Design-coaching-2.pdf

STRUCTURAL DESIGN – LECTURE

PROBLEM 6: A hanging storage bin weighing 260 kN will be supported by three circular threaded rods. Using A36 steel, Fy = 250 MPa, Fu = 400 MPa. ➀ ➁ ➂

Determine the required rod diameter of unthreaded rod. Determine the diameter of the rod before upsetting if an upset rod is used. Determine the required diameter of the upset end.

➀

Solution: Required rod diameter of unthreaded rod T = 0.33 Fu AD

260 = 86.67 kN 3 ⎛ π⎞ 86670 = 0.33(400) ⎜ ⎟ D2 ⎝ 4⎠ T=

D = 28.9 say 30 mm ➁

Diameter of the rod before upsetting if an upset rod is used T = 0.60 Fy AD

⎛ π⎞ 86670 = 0.60(250) ⎜ ⎟ D2 ⎝ 4⎠ D = 27.12 say 28 mm ø ➂

Required diameter of the upset end π AD = (25)2 4 AD = 615.75 mm2

T = 0.60(250)(615.75) T = 92362.5 N Dia. of upset end: T = 0.33 Fu AD ⎛ π⎞ 92362.5 = 0.33(400) ⎜ ⎟ D2 ⎝ 4⎠ D = 29.84 say 30 mm

28 mm

30 mm

STRUCTURAL DESIGN – LECTURE

PROBLEM: Light gage cold-formed steel channels used as purlins are simply supported on roof trusses 6 m. apart. Given: Roof slope . . . . . . . . . . . . . . . . . . . 1V:4H Dead load (purlin wt. included) D = 720 Pa Live load, L = 1000 Pa Wind pressure, W = 1440 Pa Wind pressure coefficient: At windward side, pressure = 0.18 At leeward side, suction = 0.60 Properties of C Purlins: C 200 mm x 76 mm Sx = 6.19 x 104 mm3 Sy = 1.38 x 104 mm3

fbx = fby = 207 MPa

Assume that all loads pass through the centroid of the C section. For D + L + W, a one-third increase in allowable stresses is allowed. ➀ What is the safe purlin spacing (m) due to D + L if sag rods are not used? ➁ What is the max. spacing of purlins (m) due to D + L + W at the leeward side? ➂ How much is the safe spacing of purlins (m) due to D + L + W at the windward side? S

SOLUTION: ➀ Spacing due to D + L if sag rods are not used DL = 720S LL = 1000S W = 1440S Windward pressure = 1440S(0.18) = 259.20S Leeward pressure = 1440S(0.60) = 864S 1 tan θ = 4 θ = 14.04˚

θ

1 4

y

θ DL + LL=1720S

Bending along x - axis : Wx = 1720S Cos 14.04˚ Wx = 1668.62S 2

Mx =

Wx L

8 1668.62S(6)2 Mx = 8 M x = 7508.78S N.m.

x y Wx Wy

x

STRUCTURAL DESIGN – LECTURE

PROBLEM:

fb x =

(CONTINUATION)

Mx Sx

=

7508.78S(1000) = 121.31S 6.19 x 10 4

Bending along the y - axis : Wy = 1720S Sin 14.04˚ Wy = 417.27S My =

Wy L2

(no sag rods)

8 417.27S(6)2 My = 8 M y = 1877.72S N.m

fb y =

Mx =

8 Wy L2

Sy

1877.72S(1000) 1.38 x 10 4 fb y = 136.07S fb y =

fb x Fb x

+

fb y

= 1.0

Fb y

121.31S 136.07S + =1 207 207 S = 0.80 m.

