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Reinforced Concrete Design Module 5 Subject: Reinforced Concrete Design for Architecture Students1. Title of the Module SHEAR REINFORCEMENT AND DIAGONAL TENSION 2. Introduction When beams are subjected to external loads they produced not only bending moments but also internal shear forces. The shear...

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Reinforced Concrete Design Module 5 Subject: Reinforced Concrete Design for Architecture Students

1. Title of the Module SHEAR REINFORCEMENT AND DIAGONAL TENSION 2. Introduction When beams are subjected to external loads they produced not only bending moments but also internal shear forces. The shear does not produce failure on the vertical plane on which they act but its major effect is it includes tensile stresses on diagonal planes which is called diagonal tension. Since concrete has a relatively low tensile strength it would produce cracks making an angle of 45° with the plane on which the shear acts. When this diagonal tension stresses in combination with bending stresses created by moment exceed the tensile strength of concrete, diagonal cracking will occur that can split the beam without warning. TWO TYPES OF DIAGONAL CRACKS IN REINFORCED CONCRETE BEAMS: 1) WEB-SHEAR CRACKS 2) FLEXURE-SHEAR CRACKS

To eliminate this cracks, we design steel reinforcements to resists the diagonal tension called stirrups or bent bars are provided where shear reinforcement is necessary.

LOCATION OF STIRRUPS IN A SIMPLE BEAM A) When shear reinforcement is perpendicular to the axis of member is used.

𝑉𝑢 = ∅𝑉𝑠 + ∅𝑉𝑐

𝑉𝑠 =

𝑉𝑢 ∅

− 𝑉𝑐

Where: 𝑉𝑢 = factored shear force at a critical distance “d” from the face of support

𝑉𝑠 = shear strength of shear reinforcement or stirrups 𝑉𝑐 = shear strength provided by concrete ∅ = 0.75 reduction factor for shear VALUES OF 𝑽𝒄

I. Members subjected to shear and flexure only. a) SIMPLIFIED CALCULATION: 𝑉𝑐 = 0.17λ√𝑓′𝑐 𝑏𝑤 𝑑

Modification factors λ as multiple of √𝑓′𝑐 1. λ = 0.85 for sand – lightweight concrete 2. λ = 0.75 for all lightweight concrete 3. λ = 1.0 for normal weight concrete 4. λ =

𝑓′𝑐𝑡

0.56 √𝑓′𝑐

≤ 1.0 for average splitting tensile strength of lightweight concrete 𝑓′𝑐𝑡

𝑉𝑐 = shear strength provided by concrete 𝑏𝑤 = width of beam

d = effective depth of beam b) DETAILED CALCULATION: 𝑉𝑐 = [0.17 λ √𝑓′𝑐 + 17 𝜌𝑊

𝑉𝑢 𝑑 𝑀𝑢

𝑉𝑢 𝑑 𝑀𝑢

] 𝑏𝑤 𝑑 but not greater than 𝑉𝑐 = 0.29√𝑓′𝑐 𝑏𝑤 𝑑

shall not be greater than 1.0

Vu = factored shear force at a critical distance “d” from the face of support Mu = simultaneous factored moment at a critical section for Vu 𝜌𝑊 =

𝐴𝑠 𝑏𝑑

II. Members subjected to axial compression. a) SIMPLIFIED CALCULATION:

𝑉𝑐 = 0.17 [1 +

𝑁𝑢

14𝐴𝑔

] λ √𝑓′𝑐 𝑏𝑤 𝑑

𝑁𝑢 = axial compressive force in N

𝐴𝑔 = cross sectional gross area of concrete section

b) DETAILED CALCULATION:

𝑉𝑐 = [0.17 λ √𝑓′𝑐 + 17 𝜌𝑊 𝑀𝑚 = 𝑀𝑢 - 𝑁𝑢 (

𝑉𝑢 𝑑 𝑀𝑚

4ℎ−𝑑 8

)

𝑉𝑢 𝑑 𝑀𝑢

] 𝑏𝑤 𝑑

is not limited to 1.0

𝑀𝑚 = is modified moment in kN-m

𝑁𝑢 = axial compressive force in kN h = total depth of beam in m d = effective depth of beam in m 𝑉𝑐 shall not be greater than 𝑉𝑐 = 0.29 [√1 +

0.29 𝑁𝑢 𝐴𝑔

] λ √𝑓′𝑐 𝑏𝑤 𝑑

When 𝑀𝑚 as computed from the formula 𝑀𝑚 = 𝑀𝑢 - 𝑁𝑢 ( USE 𝑉𝑐 = 0.29 [√1 +

0.29 𝑁𝑢 𝐴𝑔

4ℎ−𝑑 8

) is negative.

