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GIVEN DATA

1 Slab structure as presented in Figure 1.1

2 Length measured from the central axis of the beams and walls l 1 = 2,55m; l 2 = 6,8m Load bearing walls with thickness b t = 340mm

- Middle column: cross-sectional area b c × h c = 300x300mm

- Columns inside walls on axis 1, 6: 300x300mm

- Outer columns supporting the primary beams: 300x300mm

3 Slab for civil buildings, which consists of 4 layers as shown in Figure 1.1 Nominal live load P tc = 8,05kN/m 2 ; factor of reliability of live load n = 1,2

4 Materials: Concrete with Compressive Grade B20, plate reinforcement and stirrups class

CI, longitudinal reinforcement class CII

Figure 1.1 Plan of Slab and Slab details

CALCULATION OF PLATE

Analysis

Ribbed slab supported by beams along 2 directions The beams located on axis 2, 3, 4,

5 are primary beams, perpendicular to which are secondary beams Dimensions of plate aperture: l 1 = 2550mm, l 2 = 6800mm, l 2 > 2l 1 so it has similar features of the One-way slab.

Pre-Select components dimensions

= m =  = Assume h b = 80mm whereas D = 1,1 with moderate load, m = 35 with continuous plate

Cross-sectional dimensions of secondary beam selection:

= m  =  = Assume h dp = 500mm, b dp = 220mm;

Cross-sectional dimensions of primary beam selection: primary beam span equals to distance between 2 columns, 7650mm

= m  =  = Assume h dc = 800mm, b dc = 300mm;

One-way slab, establish a simplified model of a plate strip with the width of b 1 =1m perpendicular to the secondary beams Consider the strip as a continuous beam

Design span length of plate:

Design load

Dead load is calculated and presented in Table 2.1

Table 2.1 Calculation of dead load

Layers of plate Nominal Value

- Lining brick of 10mm, γ = 20kN/m 3 0,01×20 = 0,200 1,1 0,220

- Lining mortar of 30mm, γ = 18kN/m 3 0,03×18 = 0,540 1,3 0,702

- Reinforced concrete slab of 70mm, γ = 25kN/m 3 0,08×25 = 2,000 1,1 2,200

Strip of slab with thickness of b 1 = 1m has q b = 13,02×1 = 13,02kN/m.

Design internal forces

Following the plastic scheme model:

- Bending moments at the outer span and the second support:

- Bending moments at the middle spans and middle supports:

  The value of the maximum shear force:

Figure 2.1 Schematic and internal force diagram in slab strip

Calculation of bending moment bearing reinforcement

Concrete grade B20, has R b = 11,5MPa, reinforcement class CI, has R s = 225MPa Calculate the internal force based on plastic scheme, with coefficient of limiting compression area  pl = 0,255

Assume that a = 15mm for all section: h 0 = h b – a = 80 – 15 = 65mm

At the outer support and outer span, with M = 3,9kNm (convert into 3,9×10 6 Nmm):

According to Appendix 9 in ‘Reinforced concrete structures’, ζ = 0,930;

 = b h =   Select diameter of reinforcement 8mm, a s = 50,27mm 2 , spacing of adjacent rebars is:

At the middle supports and middle spans, with M = 4,416kNm, α m = 0,091; ζ = 0,952; A s

Select diameter of reinforcement 8mm, a s = 50,27mm 2 , design spacing of adjacent rebars is s = 158,6mm → Select ɸ8, s = 150mm

Check the working height h o with covering thickness of 10mm: h ot = 80 – 10 – 0,5×8 = 66mm > 65mm → Safety

Reinforcement under negative moment: with p b /g b = 9,66/3,36 = 2,9 < 3, value ν = 0,25; extension from edge of the secondary beam is: νl o = 0,25×2,33= 0,583m, extension from axis of the secondary beam is: νl o +0,5b dp = 0,583+0,5×0,22 = 0,6925m

Longitudinal reinforcement under negative moment is arranged alternatively, extension of the shorter reinforcement from the edge of secondary beam is:

From the axis of secondary beam:

Longitudinal reinforcement under positive moment is arranged alternatively, distance from the end of the shorter reinforcement to edge of the secondary beam is:

Q B T = 18,4kN < Q bmin Concrete achieves shear resistance!

Reinforcement under negative moment is placed perpendicular to the primary beam and wall bracing, using Φ6 bars with a spacing of s = 150 mm The cross-sectional area is 188.5 mm² per meter, which exceeds 50% of the reinforcement area at the middle support of the plate, calculated as 0.5 × 317 mm² = 159 mm² Capping reinforcement is used, with extension from the edge of the primary beam.

