• Author / Uploaded
• Frances Nicole Quito

Citation preview

STEEL AND TIMBER DESIGN CE-506 SUBMITTED BY: ESGUERRA, ALFA ROMEO C. QUITO, FRANCES NICOLE SUBMITTED TO: ENGR. BRYAN DALE P. YU, MSCE BEAM-COLUMN 1.

A steel column W 300 mm 203 kg/m section is subjected to an axial load of 2670 kN and moments from framing beams as follows: Mx at the top=108 kN.m. Mx at the bottom= -108 kN.m. Find the critical buckling load (kN)

Prop. Of the WF Column A= 2570 mm2 tr= 32 mm d= 340 mm Ix= 5.16x108 MPa br= 315 mm Iy= 1.65x108 MPa tw= 20 mm E= 200 GPa Fy= 345.6 MPa Unbraced column length= 3m

Effective length factor, K=1.0 Sidesway is prevented

P α=

π 2 EI ( KL)2

P α=

π 2 ( 200000 ) (1.65 x 108 ) (1 x 3000)2

Pα = 36188550 N Pα = 36189 kN

2.

A W12 x 106 beam-column connection shown is pin-connected at both ends. It is subjected to an axial load of 1338 kN and a mment of 122 kN.m. at the top of a moment of 163 kN.m. at the bottom. Bending is about the strong axis. Sidesway is not prevented and K=1.18. Use A36 steel with Fy= 248 MPa. Unsupported height of the column is 4.27 m. Determine the allowable bending stress about the strong axis if only bending is acting. Prop. Of W12 x 106 A= 20129 mm2 tf= 25.15 mm d= 327.41 mm Ix= 388x106 mm4 bf= 310.39 mm Iy= 125x106 mm4 tw= 15.49 mm Sx= 2376x103 mm3 rx=138.94 mm Sy= 808x103 mm4 ry=78.99 mm L b

=

4270 mm L c=

200 b f √ Fy

L c=

200(310.39) √ 248

Lc= 3942 137900 L u= Fyd bftf L u=

137900 248 (327.41) ( 310.39 ) ( 25.15)

Lu=13258 mm Lb>Lc Lb57.23

tf= 14.6 mm Ix= 177x106 mm4 Iy= 39.9x106 mm4 Sx= 1160x103 mm3 Sy= 314x103 mm3

(0.455)2 250 Fa= 1− 2 1.83

[

]

Fa=122.47 MPa allowable compressive stress

4.

A W310 x 58 beam-column carries an axial load of 1338 kN and a moment of 122 kN.m at the top and 163 kN.m. at the bottom causing a reverse curvature. Bending is about the strong axis. Sidesway is not prevented by adequate bracings. Use K=1.18 and A36 steel Fy=250 MPa. Length of column is 4.2 m. Determine the allowable compressive strength if only axial load is acting. Prop. Of W250x58 A= 20100 mm2 d= 327 mm bf= 310 mm tw= 15.5 mm rx=139 mm ry=78.9 mm

tf= 25.1 mm Ix= 386x106 mm4 Iy= 125x106 mm4 Sx= 2360x103 mm3 Sy= 805x103 mm3

Cc=

√

2 π 2( 200,000) 250

Cc=125.66 L/r 62.81 = =0.50 Cc 125.66 L 2 ( ) r Fy Fa= 1 − 2 2 Cc FS

[ ]

L L 3 3( ) ( ) 5 r r FS= + − 3 8Cc 8 Cc3 3

5 3(0.50) (0.50) FS= + − 3 8 8 FS=1.84 (0.50)2 250 Fa= 1− 2 1.84

[

]

Fa=118.89 MPa allowable compressive stress

KL 1.18(4200) = =36.65 rx 139 KL 1.18(4200) = =62.81 ry 78.9 L Use =62.81 r

5.

A W14x90 column in a building frame for a 7m. story height supports a 1160 kN axial compression load and a moment at the top Mx1=+136 kN.m. and a moment of Mx2=+163 kN.m. at the bottom ends as shown. The column is braced against sidesway in the weak bending plane of the column (x-z plane) but with possible sidesway in the strong bending plane (yz plane). Use A36 steel. Calculate the allowable bending stress

L c=

200(368.81) √ 250

Lc= 4665 137900 L u= Fyd bftf L u=

137900 250(356.11) ( 368.81 ) (18.03)

Lu=10300 mm Lb>Lc Lb58.09 KL/r 58.09 = =0.547 Cc 106.21 KL 2 ) r Fy Fa= 1 − 2 2Cc FS

