BFC 3142 – CHAPTER 2

Loading Arrangement on beam

1. Load from One way Slab

- Ly/Lx > 2

- Beam with one way slab will support half of the load from slab.

Load on beam,

AB = CD w = nLx/2 kN/m

Ly

AC = BD w = 0

Lx

n = uniformly distributed load from slab (kN/m2)

2. Load from Two way Slab

- Ly/Lx < 2

- Two methods can be used to calculate the load from two way slab to beam which is Estimation Method and Shear Force Coefficient from Table 3.15 BS 8110.

(i) Estimation Method

Load on beam,

AC = BD w = nLx/3 (kN/m)

AB = CD (kN/m)

(ii) Shear Force Coefficient from Table 3.15: BS 8110

Load on beam may be calculated from the following equations;

Shear Force at Long Span, Ly → w = Vsx = βvx.n.lx (kN/m)

1

A

B D

C

A C

DB

Ly

Lx

BFC 3142 – CHAPTER 2

Shear Force at Short Span, Lx → w = Vsy = βvy.n.lx (kN/m)

Where; βvx, βvy = shear force coefficient from Table 3.15. n = design load (kN/m2)

lx = short span (m)

- There are 9 cases of slab edge continuity that may be exist for shear force coefficient as shown in figure below.

2

BFC 3142 – CHAPTER 2

EXERCISE 2.2

Figure 2.2 shows the plan view of first floor office building. If the slab thickness is 150 mm for all slab. Calculate the dead and imposed load of beam C/1-3. Use beam size 200 mm x 450 mm.

Solution:

Load on SlabSlab Self-weight = 0.150 x 24 = 3.6 kN/m2

Slab S1Dead load, Gk = 3.6 kN/m2

Imposed load, Qk = 2.0 kN/m2

Slab S2Dead Load, Gk = 3.6 kN/m2 Imposed Load, Qk = 2.5 kN/m2

Load distribution from slab

Slab 1 Ly/Lx = 3500/2850 = 1.23 < 2.0 : Two way slabSlab Case 8: Three edges discontinuous (one short edge continuos), βvx = 0.37Vsx = βvx. n. LxGk = 0.37 x 3.6 x 2.85 = 3.8 kN/mQk = 0.37 x 2.0 x 2.85 = 2.1 kN/m

Slab 2 Ly/Lx = 5000/2850 = 1.75 < 2.0 : Two way slabSlab Case 8: Three edges discontinuous (one short edge continuos), βvx = 0.45Vsx = βvx. n. LxGk = 0.45 x 3.6 x 2.85 = 4.6 kN/mQk = 0.45 x 2.5 x 2.85 = 3.2 kN/m

3

S1S2

Brickwall 3m height

BFC 3142 – CHAPTER 2

Load on Beam C/1-3

Span 1-2Gk: From Slab S1 = 3.8 kN/m Qk: From Slab S1 = 2.1 kN/m

Brickwall = 2.6 x 3.0 = 7.8 kN/mBeam s/weight = 24 x (0.2 x (0.45-0.15)) = 1.44 kN/m

Total dead load, Gk = 13.04 kN/m Total imposed load, Qk = 2.1 kN/m

Span 2-3Gk: From Slab S2 = 4.6 kN/m Qk: From Slab S2 = 3.2 kN/m

Brickwall = 7.8 kN/mBeam s/weight = 1.44 kN/m

Total dead load, Gk = 13.84 kN/m Total imposed load, Qk = 3.2 kN/m

mmmmmmmmmmmmmmmmmmmmmmmmmmmmm

4

1 323.85 m 5 m

Gk = 13.04 kN/mQk = 2.1 kN/m

Gk = 13.84 kN/mQk = 3.2 kN/m

BFC 3142 – CHAPTER 2

EXERCISE 2.3

Figure below shows the first floor layout plan of office building. If all beams size are 200 x 500 mm, determine the following;

– Characteristic dead and imposed load act on the beam C/1-4 if all slab thickness are 150 mm and the brickwall heights is 3m.