From axis of the primary beam: 1 1

Distribution reinforcement is laid in a direction perpendicular to the load‑bearing reinforcement Use Φ6 bars with spacing s = 200 mm; the distribution reinforcement provides an area of 141 mm^2 per meter, guaranteed to be greater than 30% of the assumed reinforcement area at the span The outer span area is 0.3 × 4649 mm^2 (≈1,395 mm^2) and the middle span area is 0.3 × 317.1 mm^2 (≈95.1 mm^2).

Figure 2.2 Reinforcement arrangement in slab

SECONDARY BEAM CALCULATION

Schematic Diagram

Given secondary beam type is a symmetrical 5-span continuous beam We consider the left half of the beam (figure 1.4)

There is an overlap of S d between the beam and the wall, which is equal to the thickness of the wall, S d = 340mm

C d = min (S d /2 và l 2 /40); (S d /2) = 170mm = (l 2 /40) = 170mm So C d = 170mm

Design span length of secondary beam:

Figure 3.1 Schematic diagram and Internal force diagrams for secondary beam

Design load

- Self-weight of beam (excluding the 80mm-thick plate) g op = b dp (h dp – h b )γn = 0,22×(0,50 – 0,08)×25×1,1 = 2,541kN/m

- Dead load carried from plate: g b l 1 = 3,36×2,55 = 8,558kN/m Total Dead load value: g p = g odp +g b l 1 = 2,541+8,558 = 11,099kN/m Live load from plate: p p = p b l 1 = 9,66×2,55 = 24,633kN/m

Design Internal force

Ordinate of Bending moment Envelope (Positive branch)

+ At the outer span: M + = β 1 q p l pb 2 = β 1 ×35,732×6,65 2 = β 1 ×1580,14953kNm + At the middle span: M + = β 1 q p l p 2 = β 1 ×35,732×6,5 2 = β 1 ×1509,6686kNm Ordinate of Bending moment Envelope (Negative branch)

According to Appendix 11 in ‘Reinforced concrete structures’, with ratio p p /g p = 2,219, the values of the coefficients are k = 0,259 and β 1 , β 2 , results presented in Table 3.1 (no interpolation needed to determine k and β 2 )

Table 3.1 Bending moment envelope of secondary beam calculation

Span, cross- section β value M value (kNm) β 1 β 2 M + M -

The distance between the cross-section under negative bending moment equals to 0 and the second support is: x = kl pb = 0,259×6,65 = 1,721m

The distance between the cross-section under positive bending moment equals to 0 and the support is:

+ At the outer span: 0,15l pb = 0,15×6,65 = 0,998m

+ At the middle span: 0,15l p = 0,15×6,60 = 0,975m b Shear force

Bending moment Envelope and Shear force diagram are presented in Figure 3.1.

Longitudinal reinforcement

Concrete grade B20 with compressive strength Rb = 11.5 MPa and concrete tensile strength Rbt = 0.90 MPa, longitudinal reinforcement class CII with design strengths Rs = 280 MPa and Rsc = 280 MPa, and stirrups class CI with Rsw = 175 MPa are analyzed for a negative moment using a plastic scheme model With plastic hinge parameters ξ_pl = 0.3 and α_pl = 0.255, the internal forces are obtained by plastic equilibrium of the hinge: the concrete compression block carries up to Rb, the steel reinforcement provides tensile resistance via Rs and Rsc, and the stirrups contribute Rsw; the resulting internal force distribution yields the plastic moment capacity M_pl and the corresponding neutral axis location, satisfying compatibility between concrete strain and steel strain under the given material strengths and plastic parameters.

The beam cross section is calculated as a rectangular with b = 220mm; h = 500mm Assume a = 40mm, h 0 = 500-40 = 460mm

At the second support, with M = 112,98kNm

At third support, with M = 94,354kNm: α m = 0,176, ζ = 0,902; A s = 812mm 2 , μ% = 0,74% b For positive moment

The calculated cross section is in T-shaped having the flange in compression zone, the flange thickness h f = 80mm

The extension of flange: S f does not exceed the minimum value of:

+ A half of the clearance between the adjacent secondary beams:

0,5l o = 0,5×2,33 = 1,165m (since h > 0,1.h, with h = 500mm and the distance between transverse beams are ' f longer than those of longitudinal beams: 6,8m > 2,55m)

The effective width of flange: b f ’= b+2S f = 220+2×1083 = 2386mm

M + max = 143,794kNm < M f = 922,208kNm → Neutral axis is within flange

Calculation in rectangular b = b ' f = 2386mm, h = 500mm, a = 40mm, h 0 = 460mm

At outer spans, with M + = 143,794kNm

At second span and middle spans, with M = 94,354kNm: α m = 0,016; ζ = 0,992; A s = 739mm 2 ; μ = 0,73%.