] λ √𝑓′𝑐 𝑏𝑤 𝑑

III. Members subjected to significant axial tension.

𝑁

𝑉𝑐 = 0.17 [1 + 14𝐴𝑢 ] λ √𝑓′𝑐 𝑏𝑤 𝑑 but not less than zero 𝑔

𝑁𝑢 = axial compressive force in N (negative for tension) 𝐴𝑔 = cross sectional gross area of concrete section

VALUES OF SHEAR STRENGTH REINFORCEMENT(VS)

A) When shear reinforcement is perpendicular to the axis of member is used. where: 𝐴𝑣 = the area of stirrups within spacing “s”. It shall be taken as two times the area of bar in a circular tie, hoop or spiral at a spacing “s”. 𝑠 = spacing of stirrups which is measured in a direction parallel to longitudinal reinforcement. 𝑓𝑦𝑡 = specified yield strength of circular ties, hoop or spiral reinforcement.

𝑇 = 𝐴𝑣 𝑓𝑦𝑡 (n) (capacity of stirrups in N) Where: 𝑑

𝑛 = (number of shear reinforcement)

𝑉𝑠 =

𝑠

𝑉𝑢 ∅

− 𝑉𝑐

; ∅ = 0.75

; 𝑠 = spacing of shear reinforcement

𝑉𝑠 shall not be greater than 0.66√𝑓′𝑐 𝑏𝑤 𝑑 𝑉𝑠 = shear capacity of stirrups in N 𝑇 = 𝑉𝑠

𝐴𝑣 𝑓𝑦𝑡 (𝑛) = 𝑉𝑠 𝐴𝑣 𝑓𝑦𝑡 𝑉𝑠 =

𝑑

𝑠

= 𝑉𝑠

𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑠

B) When inclined stirrups are used as shear reinforcement.

𝑉𝑠 =

𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑠

(𝑆𝑖𝑛 𝛼 + 𝐶𝑜𝑠 𝛼)

Where: 𝛼 = angle between inclined stirrups and longitudinal axis of the member and, S is measured in a direction parallel to longitudinal reinforcement. C) When shear reinforcement consists of a single bar or a single group of parallel bars, all bent up at the same distance from the support. 𝑉𝑠 = 𝐴𝑣 𝑓𝑦𝑡 sin 𝛼 but not greater than 0.25√𝑓′𝑐 𝑏𝑤 𝑑 Where: 𝛼 = is the angle between the bent up reinforcement and longitudinal axis of the member. D) When shear reinforcement consists of a series of parallel bent up bars or group of parallel bent up bars at different distances from the support, shear strength. 𝑉𝑠 =

𝐴𝑣 𝑓𝑦𝑡 𝑑 (sin 𝛼 + cos 𝛼) 𝑠

E) Minimum area of shear reinforcement 𝐴𝑣 𝑚𝑖𝑛 =

0.062√𝑓′𝑐 𝑏𝑤 𝑆 𝑓𝑦𝑡

But shall not be less than

0.35𝑏𝑤 𝑆 𝑓𝑦𝑡

F) Min. area of shear reinforcement 𝐴𝑣 𝑚𝑖𝑛 shall be provided in all reinforced concrete flexural 1 members (pre stressed or non pre stressed) where 𝑉𝑢 exceeds ∅𝑉𝑐 , except the following 2 members:

Solid slabs and footing: 1) Hollow core units with total un-topped depth not greater than 300 mm and hollow core 1 units where vu is not greater than ∅𝑉𝑐𝑤 , 𝑉𝑐𝑤 = nominal shear strength provided by 2

concrete when diagonal crack results from high principal tensile stress.