➁ Spacing of purlins due to D + L + W at the leeward side Wy = 1720S Sin θ Wy = 417.27S Wx = 1720S Cos 14.04 – 864S Wx = 804.62S

Wx L2

My

DL+LL=1720S

θ 864S

804.62S(6)2 (1000) = = 362 x 10 4 S 8

S

417.27S(6)2 (1000) My = = = 187.8 x 10 4 S 8 8 4 M 362 x 10 S fb x = x = = 58.48S Sx 6.19 x 10 4 My

187.8 x 10 4 S fb y = = = 136.09S Sy 1.38 x 10 4 fb x Fb x

+

fb y Fb y

= 1.0

58.48S 136.09S + = 1.0 ⎛ 4⎞ ⎛ 4⎞ ⎜⎝ 3 ⎟⎠ (207) ⎜⎝ 3 ⎟⎠ (207) S = 1.40 m.

y y

864S

Wy

Wx x

Wy x

1720S

STRUCTURAL DESIGN – LECTURE

PROBLEM: (CONTINUATION) ➂ Spacing of purlins due to D + L + W at the windward side D + L = 1720S WL = 259.20S Wx = 259.20S + 1720S Cos 14.04 Wx = 1927.82S Wx L2 Mx = 8 1927.82S(6)2 (1000) Mx = N.mm 8 M x = 867.52S x 10 4

y 259.20S θ

Mx

867.52S(10)4 fb x = = = 140.15S Sx 6.19 x 10 4 Wy = 1720S Sin 14.04˚ Wy = 417.27S My =

Wy L2 8 My

417.27S(6)2 (1000) = = 187.8 x 10 4 S 8

187.8 x 10 4 S fb y = = = 136.09S Sy 1.38 x 10 4 fb x Fb x

+

fb y Fb y

= 1.0

140.15S 136.09S + = 1.0 ⎛ 4⎞ ⎛ 4⎞ ⎜⎝ 3 ⎟⎠ (206) ⎜⎝ 3 ⎟⎠ (206) S = 0.9943 Use S = 1.0 m.

1720S

y Wx

x

Wy

x

STRUCTURAL DESIGN – LECTURE

PROBLEM: From the given figure, beam EFGH and IJKL are 3-span beams supported by girders AM, BN, CO and DP. Panel BCONB only is subjected to live load of 4.8 kPa. Assume EI is constant. Due to given live loads only. A B C D ➀ Find the positive moment at span FG in kN.m. ➁ What is the maximum shear of span EF in kN? 2.5 ➂ How much is the maximum moment at span EF in kN.m.? Solution: ➀ Positive moment at span FG in kN.m. 3 MFG = wL2 40 3 MFG = (12)(6)2 40 MFG = 32.4 kN.m.

E

F

G

H

I

J

K

L

2.5

2.5 M

N

6m

O

6m

P

6m

➁ Max. shear at span EF.

wL 20 12(6) = 20 = 3.6 kN

Vmax = Vmax Vmax

M=-21.6 kN.m E

F

6m

R 12 kN/m

R

E

H

G

F 6m

➂ Max. moment at span EF wL2 M max = 20 12(6)2 M max = 20 M max = - 21.6 kN

R

6m

6m

R1

R1

36 kN -3.6 kN

-3.6 kN -36 kN +32.4 kN.m

-21.6

-21.6

STRUCTURAL DESIGN – LECTURE

PROBLEM 4: DL = 8 kN/m LL = 4 kN/m Prop. of W section: d = 350 mm I = 1.6 x 108 tw = 8 mm Fy = 248 MPa

wu

A

x = 3.33

C Assumeto be simply supported

4m

➀ Compute the max. bending stress if full length of the beam is loaded. ➁ Compute the max. bending stress considering possible live load pattern. ➂ Compute the max. shear stress of the beam. Solution: ➀ Max. bending stress if the full length of the beam is loaded 24(2) 40(3.33) ∑M C = 0 Max. M = + 2 2 6 R1 = 12(8)(4) Max. M = 42.66 kN.m. R1 = 64 kN MC fb = R 2 = 12(8) - 64 I R 2 = 32 kN 42.66 x 10 6 (350 / 2) fb = 40 - 12x = 0 1.6(10)8