Selection and arrangement of longitudinal reinforcement

Table 3.2 Longitudinal rebars arrangement of main sections of beam

Section Outer span Second support (B) Span no.2 Third support

Design A s 1131 mm 2 997 mm 2 739 mm 2 812 mm 2 739 mm 2

Rebars area 1140 mm 2 1008 mm 2 763 mm 2 823 mm 2 763 mm 2

Figure 3.2 Longitudinal rebars arrangement of main sections of beam

Select covering thickness c = 25mm, with one layer of reinforcements arranged, h 0 = h – c – ɸ/2 = 500 – 25 – 22/2 = 464mm > 460mm

Safety is ensured with a = 40mm!

Calculation of transverse reinforcement

From shear force diagram, figure 1.4 we have:

The maximum shear force Q max = Q 2 T = 142,57kN

With selected materials: R b = 11,5MPa; R bt = 0,9MPa; R sw = 175MPa

Dimensions of beam: b = 220mm, h = 500mm, h o = 464mm

Compute: Q bmin = 0,5R bt bh o = 0,5x0,9x220x455 = 44550N = 44,55kN

(Which means all of the shear force is sustained by concrete and stirrups)

Therefore, stirrups calculation is needed!

Calculation of outer-span’s stirrup (without diagonal reinforcement):

• Calculation with shear force at the left side of outer-span: Q 1 = 95,05kN

Given: Q 1 = 95,05kN = Q đb = 6 R bh bt 0 2 ( 0,75 q sw + q p − 0,5 p p ) whereas q p : total load of secondary be am, q p = g p + p p = 11,099+24,633 = 35,732kN/m q swmin = 0,25R bt b= 0,25×0,9×220 = 49,5N/mm

+ Assume that c ≤ 2h 0 = 928mm, determine the unit shear resistance of stirrups q sw :

- Determine co with q sw = 32,3N/mm:

+ Assume that 2h 0 = 928mm ≤ c ≤ 3h 0 = 1392mm, determine q sw :

Select q sw = max (q sw ; q swmin ) = 49,5N/mm

The cross-sectional area of a layer of stirrups is:

The design distance between stirrups is:

= =  Select distance between layers for arrangement

Distance between layers of stirrup S tt :

S tt = 200mm Distance between stirrups in accordance of detailing requirement s ct :

S ct = min of 0,5h o = 232mm and 300mm, so select S ct = 232mm

Maximum distance between layers of stirrup S max :

= = The design distance between stirrups is s = min{S tt ; S ct ; S max }, so s = s tt = 200mm Select ɸ6, 2 legs stirrup, distance s = 200mm for left edge of outer-span

• Calculation with shear force at the left side of support B: Q 2 T = 142,57 kN

+ Design stirrups that satisfies structural requirements: Select ɸ6, 2 legs, s = 130mm Determine q sw (the force that the stirrups can bear)

= =  = N/mm q swmin = 0,25R bt b = 0,25×0,9×220 = 49,5N/mm < q sw = 76,125N/mm

Thus Q đb = 6R bh bt 0 2 (0,75q sw +q p −0,5p p )3498N 3, 498kN

Calculation of 2 nd span and mid-span’s stirrup (without diagonal reinforcement):

Calculation with shear force at the right side of support B: Q 2 P = 116,13kN

+ Design stirrups that satisfies structural requirements: Select ɸ6, 2 legs, s = 180mm Determine q sw (the force that the stirrups can bear)

= =  = N/mm q swmin = 0,25R bt b = 0,25×0,9×320 = 49,5N/mm < q sw = 54,98N/mm

Thus Q đb = 0,5R bh bt 0 +(q p −0,5p p )3h 0 +1,5q h sw 0 6795N 6,795kN

Calculation and construction of the Material Resistance Envelope diagram

+ At the outer span, positive moment, flange of T-section under compression zone, with breadth: b = b f ’ = 2386mm, arrange reinforcement 2Φ22+1Φ22, area of reinforcement