2) Concrete joist construction

3) Beams with total depth “h” not greater than 250 mm 4) Beams integral with slabs with total depth “h” not greater than 600 mm and not greater than 2.5 times thickness of flange or 0.50 the width of web. 5) Beams constructed of steel fiber-reinforced, normal weight concrete with fc’ not exceeding 40 MPa and “h” not greater than 600 mm and 𝑉𝑢 not greater than ∅0.17√𝑓′𝑐 𝑏𝑤 𝑑

SPACING LIMITS FOR SHEAR REINFORCEMENT(S) 1) When Vs < 0.33√𝑓′𝑐 𝑏𝑤 𝑑 max spacing of shear reinforcement placed perpendicular to axis of member. Max. S = 0.75h or 600 mm

(pre stressed members)

2) When Vs > 0.33√𝑓′𝑐 𝑏𝑤 𝑑 Max spacing of shear reinforcement. Max. S = 0.375h or 600 mm (pre stressed members) 3) Inclined stirrups and bent longitudinal reinforcement shall be so spaced that every 45° line, extending toward the reaction from mid-depth of member d/2 to longitudinal tension reinforcement, shall be crossed by at least one line of shear reinforcement. A) Shear force for beams with minimum amount of web reinforcement. 𝑉𝑢 = ∅𝑉𝑐

B) Shear force for beams with no web reinforcement is needed. 𝑉𝑢 =

∅𝑉𝑐 2

NSCP 2015 SPECS: 1) Members subjected to shear and flexure only. TABLE 422.5.5.1 DETAILED METHOD FOR CALCULATING 𝑉𝑐 LEAST OF (a), (b), or (c): (a) 𝑉𝑐 = (0.16𝜆√𝑓′𝑐 + 17𝜌𝑤

𝑉𝑢 𝑑

𝑀𝑢

) 𝑏𝑤 𝑑

(b) 𝑉𝑐 = (0.16𝜆√𝑓′𝑐 + 17𝜌𝑤 )𝑏𝑤 𝑑 (c) 𝑉𝑐 = (0.29𝜆√𝑓′𝑐 )𝑏𝑤 𝑑

Mu occurs simultaneously with Vu at the section considered 2) Members subjected to axial compression. TABLE 422.5.6.1 DETAILED METHOD FOR CALCULATING 𝑉𝑐 FOR NON- PRESTRESSED MEMBERS WITH AXIAL COMPRESSION LESSER OF (a), (b), OR (c): (a) 𝑉𝑐 = (0.16𝜆√𝑓′𝑐 + 17𝜌𝑤

𝑉𝑢 𝑑

𝑀𝑢 −𝑁𝑢

4ℎ−𝑑 8

(b) Equation not applicable if 𝑀𝑢 − 𝑁𝑢 (c) 𝑉𝑐 = (0.29𝜆√𝑓′𝑐 𝑏𝑤 𝑑)√1 +

0.29𝑁𝑢

) 𝑏𝑤 𝑑

4ℎ−𝑑 8

≤0

𝐴𝑔

Mu occurs simultaneously with Vu at the section considered SUMMARY OF PROVISIONS FOR WEB REINFORCEMENT (2010/2015 NSCP)

1.) when 𝑉𝑢 ≤ 2.) when

∅𝑉𝑐 2

∅𝑉𝑐 2

,

< 𝑉𝑢 ≤ ∅𝑉𝑐 ,

3.) when 𝑉𝑠 ≤ 0.33√𝑓`𝑐𝑏𝑤 𝑑, 𝑉𝑢 ≤ 3∅𝑉𝑐

stirrups are not required 𝑠≤

𝑠≤

𝑠≤

𝑠≤

𝑠≤ 𝑠≤ 𝑠≤

𝐴𝑣 𝑓𝑦𝑡

0.062√𝑓`𝑐𝑏𝑤 𝐴𝑣 𝑓𝑦𝑡

0.35𝑏𝑤

𝑑

2

or 600mm 𝐴𝑣 𝑓𝑦𝑡

0.062√𝑓`𝑐𝑏𝑤

5.) when 𝑉𝑠 > 0.66√𝑓`𝑐𝑏𝑤 𝑑, 𝑉𝑢 > 5∅𝑉𝑐

Base on minimum shear area requirement

or

𝐴𝑣 𝑓𝑦𝑡

0.35𝑏𝑤

𝐴𝑣 𝑓𝑦𝑡𝑑 𝑑

2

𝑉𝑠

Base on strength requirements

or 600mm

4.) when 0.33√𝑓`𝑐𝑏𝑤 𝑑 < 𝑉𝑠 ≤ 0.66√𝑓`𝑐𝑏𝑤 𝑑, 3∅𝑉𝑐 < 𝑉𝑢 ≤ 5∅𝑉𝑐

Base on minimum shear area requirement

or

𝑠≤ 𝑠≤

𝐴𝑣 𝑓𝑦𝑡𝑑 𝑑

4

𝑉𝑠

or 300mm

Beam is too small to accommodate shear reinforcements. Revise by increasing beam dimensions.

