq < 123.70 kips/connection P s,eq

bgscmc04exprob

Documents