Select covering thickness of 25mm, a = 22/2+25 = 36mm, D h o = 500 – 36 = 464mm

 = =    x = ξ.h 0 = 0,0251×464 = 11,65mm < h’ f = 80mm - neutral axis is within the flange ζ = 1 – 0,5ξ = 1 – 0,5×0,0251 = 0,9875

+ At second support, negative moment, section is a rectangle with b×h = 220×500, arrange reinforcement 2Φ20 + 1Φ22, area is A s = 1008mm 2

Select covering thickness of 25mm, a = 25+22/2 = 36mm, h 0 = 500- 36 = 464mm

The results of bearing capacity are shown in Table 3.3, all sections are calculated according to the case of single reinforcement section (section under positive moment, b is replaced by b f ’)

Table 3.3 Bearing Capacity of cross sections

Section Amount and Reinforcement area (mm 2 ) h 0

Cut 1Φ18, remain: 2Φ18, A s 509 466 0,0112 0,9944 66,04 b Determination of theoretical position for cutting of bars

Figure 3.3 Theoretical section for cutting of No.4 rebar

After cutting the No 4 rebar, the bearing capacity of the remaining reinforcement (2Φ20) is 75.69 kNm The theoretical position for cutting the No 4 rebar is defined by the intersection of the bending moment envelope and the material resistance envelope for the 2Φ20 reinforcement.

Determine the extension length: Shear force Q corresponding to the theoretical section is Q,47kN Stirrup at this section is ɸ6a180, which has:

Because at the cutting section of reinforcement no.2 there is no arranged diagonal reinforcement, so Q s,inc = 0;

Proceed with the same calculation, the results are presented in Table 3.4

Table 3.4 Theoretical section of reinforcements

Reinforcement Theoretical cutting section Extension length

Outer pin reinforcement no.2 (left) Distance to right-edge of pin 1: 1242mm W2 l= 625mm

Outer span reinforcement no.2 (right) Distance to left-edge of pin 2: 2247mm W2 r= 520mm

Second pin reinforcement no.3 (left) Distance to left-edge of pin 2: 1721mm W3 l c5mm

Second pin reinforcement no.3 (right) Distance to right-edge of pin 2: 932mm W3 r 0mm

Second pin reinforcement no.4 (left) Distance to left-edge of pin 2: 571mm W4 l = 910mm

Second pin reinforcement no.4 (right) Distance to right-edge of pin 2: 690mm W4 r = 945mm

Middle pin reinforcement no.6 (left) Distance to right-edge of pin 2: 1815mm W6 l = 560mm

Third pin reinforcement no.8 (left) Distance to left-edge of pin 3: 689mm W8 l 0mm

Constructive reinforcement

Reinforcement no.8 (2Φ12): this reinforcement used as steel strip reinforcement at outer span, the part where there is no negative moment

Reinforced area is 226mm 2 , no less than 0,1%bh 0 = 0,001×220×465 = 102,3mm 2 Secondary beam envelope diagram of material and rebars arrangement are presented in

CACULATION PRIMARY BEAM

Schematic diagram

The primary beam is a continuous beam with four spans, featuring a cross-section of h_dc = 800 mm and b_dc = 300 mm, with columns sized 300 × 300 mm The design span length for the outer and middle spans equals l = 7,650 m, and the schematic diagram illustrating the arrangement is shown in Figure 4.1.

Figure 4.1 Primary beam schematic diagram

Design load

Self-weight of beam is converted into the concentrated loads:

G 0 = b dc (h dc - h b )γnl 1 = 0,30×(0,8 - 0,08)×25×1,1×2,55 = 15,147kN Dead load transmitted from secondary beam: G 1 = g dp l 2 = 11,099×6,8 = 75,4718kN Concentrated dead load: G = G 0 +G 1 = 15,147 + 75,4718 = 90,619kN Concentrated live load transmitted from secondary beam:

Design internal forces

a Determination the envelope diagram of bending moment:

In this case, internal forces in primary beam are calculated using elastic model

Find the cases where the applied load is detrimental to the beam (Figure 4.3)

Determine the bending moment diagram due to dead load G:

M G = αGl = α×90,619×7,65 = α×693,234 (kNm) The coefficient α is taken from Appendix 12 in ‘Reinforced concrete structures’

Determine the bending moment diagrams due to live load P i :

Consider 6 adverse cases of live load, Figure 4.3.c, d, e, f, g, h

In the M P3 schematic, alpha is not provided for determining the moments at sections 1 through 4 To compute them, divide AC into two segments, AB and BC Since spans 1 and 2 carry loads, the moment M0 of the simply supported beam is M0 = P l1 = 167.5044 × 2.55 ≈ 427.136 kNm The moment is constructed using the Hanging Diagram Method and the relation of similar triangles, as shown in Figure 4.2.