3. Learning Outcomes At the end of the course, the students should be able to: a. be familiar with the NSCP 2015 provisions regarding shear reinforcement and compare it with previous specifications b. apply NSCP 2015 specs correctly in the design of strirrups or shear reinforcements

4. Learning Content These are the application of the latest NSCP provisions of in design and analyses of shear reinforcements. EXAMPLE 1: A rectangular beam 350 mm width and effective depth of 600 mm, 𝑓′𝑐 = 20.7 MPa, 𝑓𝑦 = 𝑓𝑦𝑡 = 414 MPa. Determine the required spacing of stirrups if; a) Vu = 58 KN

b) Vu = 350 KN c) Vu = 710 KN Solution: A.) Vu = 58 KN Vc = 0.17√𝑓′𝑐 𝑏𝑤 𝑑 Vc = 0.17(√20.7)(350)(600) Vc = 162.425KN 𝛷𝑉𝑐 = 60.909 𝐾𝑁 2 Vu <

𝛷𝑉𝑐 2

(no stirrups needed)

B.) Vu = 350 KN Vc = 0.17√𝑓′𝑐 𝑏𝑤 𝑑 Vc = 0.17(√20.7)(350)(600) Vc = 162.425KN 𝛷𝑉𝑐

= 60.909 𝐾𝑁

2

Vs =

𝑉𝑢 𝛷

− 𝑉𝑐

Vs = 304.24 KN Vs < 0.33√𝑓𝑐 ′𝑏𝑤 𝑑 304.24 KN < 315.296 KN (Use case 3) 𝐴𝑣𝑓𝑦𝑡

S=

0.062√𝑓𝑐 ′𝐵𝑤

S= S=

𝜋/4(10)^2(2)(414) = 0.062√20.7(350)

658.681 say 655 mm

𝐴𝑣𝑓𝑦𝑡

0.35𝐵𝑤

S=

𝜋/4(10)^2(2)(414) = 0.35(350)

530.865 say 530 mm

S=

𝐴𝑣𝑓𝑦𝑡𝑑

S=

𝜋/4(10)^2(2)(414)(600)

𝑉𝑠

𝑑

304240

S= 𝑜𝑟 600 𝑚𝑚 2

= 128.249 say 125 mm

S=

600 2

= 300𝑚𝑚

Use 10mm stirrups @ 125mm O.C

C.

Vu = 710 KN Vc = 0.17√𝑓′𝑐 𝑏𝑤 𝑑 Vc = 0.17(√20.7)(350)(600) Vc = 162.425KN 𝛷𝑉𝑐 2

= 60.909 𝐾𝑁 𝑉𝑢

Vs = 𝛷 − 𝑉𝑐 Vs = 784.242 KN

Vs > 0.66√𝑓′𝑐 𝑏𝑤 𝑑 784.242 KN > 630.592 KN (Beam is too small to accommodate shear reinforcement. Revise by increasing beam dimension)

EXAMPLE 2: A rectangular beam is to be designed to carry a factored shear force 𝑉𝑢 = 134 KN. 𝑓′𝑐 = 27.6 MPa, 𝑓𝑦𝑡 = 414.6 MPa. Use 𝜆 = 1.0 for normal weight concrete. a) What is the minimum dimension for a beam with no web reinforcement? b) Determine the size of the beam if the minimum amount of web reinforcement is needed. c) Determine the location from the support of the points where min. web reinforcement is required if it has a span of 8 m.