[ ] (

KL KL 3 3( ) ( ) 5 r r FS= + − 3 8 Cc 8 Cc3 Prop. Of W360x134 d=369 kk bf=369 mm tf=28 mm tw=11.2 mm ry= 94 mm

3

2

A=17100 mm Sx=2330x103 mm3 Sy=817x103 mm3 rx=156 mm

Determine the allowable bending stress if only bending moment alone existed. Fb=0.75Fy (along weak axis) Fb=0.75(350) Fb=263.50 MPa Determine the allowable bending stress if only axial load is acting. KL 1.3(4200) = =35 rx 156

5 3(0.547) (0.547) FS= + − 3 8 8 FS=1.85

[

Fa= 1−

(0.547)2 350 2 1.85

]

Fa=160.89 MPa allowable compressive stress

1.

BOLTED CONNECTIONS An eccentrically loaded bolted connection is shown. Diameter of A325 bolts in standard holes is 20 mm. The applied load is 100 kN. Determine the polar moment of inertia of the group of bolts.

2. A lapped, bolted tension member is shown. Diameter of bolts are 18 mm ø and the plate material is A36 steel Fy=250 MPa., Fu=400 MPa. Assume the fasteners are adequate and do not control the tensile capacity. Diameter of hole is 3mm bigger than the diameter of the bolt.

a. Determine the tensile capacity of the lapped joint based on gross area. P=0.60FyAg Ag=Width x thickness P=0.60 (250) (300) (12) P=540kN b. Determine the tensile capacity of the lapped joint based on net area. P=0.50FuAn An= 300(12)-3(21)(12) An= 2844 mm2 P=0.50(400)(2844) P=568.8 kN ∑(x2+y2) =6(75)2+4(100)2 ∑(x2+y2) =73750 mm2

3.

A double shear butt connection is connected by 9-22mm dia. Bolts Thickness of the outer plate is 12.5 mm while the main plate is 22 mm. Spacing of bolts is 75 mm with an edge distance of 37.5 mm. A36 steel is used. Fy=248 MPa, Fu= 400 MPa. Allowable shear stress of bolts= 117 MPa Allowable bearing stress of bolts= 1.2 Fu Dia. Of hole= 3mm greater than bolt diameter.

a.

4. An eccentrically loaded connection is subjected to an axial vertical load P acting at an eccentricity “e”. The bracket is a structural tee having a web thickness of 6 mm. A325 high strength bolts is connected in two vertical rows and has a diameter of 20 mm. The bolts are in standard holes having a diameter of 22 mm ø and are spaced at 90 mm vertically. Assume that the column and brackets are adequate. Allowable shear stress of bolts= 145 MPa

Determine the capacity of the connection based on the strength of the bolts SHEAR OF BOLTS

a.

Compute the value of P based on the allowable shear stress. fv= P/A T=AvFv π T = ( 22 )2( 2)¿ 4 T= 800559 N BEARING OF BOLTS T= AbFp Fp=1.2Fu T= 22(22)(9)(1.2)(400) T=2090880 N Use T= 800.56 kN

fv=

P

π ( 20 )2 (8) 4 P=328.89 kN

5.

Two plates each with thickness t=18mm are bolted together with 6 – 22 mm ø bolts forming a lap connection. Bolts spacing are as follows: S1= 40 mm, S2= 80 mm, S3=100 mm. Bolt hole diameter= 25 mm. Allowable stress: Shear stress of the bolt: Fv=100 MPa Bearing stress of the bolt= Fp=1.2Fu

a.

Calculate permissible tensile load P based on shear capacity of bolts π P= ( 22 )2 (2)(6)(100) 4 P= 228.08 kN

b.

Calculate permissible tensile load P based on bearing capacity of bolts P= AbFp P= 18(22)(6)(1.2)(400) P=1140.48 kN

6. The bracket shown is a structural tee. It is subjected to an eccentric load of 112 kN at an eccentricity of 200 mm. Diameter of A325 N high strength bolts is 19 mm. The bolts are placed in two vertical rows and are spaced 75 mm vertically. Assume the column and brackets are adequate. Allowable shear stress is 145 MPa.

Determine the max. tensile stress of the bolt. I=Ad2 π 2 2 2 I¿ ( 19 ) [ 4(112.5) + 4(37.5) ] 4 I=15.95x106 M=112000(200) M=22.4 x106 N.mm

Mc I 22.4 x 106 (112.5) f t= 15.95 x 10 6 ft =158 MPa WELDED CONNECTIONS Two plates 300 mm x 16mm is to be connected using butt weld connections. Use A36 steel. Fy=250 MPa, Fu= 400 MPa and electrodes E70 with Fu=485 MPa. f t=

1.

a. Determine the capacity of the plate connection assuming the butt weld is adequate. Use a reduction factor of 0.85 for the Net Area.