Figure 4.2 Schematic diagram to calculate moment in some sections

Apply the same formula to calculate the values of M P4 ; M P5 ; M P6 schematic diagram: + M P4 :

The results are calculated and shown in Table 4.1

Table 4.1 Calculate and Combine the Bending Moment

Figure 4.3 The bending moment diagram of load cases The envelope diagram of Bending Moment:

Ordinate of the envelop diagram of Bending Moment:

M max = M G +max (M Pi ); M min = M G +min (M Pi )

M max and M min for each section are calculated and presented in Table 4.1

Figure 4.4 shows details of M max and M min for half of the beam (2 spans, given the beam is symmetrical) Use Figure 4.5 to determine moment at the edge of support M mg

Figure 4.4 Bending moment envelope diagram obtained from the Combination Method

To determine the moment at the edge of the support, analyze the envelope of the bending moment diagram at the section immediately to the right of support B The M_min envelope shows a gentler slope than the left side, indicating that the magnitude of the moment at the right edge of support B is greater than at the left edge.

Figure 4.5 Schematic diagram to determine M mg

Similarly, at support C: M mg C = 456,00kNm b Determine the envelope diagram of Shear Force:

Ordinate in the envelope diagram of shear force:

- Due to the effect of dead load G: Q G = βG = β×90,619 (kN)

- Due to the effect of live load P i : Q Pi = β i P= β i ×167,504 (kN)

Whereas the coefficient β is according to Appendix 12 in ‘Reinforced concrete structures’, cases are shown in Figure 4.3, the result are shown in Table 4.2

Table 4.2 Calculation and Combination of the Shear Forces

Middle of the outer span

At mid-span of the spans, the shear force Q is determined by the method of sections, applying the equilibrium equations to the spans This involves isolating a portion of the structure around the mid-span and solving for Q from the balance of forces acting on that section For example, at the mid-span of the outer span, the equilibrium of the chosen section yields the shear force Q needed for analysis and design.

The envelope diagram of shear force is shown in Figure 4.6

Figure 4.6 Envelope diagram of shear force

Longitudinal reinforcement

Concrete with Compressive Grade B20 with R b = 11,5MPa; R bt = 0,9MPa; reinforcement class CII with R s = 280MPa, R sc = 280MPa

According to Appendix 8, with working condition coefficient of concrete γ b2 = 1,0; coefficient of limiting compression area when internal forces are determined using elastic diagram is ξ R = 0,623; α R = 0,429 a For negative moment:

For the rectangular section with dimensions b = 300mm, h = 800mm

At supports, reinforcement of primary beam must be put underneath the top layer of reinforcement in secondary beam, so a is considerably large

At support B, with M mg = 562,73kNm

At support C, with M mg = 456,00kNm; α m = 0,255; ζ = 0,850; A s = 2661mm 2 , μ = 1,23% b For positive moment:

For T-section when the flange is compressed with h f = 80mm

Effective breadth S f is minimum value of following value:

+ (1/6)l d = (1/6)×7,65 = 1,275m + Half of clearance between two adjacent primary beams: 0,5l=0,5×6,5= 3,25m (due to h’ f > 0,1h; with h = 800mm and transverse beams are secondary beams with clearance of 2,55m)

M max = 531,473kNm < M f = 1678.08kNm → Neutral axis is within the flange

For rectangular section with dimensions b = b’ f = 2850mm, h = 800mm, a = 70mm, h 0 720mm

At outer span, with M = 531,473kNm

At middle span, with M = 361,42kNm; α m = 0,021, ζ = 0,990, A s = 1787mm 2 , μ = 0,82%

Table 4.3 Selection of longitudinal reinforcement for primary beam

Section Outer span Support B Second span Support C

Design A s 2641mm 2 3470mm 2 1787mm 2 2661mm 2

Rebars area 2704mm 2 3695mm 2 1722mm 2 2723mm 2

Longitudinal reinforcement arrangement at the main sections is presented in Figure 4.7