Solution: a) Minimum dimension for a beam with no web reinforcement. 𝑉𝑐 = 0.17𝜆√𝑓′𝑐 𝑏𝑤 𝑑 𝑉𝑐 = 0.17(1)√27.6𝑏𝑤 𝑑 𝑉𝑐 = 0.893𝑏𝑤 𝑑

1 𝑉𝑢 = ∅𝑉𝑐 2

1

134000 = (0.75)0.893 𝑏𝑤 𝑑 2 𝑏𝑤 𝑑 = 400149

Try 𝑏𝑤 = 450 𝑚𝑚 𝑑(450) = 400149 𝑑 = 889 𝑠𝑎𝑦 890 𝑚𝑚 Use 𝑏𝑤 = 𝟒𝟓𝟎 𝒎𝒎 𝑑 = 𝟖𝟗𝟎 𝒎𝒎

b) Size of the beam if the minimum amount of web reinforcement is needed. 𝑉𝑢 = ∅𝑉𝑐 134000 = (0.75)0.893𝑏𝑤 𝑑 𝑏𝑤 𝑑 = 200075 Try 𝑏𝑤 = 300 𝑚𝑚 𝑑(300) = 200075 𝑑 = 667 𝑠𝑎𝑦 670 𝑚𝑚 Use 𝑏𝑤 = 𝟑𝟎𝟎 𝒎𝒎 𝑑 = 𝟔𝟕𝟎 𝒎𝒎

c) Location from the support of the points where min web reinforcement is required if it has a span of 8 m. ∅𝑉𝑐 = 0.11𝜆√𝑓′𝑐 𝑏𝑤 𝑑 ∅𝑉𝑐 = 0.11(1)√27.6(300)(670) ∅𝑉𝑐 = 116156 𝑁 ∅𝑉𝑐 2

= 58.1 𝑘𝑁

By ratio and proportion 134 116.2 = 𝑥1 3.33

𝑥1 = 2.89 𝑚

𝑥2 = 1.44 𝑚

4 − 2.89 = 1.11 𝑚

1.11 + (2.89 − 1.44) = 2.56 𝑚 Location from the support of the points where min web reinforcement is 1.44 m to 2.56 m.

EXAMPLE 3: The figure shows a floor plan which is design to carry a live load of 4.8 kPa and a dead load of 4.9 kPa (including weight of slab, beam, ceiling, partition, floor, etc.). the slab thickness is 100 mm. the beam has a width of 250 mm and has a depth of 350 mm below the slab. the girder has a width of 350 mm and a depth of 400 mm. use 70 mm covering from center of steel reinforcements using 𝜆 = 1.0 for normal weight concrete. 𝑓′𝑐 = 20.7 MPa, 𝑓𝑦 = 415 MPa, 𝑓𝑦𝑡 = 275 MPa.

a) Assuming the beam to be simply supported, determine the critical factored shear force at the critical section for beam be. b) Determine the spacing for the two legs of 10 mm 𝜙 stirrups. c) Using NSCP code, determine the maximum spacing of stirrups.

Solution: a) Critical factored shear force at the critical section for the beam BE. Using clear span: 𝐿𝑛 = 6 − 0.35

𝐿𝑛 = 5.65 𝑚

𝑊𝑢 = 1.2𝐷𝐿 + 1.6𝐿𝐿

𝑊𝑢 = 1.2(4.9) + 1.6(4.8)

𝑊𝑢 = 13 .56 𝑘𝑁⁄𝑚2

𝑊𝑢 = 13 .56(2.8)

𝑊𝑢 = 37 .968 𝑘𝑁⁄𝑚

2𝑅 = 37 .968(5.65)

𝑅 = 107 .26 𝑘𝑁

Shear force Vu: 𝑉𝑢 = 107.26 − 37.968 (0.38)

𝑉𝑢 = 𝟗𝟐. 𝟖𝟑 𝒌𝑵

b) Spacing for the two legs of 10 mm ᴓ stirrups

𝑉𝑐 = 0.17𝜆√𝑓′𝑐 𝑏𝑤 𝑑 𝑉𝑐 = 0.17(1)√20.7(250)(380) 𝑉𝑐 = 73478 𝑁 Nominal shear strength provided by shear reinforcement 𝑉𝑢 = ∅𝑉𝑠 + ∅𝑉𝑐 𝑉𝑢 𝑉𝑠 = − 𝑉𝑐 ∅ 92.83 − 73 .478 𝑉𝑠 = 0.75 𝑉𝑠 = 50.30 𝑘𝑁 Spacing of the two legs of 10 mm ᴓ stirrups 𝜋

𝐴𝑣 = (102 )(2) 4

𝐴𝑣 = 157 𝑚𝑚2 𝑆= 𝑆=

𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑉𝑠

157(275 )(380) 50297.68

𝑆 = 326.19 𝑚𝑚

Say 𝑆 = 𝟑𝟐𝟎 𝒎𝒎

c) Maximum spacing of stirrups.