Based on gross area: T=0.60FyAg T=0.60(250)(300)(16) T=720 000 N Based on net area: T= 0.50FuAe T=0.50(400)(0.85)(300)(16) T=816 000 N T=720 kN 2.

A welded bracket shown is required to assist a load of 90 kN. The steel is A36 (Fy= 250 MPa) and the welding is to be

R=√ (470.55)2 +(352.91)2 R=588.19 N/mm

performed by using E70 electrodes (SMAW) Fu= 485 MPa. J=10.4 x 106 mm4 Determine the force on the weld due to torsional moment 3.

A plate used as a tension member is connected to a gusset plate, as shown in Figure 7.40. The welds are 3/16 – inch fillet welds made with E70XX electrodes. The connected parts are A36 steel/ Assume that the tensile strength of the member is adequate, and determine the availability strength of the welded connection.

37.5+e=150+250 e=362.50

ɵ=0°

M=90000(362.5)

Fnw = 0.60FEXX

M=32.625x106N.mm My J 32.625 x 106 (112.50) f x= 10.4 x 106 fx=470.55 N/mm f x=

Rn= 0.707wFnw = 0.707(3/16)(0.6 x 70) = 5.568 kips/in.

The allowable strength of the weld is Rn/Ω = 5.568 / 2.00 = 2.784 kips/in.

The shear yield strength is

Rn/Ω = 0.6Fyt / 1.50 = 0.6(36)(1/4) / 1.50 = 3.6 kips/in.

The shear rupture strength is Rn/Ω = 0.6Fut / 2.00 = 0.6(58)(1/4) / 2.00 = 4.35 kips/in.

The weld shear strength controls. For the connection, Rn/Ω = 2.784 kips/in. x (4 + 4)in = 22.3kips Allowable strength of the weld is 22.3 kips

4.

A double fillet lap joint is shown in the figure. The steel is A36 steel Fy=250 MPa and the electrode used was E70 (Fu=480 MPa) The fillet weld is 8 mm and the shielded metal arc welding (SMAW) as used.

x=47.5 mm 5. The 150mm x 18 mm plate is welded to a 300 mm x 18 mm plate by a 10 mm fillet weld. The steel is A36 and the electrode use was E70 (SMAW) shielded metal arc welding. For welding: Fu=483.8 MPa For steel plate: Fy= 248 MPa, Fu= 400 MPa

a. Determine the allowable tensile load that may be applied to the connection based on shear capacity of the fillet weld. T=0.707tL0.30Fu T=0.707(8)(200+200)(0.3)(480) T=325785.6 N T=325.8 kN b. Determine the allowable tensile load that may be applied to the connection based on gross area of the plate T= 0.60FyAg T=0.60(250)(200)(9.5) T=285 000 N T= 285 kN c. Determine the minimum length of the lap x= 5(9.5) (5x the thinner part joined but not less than 25 mm)

a. Determine the tensile load of the welded lap joint based on shear capacity of fillet T=0.707tL0.30Fu T=0.707(10)(125+125)(0.3)(483.8) T=410 456 N T=410.46 kN a. Determine the tensile load of the welded lap joint based on gross area of plate. T= 0.60FyAg T=0.60(248)(150)(18 ) T=401 760 N T= 401.8 kN

6.

The bracket carrying a load P is 14 mm thick. It is welded. Allowable weld stress, Fv= 145 MPa, weld thickness=9.5mm

a.

If P =200 kN and it acs at the centroidal axis of the weld group, calculate the force per unit length (kN/m) on the weld.

F=P/L F=200/ (0.125(2)+(0.25)) F= 400 kN/m

SHEAR IN BEAMS

1.

A W 420 x 85 steel is fully restrained with a uniformly distributed superimposed load of 25 kN/m. The beam has a span of 10 m. Properties of W420x85 A = 10839 mm2 d = 420 mm tw = 11 mm Ix = 310 x 10⁶ mm⁴ tf = 18 mm bf = 180 mm a. Compute the bending stress in MPa. Bending stress in MPa. 85(9.81) 1000 W = 25.834 KN/m W = 25 +

W L2 12 25.834(10)2 Ma = 12 Ma = 215.28 KN.m Ma =

fb =

MC I

215.28 x 106 ( 420) 2(310)106 fb = 145.83 MPa fb =

b. Compute the maximum web shear stress in MPa. V dtw 2R = 25.834(6) R = 77.50 KN V = 77.50 KN V fv = dtw 77500 fv = 420(11) fv = 16.77 MPa fv =

2.