Figure 4.7 Longitudinal reinforcement arrangement at the main sections

+ For reinforcement at the outer span: select covering thickness c = 30mm > ɸmax, arrange

2 layers with vertical spacing t = 30mm:

- Bottom layer of 2ɸ25+1ɸ25, A s1 = 1472,6mm 2 ; a 1 = c + 0,5d 1 = 30 + 0,5×25 = 42,5mm

- Upper layer of 2ɸ28, A s2 = 1231,5mm 2 ; a 2 = c + d 1 + t + 0,5d 2 = 30 + 25 + 30 + 0,5×28 = 99mm

→ h ot = 800 – 68,5 = 731,5mm > h o = 730mm (assumed value for calculations above)

So, it is safe with a = 70mm!

+ For reinforcement at the middle span: select covering thickness c = 30mm, arrange 1 layer of reinforcement: 2ɸ25+2ɸ22, A s = 1742mm 2 : a = c + 0,5d = 30 + 0,5×25 = 42,5mm

So, it is safe with a = 70mm!

For reinforcement under a negative bending moment, specify the covering thickness c = 47 mm measured to the inner edge of the reinforcement arranged in the secondary beam This accounts for the secondary beam cover of 25 mm and a maximum reinforcement diameter of Ø22.

- At support C, arrange 2 layers with vertical spacing t = 30mm:

● Top layer of 4ɸ25, A s1 = 1963,6mm 2 ; a 1 = c + 0,5d 1 = 47 + 0,5×25 = 59,5mm

● Lower layer of 2ɸ22, A s2 = 760,3mm 2 ; a 2 = c + d 1 + t + 0,5d 2 = 47 + 25 + 30 + 0,5×22 = 113mm

→ h ot = 800 – 74,5r5,5mm > h o = 720mm, → So, it is safe with a = 80mm!

- At support B, arrange 2 layers with vertical spacing t = 30mm:

● Top layer of 4ɸ28, A s1 = 2463mm 2 ; a 1 = c + 0,5d 1 = 47 + 0,5×28 = 61mm

● Lower layer of 2ɸ28, A s2 = 1231,5mm 2 ; a 2 = c + d 1 + t + 0,5d 2 = 47 + 28 + 30 + 0,5×28 = 119mm;

Here, h_t is 800 minus 80.5, which equals 719.5 mm, and it is slightly less than h0, which is 720 mm Because the difference between h0t and h0 is very small and the chosen rebar area already exceeds the designed rebar area, recalculating h0 is unnecessary.

+ Check spacing of reinforcement: (1 row 4ɸ25)

Select covering at the side is c = 30mm > ɸ Reinforcement is arranged equidistantly, horizontal spacing is:

3 3 b c t = − −  = −  −  = mm(Larger than the requirement of

The inner bars should be arranged near the bars in the corner to widen the gap between them, which makes it easier to use compactors in concrete casting phase.

Calculation of reinforcement for shear force

From shear force diagram of primary beam:

Right edge of support A: Q A P = 208,25kN, shear force is constant through the length l 1

Left edge of support B: Q B T = 337,81kN, shear force is constant through the length l 1

Right edge of support B: Q B P = 304,47kN, shear force is constant through the length l 1 Left edge of support C: Q C T = 289,50kN, shear force is constant through the length l 1

❖ Design stirrups for shear force at the right of support A: Q A P = 208,25kN

Beam dimensions: b = 300mm, h = 800mm, h o = 744mm

Determine: Q bmin = 0,5R bt bh o = 0,5×0,9×300×731,5 = 98753N = 98,753kN

Q bmin = 98,753kN < Q = 208,25kN < 0,3R b bh o = 757,102kN

Calculation of stirrups (without diagonal reinforcement) :

Given: Q = 208,25kN = Qđb = 4,5R bh q bt 0 2 sw

(Which also means that all shear forces are applied to concrete and stirrups)

= R bh =    N/mm q swmin = 0,25R bt b= 0,25×0,9×300 = 67,5N/mm q swmin = 66,7N/mm < q sw = 67,5N/mm

C o = 2069mm > 2h o = 2×742,5mm = 1485mm Assume C o = 2h o 85mm

 Clearly, q sw < q swmin = 67,5N/mm Design stirrups based on q swmin = 67,5N/mm

+ Select the diameter of the stirrups 10mm (ɸ w = 10mm), 2 legs (n = 2)

Area of a layer of stirrups:

= = The design spacing according to calculations of S tt between the stirrup layers:

= =  The spacing between layers of stirrups according to structure:

S ct = min (0,5h o = 372mm; 300mm) →select S ct = 300mm

The greatest spacing between layers of stirrups Smax :

= = The stirrups is arranged with spacing s = min (S tt ; S ct ; S max ); select s = S ct = 300mm

❖ Calculation with shear force at the left side of supportB: Q B T = 337,81kN

Dimensions of beam: b = 300mm, h = 800mm, h o = 723mm.