– Shear force and bending moment envelope of beam C/1-4.

Solution:

Load on SlabSlab Self-weight = 0.150 x 24 = 3.6 kN/m2

Slab S1, S2, S3, S4 & S5Dead load, Gk = 3.6 kN/m2

Imposed load, Qk = 2.5 kN/m2

Load distribution from slab

Slab S1 Ly/Lx = 7500/3000 = 2.5 > 2.0 : One way slabw = n. Lx/2Gk = 3.6(3.0)/2 = 5.4 kN/mQk = 2.5(3.0)/2 = 3.75 kN/m

Slab S2 Ly/Lx = 6000/3000 = 2.0 ≤ 2.0 : Two way slabSlab Case 2: One short edge discontinuos, βvx = 0.52Vsx = βvx. n. LxGk = 0.52 x 3.6 x 3.0 = 5.6 kN/mQk = 0.52 x 2.5 x 3.0 = 3.9 kN/m

5

1 2

B

A

3 4

C

D

6 m 7.5 m 6 m

6 m

6 m

3 m S1 S2

S3 S4 S5

BFC 3142 – CHAPTER 2

Slab S3 Ly/Lx = 6000/6000 = 1.0 < 2.0 : Two way slabSlab Case 7: Three edges discontinuous (One long edge continuous) , βvy = 0.29Vsy= βvy. n. LxGk = 0.29 x 3.6 x 6.0 = 6.3 kN/mQk = 0.29 x 2.5 x 6.0 = 4.4 kN/m

Slab S4 Ly/Lx = 7500/6000 = 1.25 < 2.0 : Two way slabSlab Case 2: One long edge discontinuos, βvx = 0.46Vsx = βvx. n. LxGk = 0.43 x 3.6 x 6.0 = 9.3 kN/mQk = 0.43 x 2.5 x 6.0 = 6.5 kN/m

Slab S5 Ly/Lx = 6000/6000 = 1.0 < 2.0 : Two way slabSlab Case 4: Two adjacent edges discontinuos, βvx = 0.40Vsx = βvx. n. LxGk = 0.40 x 3.6 x 6.0 = 8.6 kN/mQk = 0.40 x 2.5 x 6.0 = 6.0 kN/m

Load on Beam C/1-4

Span 1-2Gk: From Slab S3 = 6.3 kN/m Qk: From Slab S3 = 4.4 kN/m

Brickwall = 2.6 x 3.0 = 7.8 kN/mBeam s/weight = (24)(0.20)(0.50-0.150) = 1.68 kN/m

Total dead load, Gk = 15.78 kN/m Total Imposed load, Qk = 4.4 kN/m

Span 2-3Gk: From Slab S1 = 5.4 kN/m Qk: From Slab S1 = 3.75 kN/m

From Slab S4 = 9.3 kN/m From Slab S4 = 6.5 kN/mBrickwall = 7.8 kN/mBeam s/weight = 1.68 kN/m

Total dead load, Gk = 24.18 kN/m Total Imposed load, Qk = 10.25 kN/m

Span 3-4Gk: From Slab S2 = 5.6 kN/m Qk: From Slab S2 = 3.9 kN/m

From Slab S5 = 8.6 kN/m ,m From Slab S5 = 6.0 kN/mBrickwall = 7.8 kN/mBeam s/weight = 1.68 kN/m

Total dead load, Gk = 23.68 kN/m Total Imposed load, Qk = 9.9 kN/m

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

61 2 4

3

6 m 7.5 m 6 m

Gk = 15.78 kN/mQk = 4.4 kN/m

Gk = 24.18 kN/mQk = 10.25 kN/m

Gk = 23.68 kN/mQk = 9.9 kN/m

BFC 3142 – CHAPTER 2

Critical Load arrangement on beam

(i) Load Case 1 : All span loaded with maximum load 1.4Gk + 1.6Qk

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

(ii) Load Case 2 : Alternate span loaded with maximum load 1.4Gk + 1.6Qk and minimum load 1.0Gk