𝑑

Max. S= nor 600 mm if Vs < 0.33√𝑓′𝑐 𝑏𝑤 𝑑 2 Vs < 0.33√𝑓′𝑐 𝑏𝑤 𝑑

50690 < 0.33√20.7(250)(380)

50690 < 142634 𝑑

Use Max. S= 380

Max. S=

2

2

Max. S=190 mm O.C. NOTE: 𝑑

Max. S= 4 or 300 mm if Vs > 0.33√𝑓′𝑐 𝑏𝑤 𝑑 USE Max. S=190 mm O.C.

EXAMPLE 4: The hollow box beam of a certain building construction has a span of 9 m. it is reinforced for shear with 10 mm 𝜙 bar. the beam carries a factored concentrated load of P U = 134 kN at a point 3 m. from the left end of the simply supported beam, and a total uniform factored load of WU = 14 kN/m distributed throughout the entire span. use 𝜆 = 1.0 for normal weight concrete. f `c= 20.7 MPa, fy =414.6 MPa, fyt= 270 MPa. a) Determine the distance from the left support where the stirrups can be omitted using simplified calculations. b) Determine the shear strength of the stirrups spaced at d/2 mm apart. c) Determine the factored shear-force of the beam with stirrups spaced at d/2 mm apart.

Solution:

a) Distance from the left support where the stirrups can be omitted ∑ 𝑀𝑅2 = 0

9𝑅1 = 134(6) + 14(9)(4.5) 𝑅1 = 152.333 𝑘𝑁 ∑ 𝐹𝑣 = 0

𝑅1 + 𝑅2 = 134 + 14(9) 𝑅2 = 107.667 𝑘𝑁 Shear capacity of concrete beam 𝑏𝑤 = 500 − 300 𝑏𝑤 = 200 𝑚𝑚 𝑑 = 425 𝑚𝑚 𝑉𝑐 = 0.17𝜆√𝑓′𝑐 𝑏𝑤 𝑑 𝑉𝑐 = 0.17(1)√20.7(200)( 425) 𝑉𝑐 = 65744 𝑁

No stirrups are needed if 𝑉𝑢 = ∅𝑉𝑐 𝑉𝑢 = 2 0.75(65.74) 𝑉𝑢 = 2 𝑉𝑢 = 24.65 𝑘𝑁

∅𝑉𝑐 2

𝑦 = 24.67 − 23 .67 𝑦 = 0.98

ℎ = 107 .67 − 23 .67 ℎ = 84 By ratio and proportion: ℎ 𝑦 = 9 𝑥 84 0.98 = 𝑥 9

𝑥 = 0.11 𝑚 Distance from left support where stirrups can be omitted =3+𝑥 = 𝟑. 𝟏𝟏 𝒎

b) Shear strength of the stirrups spaced at d/2 mm apart. 425 2 S = 212.5 mm 𝜋 𝐴𝑣 = (102 )(2) 4 Av = 157 mm2 𝑆=

𝑆=

𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑉𝑠

(157)(270 )(425) 𝑉𝑠 𝑉𝑠 = 𝟖𝟒𝟕𝟖𝟎 𝑵 212.5 =

c) Factored-shear force of the beam with stirrups spaced at d/2 mm apart. 𝑉𝑢 = ∅𝑉𝑠 + ∅𝑉𝑐 𝑉𝑢 = 0.75(𝑉𝑠 + 𝑉𝑐 ) 𝑉𝑢 = 0.75(84.78 + 65.74) 𝑉𝑢 = 𝟏𝟏𝟐. 𝟖𝟗 𝒌𝑵 EXAMPLE 5: A simply supported concrete beam having a width of 350 mm and an effective depth of 520 mm carries a concentrated load of PU = 370 kN at a point 2 m. from the left support, and a uniformly distributed factored load including its own weight WU = 7 kN/m. USE 𝜆 = 1.0 for normal weight concrete with f ′c= 20.7 MPa, fy = fyt = 415 MPa. using simplified calculations. a) Determine the maximum factored shear force at the critical section from C. b) Determine the spacing of the 10 mm 𝜙 stirrups between A-B. c) Determine the spacing of the 10 mm 𝜙 stirrups between B-C.