Determine: Q bmin = 0,5R bt bh o = 0,5×0,9×300×723 = 97605N = 97,605kN

Q bmin = 0,5R bt bh o = 97,605kN < Q = 337,81kN < 0,3R b bh o = 748,305kN Satisfied the conditions (12)!

+ Select stirrups that satisfies structural requirements: Select ɸ10, 2 legs, s = 150mm Determine q sw (the force that the stirrups can bear)

= =  = N/mm q swmin = 0,25R bt b = 0,25×0,9×300 = 67,5N/mm < q sw = 182,21N/mm

Q đb = 4,5R bh q bt 0 2 sw = 4,5 0,9 300 719,5   2 182, 2138536N 38,536kN

Therefore, no diagonal reinforcement needed

❖ Calculation with the shear force at the right side of support B: Q B P = 304,74kN

Arranged stirrups with ɸ10, 2 legs, s = 180mm:

= =  = N/mm q swmin = 0,25R bt b = 0,25×0,9×300 = 67,5N/mm < q sw = 152,72N/mm

Q đb = 4,5R bh q bt 0 2 sw = 4,5 0,9 300 719,5   2 152, 72 09932N 09,932kN

❖ Calculation with the shear force at the left side of support C: Q C T = 289,50kN

= =  = N/mm q swmin = 0,25R bt b= 0,25×0,9×300 = 67,5N/mm < q sw = 137,445N/mm

Q đb = 4, 5R bh q bt 0 2 sw = 4, 5 0, 9 300 725   2 137, 445 )6272N )6, 272kN

Calculation of hanging reinforcement

At the overlap where secondary beams meet primary beams, hanging reinforcement must be installed to strengthen the primary beams The concentrated load from the secondary beams is transferred transversely to the primary beams, ensuring effective distribution of shear and bending moments at the joint.

Hanging reinforcement is stirrups with area of:

Use stirrups ɸ10, a sw = 78,5mm 2 , number of legs n s = 2, number of stirrups necessary:

= n a =  Select 8 stirrups, arrange 4 on each side of the secondary beam

Spacing between stirrups is 60mm, the distance from innermost stirrups to the edge of the secondary beam is 50mm.

Construct the Material Resistance Envelope

For positive moment, cross-section is T-section with neutral axis at the flange, flange effective breath b = b’ f = 2850mm, arranged reinforcement of 2Φ28+3Φ25, reinforcement area A s = 2704mm 2 , h 0 = 735mm as calculated

 = R bh =    x = ξh 0 = 0,0313×731,5 = 23,1mm < h’ f = 80mm – Neutral axis is within the flange ζ = 1 – 0,5ξ = 1 – 0,5×0,0316 = 0,9842

Negative moment, rectangular section b× h = 300mm×800mm, arranged top layer of 4Φ28, lower layer of 2Φ28; area A s = 3694,5mm 2 ; h 0 = 719,5mm as calculated

Results of bearing capacity are presented in Table 4.4, all sections are calculated following the singly-reinforced beam case (section with positive moment is b’ f instead of b)

Table 4.4 Bearing capacity of sections

Section Amount and area of reinforcement (mm 2 ) h 0

Edge of outer span Cut 2Φ28, remains 2Φ25+1Φ25 –

A s = 1472,6 757,5 0,0166 0,9917 309,745 Edge of outer span Cut 1Φ25, remains 2Φ25 – A s 981,8 757,5 0,0111 0,9945 207,087

Edge of support B Cut 2Φ28, remains 4Φ28 – A s 2463 739 0,2705 0,865 440,716

Edge of support B Cut 2Φ28, remains 2Φ28 – A s 1231,5 739 0,135 0,932 237,590

Edge of second span Cut 2Φ22, remains 2Φ25 – A s 981,8 757,5 0,0111 0,9945 207,087

Edge of support C Cut 2Φ22, remains 4Φ25– A s

Edge of support C Cut 2Φ25, remains 2Φ25 – A s 981,8 740,5 0.1076 0,9462 192,614 b Determine the theoretical sections for cutting of reinforcement

Reinforcement No.4, located at the left edge near support B, remains in place after cutting reinforcement No.3; the mid-span of the second span still contains reinforcement No.3 (2Φ25) at the lower fiber, giving a lower-fiber bearing capacity of 207,087 kNm The material diagram intersects the envelope diagram of moment at point H, which defines the theoretical cutting section for reinforcement No.8 By the geometric relations of the congruent triangles OBD, OGH, and OEF, the distance from H to B is 2143 mm (Figure 4.8).