B

D 2m

6m

wu= 8 + 4 = 12

A

B

C

2m

6m R1

R2

40 2 x -24 Max. M

-32

fb = 44.97 MPa M1

➁ Max. bending stress considering live load pattern

∑M C = 0 6 R1 = 8(8)(4) + 4(6)(3) R1 = 54.67 R 2 = 8(8) + 4(6) - 54.67 R 2 = 33.33 kN 38.67 - 12x = 0 x = 3.22

16(2) 38.67(3.22) Max. M = + 2 2 Max. M = 46.26 kN.m. MC fb = I 46.26(350)(10)6 fb = 1.6 x 10 8 (2) fb = 48.77 MPa

➂ Max. shear stress

Fv =

V dt w

V = 40 kN

Fv =

40000 350(8)

Fv = 14.29 MPa

LL=4 DL=8 A

B

C

2m R1=84.67

6m R2

38.67 2 x -16 Max. M

-33.33

STRUCTURAL DESIGN – LECTURE

PROBLEM 2: Given: Superimposed DL = 3.2 kPa (includes floor finish, ceiling, fixtures, etc.) Live load = 3.6 kPa Beam b x h = 400 x 600 3 Concrete weighs = 24 kN/m Slab thickness = 100 mm Columns E and H are omitted such that the girder BEHK supports beam GHI at H and DEF at E. Assume E is hinged and D and F are fixed. ➀ Compute the ultimate uniform load of beam DEF. ➁ Compute the shearing stress of beam DEF if it has an effective depth of 530 mm. ➂ Compute the max. negative moment of beam DEF. ➃ Compute the max. positive moment of beam DEF.

3m

3m 3m

Solution: ➀ Ultimate uniform load Super-imposed DL = 3.2(3) = 9.6 kPa Slab = 0.10(3)(24) = 7.2 Beam = 0.4(0.6)(24) = 4.8 DL = 9.6 + 7.2 + 4.8 = 21.6 kN.m LL = 3.6(3) = 10.8 kN/m wu = 1.4 DL + 1.7 LL wu = 1.4(21.6) + 1.7(10.8) = 48.6 kN/m

A

B

C

D

E

F

G

H

I

J

K

L

7.5 m

MD

7.5 m

MF

w u =48.6 kN.m

D

7.5

E Hinged

R1

F

7.5

R

R1

L = 15

➁ Shearing stress of beam DEF 3 R = wL 8 Shearing stress : 3 R = (48.6)(15 V 8 υ= u øbd R = 273.375 kN 227810 2R1 + R = 48.6(15) υ= 0.75(400)(530) R1 = 227.81 υ = 1.43 MPa Vu = 227.81 ➂ Max. negative moment wL2 M =32 48.6(15)2 M =32 M = - 341.72 kN.m.

227.81 136.69

-136.69 -227.81

M=wL 2 /57

-M=-wL 2 /32

M=wL 2 /57

-M=-wL 2 /32

➃ Max. positive moment of beam DEF wL2 M =+ 57 48.6(15)2 M =+ = +192 kN.m. 57

STRUCTURAL DESIGN – LECTURE

PROBLEM 1: Given: Superimposed DL = 3.2 kPa Beam b x h = 400 x 600 LL = 3.6 kPa Slab thickness = 100 mm 3 Wt. of concrete = 24 kN/m The columns at E and H are deleted, thus girder BEHK supports beam DEF at E and beam GHI at H. Assume all spans are loaded. Based on tributary area of the beams. ➀ Calculate the ultimate uniformly distributed load at beam DEF. ➁ Calculate the shearing stress in beam DEF. Assume fixed support at D and F and it has an effective depth of 530 mm. ➂ Calculate the moment at E. ➃ Calculate the max. positive moment in span DE assuming fixed supports at D and F. Solution: ➀ Ultimate uniformly distributed load at beam DEF Super-imposed DL = 3.2(3) = 9.6 kN/m Slab = 0.10(3)(24) = 7.2 kN/m Beam = 0.4(0.5)(24) = 4.8 kN/m Total DL = 9.6 + 7.2 + 4.8 = 21.6 kN.m LL = 3.6(3) = 10.8 kN/m Ultimate load: wu = 1.4(21.6) + 1.7(10.8) = 48.6 kN/m ➁ Shear stress of beam DEF wL 48.6(15) R= = = 364.5 kN 2 2 2R1 + R = 48.6(15)