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

(iii) Load Case 3 : Alternate span loaded with minimum load 1.0Gk and maximum load 1.4Gk + 1.6Qk

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

7

1 2 43

6 m 7.5 m 6 m

29.13 kN/m 50.3 kN/m 59 kN/m

1 2 43

6 m 7.5 m 6 m

29.13 kN/m 24.18 kN/m 59 kN/m

1 2 43

6 m 7.5 m 6 m

15.78 kN/m 50.13 kN/m 23.68 kN/m

BFC 3142 – CHAPTER 2

EXERCISE 2.4

Design the ultimate bending moments and shear forces for beam 2/A-D by using Table 3.5 BS 8110. The loading criteria are given as follows:

Slab thickness = 125 mmFinishing, partition etc. = 2.95 kN/m2

Characteristic Imposed load = 3.00 kN/m2 Beam size = 250 x 450 mm

Solution:

Load on slabSlab Self-weight = 0.125 x 24 = 3.kN/m2 Finishing, partition etc. = 2.95 kN/m2

Total dead Load, Gk = 5.95 kN/m2 Total Imposed load, Qk = 3.00 kN/m2

Load on Beam 2/A-DGk: From Slab = 5.95(3.0/2)(2) = 17.85kN/m

Beam s/weight = (24)(0.25)(0.45-0.125) = 1.95 kN/m

Total dead load, Gk = 19.8 kN/m

Qk: From Slab = 3.0(3.0/2)(2) = 9.00 kN/m

Total Imposed load, Qk = 9.00 kN/m

Term and condition– Characteristic Imposed load, Qk < Characteristic Dead load, Gk– Uniform distributed load all over 3 span– Equal beam span

→ Therefore moment and shear force can be calculate using coefficient from table 3.5, BS 8110.

8

BFC 3142 – CHAPTER 2

Table 3.5 Design ultimate bending moments and shear forcesAt outer support

Near middle of end span

At first interior support

At middle of interior spans

At interior support

Moment 0 0.09 Fl 0.11 Fl 0.07 Fl -0.08 FlShear 0.45 F - 0.6 F - 0.55F

l - the effective span F – total design ultimate load (1.4Gk + 1.6Qk)

Total design load, w = 1.4Gk + 1.6Qk= 1.4(19.8) + 1.6(9)= 42.12 kN/m

F = wL = 42.12(8) = 336.96 kN

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

9

A B DC

8 m 8 m 8 m

42.12 kN/m

0.45F = 151.6 kN 0.55F = 185.3 kN

0.6F = 202.8 kN 0.55F = 185.3 kN 0.45F = 151.6 kN

0.6F = 202.8 kN

0.11FL = 296.5 kNm 0.11FL = 296.5 kNm

0.09FL = 246.2 kNm0.09FL = 246.2 kNm0.07FL = 188.7 kNm

SFD

BMD

BFC 3142 – CHAPTER 2

Moment distribution methods

Fixed end moment:

MAB = MBC = MCD =

Stiffness : KBA = KBC =

KBC = KCB =

Distribution factor: BA : BC = CD : CB

= 0.43 : 0.57

  0.43 0.57   0.57 0.43  -224.64 224.64 -224.64 2224.64 -224.64 224.64224.64 0.00 0.00 0.00 0.00 -224.64

112.32 0.00 0.00 -112.32

-48.3 -64.24 64.02 48.30.00 32.01 -32.01 0.00

-13.76 -18.25 18.25 13.760.00 9.12 -9.12 0.00

-3.92 -5.2 5.2 3.920.00 2.6 -2.6 0.00

-1.12 -1.48 1.48 1.120.00 0.74 -0.74 0.00

-0.32 -0.42 0.42 0.320.00 0.21 -0.21 0.00

-0.09 -0.12 0.12 0.090.00 0.06 -0.06 0.00

-0.03 -0.03 0.03 0.03

0.00 269.42 -269.42 269.42 -269.42 0.00

10

A B DC

8 m 8 m 8 m

42.12 kN/m

A B C D