Figure 4.8 Schematic diagram to determine the theoretical cut-off section

Determine the extension length of reinforcement no.4 (at the left) – W 4 t :

Q is the slope of moment diagram: 318,74 207,08

In the cutting section of reinforcement no.4 there is no diagonal reinforcement, so

In this area, stirrup is ɸ10, s = 150, so 175 157,08

Similar calculation applied for other rebars, results are presented in Table 4.5

Table 4.5 Theoretical section of reinforcement

Reinforcement Theoretical position for cutting of reinforcement

Reinforcement no.1a (left) Distance to support A: 994mm W 1a l = 1260mm Reinforcement no.1a (right) Distance to support B: 1746mm W 1a r = 1070mm Reinforcement no.2 (left) Distance to support A: 1486mm W 2 l = 1280mm

Reinforcement no.2 (right) Distance to support B: 2087mm W 2 r = 980mm

Reinforcement no.4 (left) Distance to support B: 2143mm W 4 l = 1010mm

Reinforcement no.4 (right) Distance to support C: 1910mm W 4 r = 1020mm Reinforcement no.5 Distance to support B: 3007mm W 5 = 500mm

Reinforcement no.5a (left) Distance to support B: 1101mm W 5a l = 1060mm Reinforcement no.5a (right) Distance to support B: 1389mm W 5a r = 595mm

Reinforcement no.6(left) Distance to support B: 500mm W 6 l = 1060mm

Reinforcement no.6(right) Distance to support B: 540mm W 6 r = 1135mm Reinforcement no.7a (left) Distance to support C: 1104mm W 7a = 1180mm

Reinforcement no.8 (left) Distance to support C: 479mm W 8 = 1165mm

The left-hand extension of reinforcement 1a is excessive, extending through the bearing wall, and to preserve the bearing capacity of the diagonal section, cutting the reinforcement at this location is not permitted.

After the bottom reinforcement of a beam is cut, the remaining reinforcement pulled toward the support must have an area greater than one-third of the reinforcement area at midspan.

The outer span reinforcement starts as 2Φ25 + 1Φ25 + 2Φ28; after cutting 2Φ28, the remaining area of the 2Φ25 + 1Φ25 portion is 54.46% when pulled into the support Then, from the 2Φ25 + 1Φ25 segment, 1Φ25 is cut, leaving the 2Φ25 portion with a remaining area of 36.04% when pulled into the support The middle span reinforcement is 2Φ25 + 2Φ22; after cutting 2Φ22, the remaining area of the 2Φ25 portion is 56.40% when pulled into the support.

QA = 208,25kN > Q bmin = 0,5R bt bh o = 98,753kN

Anchor length of longitudinal reinforcement = l an = 28Φ = 28×25 = 700mm (with bar no.1) + At support B:

Anchor length of longitudinal reinforcement = 20Φ = 20×25 = 500mm (with bar no.1, reinforcement is not under compression – except for double reinforcement)

Anchor length of longitudinal reinforcement = 20Φ = 20×25 = 500mm (with bar no.3, reinforcement is not under compression – except for double reinforcement)

Anchor length of longitudinal reinforcement = 20Φ = 20×25 = 500mm with bar no.3, reinforcement is not under compression – except for double reinforcement)

This reinforcement is used as steel strip reinforcement at the outer span, where there is no negative moment

The area is 308mm 2 , no less than 0,1%bh 0 = 0,001×300×758 = 227,2mm 2 b Reinforcement no.11 (2ɸ14):

In reinforced concrete beam design, additional longitudinal reinforcement is required for taller sections: when the beam height exceeds 700 mm, two bars of constructive longitudinal reinforcement should be placed at the upper fiber along the beam length In practice, however, this reinforcement is commonly provided when the beam height is 600 mm or more (h ≥ 600 mm).

Primary beam Envelope diagram of material and rebars arrangement are presented in

Figure 3.9 Resistance Envelope diagram of material and rebars arrangement of secondary beam