3m

3m 3m

A

B

C

D

E

F

G

H

I

J

K

L

7.5 m

7.5 m

MD

MF

wu=48.6 kN.m

D

7.5

F

7.5

E R

R1

R1

L = 15 182.25

182.25 3.75

3.75

3.75 3.75

-182.25

R1 = 182.25

-wL2/96

-182.25

-wL2/96

Vu = 182.25 kN υ=

Vu øbd

=

182250 = 1.15 MPa 0.75(400)(530)

➂ Moment at E wL2 M =48 wL2 48.6(15)2 ME = == 227.81 kN.m. 48 48

-wL2/48

-wL2/48

-wL2/48

➃ Max. positive moment at span DE wL2 +M = 96 48.6(15)2 +M = = 113.91 kN.m. 96

STRUCTURAL DESIGN – LECTURE

PROBLEM 23: Superimposed DL = 3.2 kPa Beam: b x h = 400 x 600 mm Live load (LL) = 3.6 kPa Slab thickness = 100 mm 3 Unit wt. of concrete = 24 kN/m A Beam DEF is simply supported at D, E, F. 2.5 m For 2 spans both loaded, the negative moment at the interior D support is wL2/8. 2.5 m For 1 span loaded, the negative moment at the interior support G 2 is wL /16. 2.5 m

For maximum stresses apply the following: 1. Pattering loading for live load. 2. Ultimate load combination: U = 1.4 DL + 1.7 LL

J

➀ What is the maximum moment at the interior support at E of beam DEF? ➁ What is the maximum reaction at the interior support at E? ➂ If the loads at ultimate conditions are as follows: Total DL (wu) = 24 kN/m LL (wu) = 12.2 kN/m Find the max. positive moment at span DE.

H

I

K

L 7.5 m

600

Beam

400

wL

L

L

L

1.25

L

0.375

R2=10/8wL

=01.250wL

R1=3/8wL

=0.375wL

R3=7/16wL

R2=10/16wL

0.4375 0.625

0.625 0.375 0.4375L

-0.375

-0.5625

+M=0.0957

-0.625 0.0703 M=0.0703wL2

F

For live load only

w1

=0.375wL

E

Slab

500

For DL and LL

R1=3/8wL

C

7.5 m

100

0.375

B

0.0703 M=0.0703wL2 -wL2/16=0.0625wL2

M=wL2/8=0.125wL2

R4=1/16wL

STRUCTURAL DESIGN – LECTURE

PROBLEM 23: Continuation ➀ Max. moment at the interior support at E of beam DEF Superimposed DL = 3.2(2.5) = 8 kN/m

Beam = 0.4(0.5(24) = 48 kN/m Slab = 0.10(2.5)(24) = 6 kN/m Total DL = 8 + 4.8 + 6 = 18.8 kN/m LL = 3.6(2.5) = 9 kN/m Max. moment at E for DL: (18.8)(7.5)2 MDL = 8 MDL = (18.8)(7.5)2 (0.125) MDL = 132.19 kN.m.

➁ Max. reaction at E of beam DEF Due to dead load: RE = 1.25 wL RE = 1.25(18.8)(7.5) RE = 176.25 kN Due to live load: RE = 1.25 wL (biggest value) RE = 0.625 wL RE = 1.25(9)(7.5) RE = 84.375 kN RE = 1.4 DL + 1.7 LL RE = 1.4(176.25) + 1.7(84.375) RE = 390 kN

Max. moment at E for LL: wL2 wL2 M= or 8 16 2 wL Use M = = 0.125wL2 8 9(7.5)2 ML = 8 ML = 9(7.5)2 (0.125) ML = 63.28 kN.m. M u = 1.4 MD + 1.7 ML M u = 1.4(132.19) + 1.7(63.28) M u = 293 kN.m.