A W 310 x 74 of A36 steel Fy=250 MPa is on a simple span of 9m. Assume full lateral support. The beam supports a uniformly distributed load of 3.6 kN/m

Determine the maximum concentrated load that can be placed at the mid-span so that it will not exceed the allowable shearing stress fv= 0.40Fy fv =

V dtw

0.40Fy =

V dtw

0.40(250)=

V 305(17.4)

V=530.7 kN P WL V= + 2 2 P 3.6(9) 530.7= + 2 2 P=1029 kN

3. A fixed ended WF beam has a span of 6m

it carries a load of 135 kN acting at 2m from B. I=91 x106 mm4, Vmax= 100 kN Prop. Of WF section d = 240 mm bf = 160 mm tw = 20 mm tf = 20 mm a. Determine the max. vertical shearing stress of the beam Q=160(20)(110)+100(20)(50) Q=452000 mm3 VQ Max. τ = Ib Max. τ = 100000(452000) 91 x 10 2( 20) Max. τ= 24.84 MPa

b. Ave. shearing stress in the web Q=160(20)(110) Q=352000 VQ τ= Ib 100000 ( 352000 ) τ= 91 x 10 2 (20 ) τ=19.34 MPa τave=19.34 + 2/3 (5.5)

1457 x 106 10958 x 103 f b=134.60 MPa

τave= 23 MPa

f b=

4. For the section shown, the shear force acting is 670 kN and the bending moment is 1475 kN-m

b. Compute the total shear carried by the web and the flanges.

F v= a. Compute the max. bending stress of the beam shown. I= + +

[ [

10 ( 1050 )3 12 3

300 ( 16 ) + 300 ( 16 )( 533 )2 2 12

] ]

250 ( 16 )3 + 250 ( 16 )( 549 )2 2 12

I =6103.5 x 106 I S= C 6103.5 x 106 S= 557 S=10958 x 103 f b=

M S

VQ Ib

For the flanges: Q=250 ( 16 )( 549 ) +300 ( 16 ) ( 533 ) Q=4754400 VQ Ib 4754400 ( 670000 ) F v= 6103.5 x 10 6 ( 10 ) F v =52.19 MPa F v=

Average stress: V d tw 670000 f v= 1114 ( 10 ) f v =60.14 MPa f v=

Shear force carried by the two flanges:

1.74 ( 2 ) [ 250 ( 16 ) +300 ( 16 ) ] 2 F f =15312 N Ff=

5.

A W 24 x 94 beam is subjected to a shear force of 900 KN.

900000(1.52 x 106 ) fv = 1119.7 x 106 (13.11)

Properties of W sections: A = 17870 mm2 tf = 22.15 mm bf=230.15mm tw = 13.11 mm d = 616.97 mm Ix=1119.7 x 10⁶ mm⁴

fv = 93.19 MPa b. Compute the shear stress at neutral axis. Q = 230.15(22.15)(297.41) + 286.335(13.11)(143.1675) Q = 2.05 x 10⁶ mm3 VQ fv = Ib 900000(2.05 x 106 ) fv = 1119.7 x 106 (13.11) fv = 125.69 MPa

a. Compute the shear stress at junctions of flange and web. VQ Ib Q = 230.15(22.15)(297.41) Q = 1.52 x 10⁶ mm3 fv =

c. Compute the shear force carried by the flange.

( 2 ) 5.31 (230.15)(22.15) 2 F = 27069 N F=

d. Compute the shear force carried by the web. F = 900000 – 27069 F = 872931 N e. Compute the average shear stress on web.

b = 230.15 VQ Ib b = 13.11 mm fv =

fv =

V dtw

fv =

900000 616.97(13.11)

fv = 111.27 MPa

6.

A wide flange section is subject to a

b. Determine the max shearing stress at

vertical of 356 kN. The moment of inertia

the neutral axis.

at the neutral axis 619 x 106. Fy= 250 MPa.

a.

Determine

the shearing stress in the flange.

Q = 265(25)(203) + 190.5(14.9)(95.25) Q = 1615237.4mm3 F v=

VQ IB

F v=

356 000(1615237.4) 619 x 10 6 (14.9)

Fv= 62.35 MPa

F v=

VQ IB

Q = 265(25)(203) Q = 1344875 mm3 F v=

VQ IB

F v=

356000(1344875) 619 x 106 (265)

Fv= 2.92MPa