➂ Max. positive moment at span DE Due to DL: (ultimate values) MDL = 0.0703 wL2 MDL = 0.0703(24)(7.5)2 MDL = 94.91 kN.m. Due to live load: MLL = 0.0957 wL2 MLL = 0.0957(12.2)(7.5)2 MLL = 65.67 kN.m. Total +M at span DE: +M = 65.67 + 94.91 +M = 160.58 kN.m.

STRUCTURAL DESIGN – LECTURE

PROBLEM 24: Longitudinal beams EFGH and IJKL are simply supported at E, F, G and H, and at I, J, K, L. Ultimate load: U = 1.2 DL + 1.0 LL Given: S = 2.4 m L1 = L2 = L3 = 7.5 m Total DL = 4.9 KPa LL = 3.6 KPa ➀ Calculate the max. reaction at K. ➁ What is the max. shear at span KL? ➂ Find the max. negative moment at K. A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

P

2.4 m

2.4 m

2.4 m

7.5 m

7.5 m

7.5 m

wu I

L

0.40wL

J

K

L

1.10wL

L

L

1.10wL

0.5wL

0.4wL

wu

0.40wL

I

L

+0.08wL2

0.20wL

0.383wL

0.583wL

K

L

L

0.45wL

-0.033wL

0.6wL

-0.4wL

-0.5wL

+0.025wL2

-0.10wL2

L

0.383wL

A -0.6wL

J

+0.08wL2

-0.10wL2

Dead and Live Load Pattern

B

C

-0.687wL

+0.0735wL2

0.033wL

-0.417wL

+0.0534wL2

-0.1167wL2

-0.0333wL2

Live Load Pattern

D

STRUCTURAL DESIGN – LECTURE

PROBLEM 24:

Continuation

SOLUTION: ➀ Max. reaction at K DL = 4.9(2.4) = 11.76 KN/m LL = 3.6(2.4) = 8.64 KN/m Due to dead load: RK = 1.10wL RK = 1.10(11.76)(7.5) RK = 97.02 Due to live load: (select the biggest value of the 3 reactions at K) RK = 0.45wL RK = 0.55wL RK = 1.10wL Use RK = 1.10wL RK = 1.10(8.64)(7.5) RK = 71.28 Max. reaction at K: Rmax = 7.2 DL + 1.0 LL Rmax = 1.2(97.02) + 1.0(71.28) Rmax = 187.70 KN ➁ Max. shear at span KL Due to dead load: VKL = 0.60wL VKL = 0.60(11.76)(7.5) VKL = 52.92 KN Due to live load: (select the biggest value) VKL = 0.60wL VKL = 0.033wL VKL = 0.55wL Use VKL = 0.60wL VKL = 0.60(8.64)(7.5) VKL = 38.88 KN Max. shear at span KL: Vmax = 1.2 DL + 1.0 LL Vmax = 1.2(52.92) + 1.0(38.88) Vmax = 102.38 KN

wu

wu J

I

L 0.45w L

K

L

L 0.55w L

L 0.55w L

0.45w L

0.55w L

0.45w L

-0.45w L

-0.55w L 0.1013w L2

0.1013w L2 -0.05w L2

Live Load Pattern

➂ Max. negative moment at K Due to dead load: MK = - 0.10wL2 MK = - 0.10(11.76)(7.5)2 MK = - 66.15 KN.m. Due to live load: (select the biggest value) MK = - 0.10wL2 MK = - 0.033wL2 MK = - 0.05wL2 Use MK = - 0.10(8.64)(7.5)2 MK = - 48.6 KN.m. Max. negative moment at K: MKmax = 1.2 DL + 1.0 LL MKmax = 1.2(66.15) + 1.0(48.6) MKmax = - 127.98 KN.m.