Solution: a) Maximum factored shear force at the critical section from C. ∑ 𝑀𝑅2 = 0

9𝑅1 = 370(3) + 7(5)(2.5)

𝑅1 = 239.5 𝑘𝑁 ∑ 𝐹𝑣 = 0

𝑅1 + 𝑅2 = 370 + 7(5) 𝑅2 = 165.5 𝑘𝑁

𝑉𝑢 = 165.5 − 7(0.52)

𝑉𝑢 = 𝟏𝟔𝟏. 𝟖𝟔 𝒌𝑵 from C

b) Spacing of the 10 mm ᴓ stirrups between A to B. 𝑉𝑢 = 239.5 − 7(0.52) 𝑉𝑢 = 235.86 𝑘𝑁

Shear strength of the beam: 𝑉𝑐 = 0.17𝜆√𝑓′𝑐 𝑏𝑤 𝑑 𝑉𝑐 = 0.17(1)√20.7(350)( 520) 𝑉𝑐 = 140769 𝑁 Shear capacity of reinforcement: 𝑉𝑢 𝑉𝑠 = − 𝑉𝑐 ∅ 235.86 𝑉𝑠 = − 140 .77 0.75

the

shear

𝑉𝑠 = 173.712 𝑘𝑁 Spacing of 10 mm ᴓ stirrups between A and B: 𝜋 𝐴𝑣 = (102 )(2) 4 Av = 157 mm2 𝑆=

𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑉𝑠

(157)(415)(520) 173712 𝑆 = 195.04 Say 𝑆 = 190 𝑚𝑚 𝑆=

𝑀𝑎𝑥 𝑆 =

𝑑

2

𝑜𝑟 600 𝑚𝑚 if 𝑉𝑠 < 0.33√𝑓′𝑐 𝑏𝑤 𝑑

𝑉𝑠 < 0.33√20.7(350)(520) 173712 < 273256 ok

𝑑 2 520 𝑀𝑎𝑥 𝑆 = 2 𝑀𝑎𝑥 𝑆 = 260 𝑚𝑚 > 190 𝑚𝑚 Use 𝑆 = 𝟏𝟗𝟎 𝒎𝒎 O.C.

𝑀𝑎𝑥 𝑆 =

c) Spacing of the 10 mm ᴓ stirrups between B and C. 𝑉𝑢 = 161.86 𝑘𝑁 𝑉𝑢 − 𝑉𝑐 ∅ 161.86 − 140 .77 𝑉𝑠 = 0.75 𝑉𝑠 = 75.043 𝑘𝑁 𝑉𝑠 =

Spacing of 10 mm ᴓ stirrups between B and C: 𝜋 𝐴𝑣 = (102 )(2) 4 Av = 157 mm2

𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑉𝑠 (157)(415)(520) 𝑆= 75043 𝑆 = 451.48 Say 𝑆 = 450 𝑚𝑚 𝑆=

𝑀𝑎𝑥 𝑆 =

𝑑

2

𝑜𝑟 600 𝑚𝑚 if 𝑉𝑠 < 0.33√𝑓′𝑐 𝑏𝑤 𝑑

𝑉𝑠 < 0.33√20.7(350)(520) 75043 < 273256 ok

𝑑 2 520 𝑀𝑎𝑥 𝑆 = 2 𝑀𝑎𝑥 𝑆 = 260 𝑚𝑚 < 450 𝑚𝑚 Use 𝑆 = 𝟐𝟔𝟎 𝒎𝒎 O.C.

𝑀𝑎𝑥 𝑆 =

EXAMPLE 6: A reinforced concrete beam having a simple span of 8 m carries a uniformly distributed uniform load Wu which includes the weight of the beam. over a distance extending 2 m out from each supports a 10 mm 𝜙 stirups are uniformly spaced at 100 mm on centers, in the middle half of the span, the stirrups spacing is also uniform but increases to 225 mm on centers. Use 𝜆 = 1.0 for normal weight concrete with f ′c= 20.7 MPa, fy = fyt = 415 MPa. a) Determine the value of Wu based on the flexural capacity of the beam. b) Determine the value of Wu based on the shear capacity of the stirrups from A-B. c) Determine the value of Wu based on the shear capacity of the stirrups from B-C.

Solution: a) Value of WU based on the flexural capacity of the beam.

𝐴𝑠 =

𝜋 (252 )(3) = 1472.622 𝑚𝑚2 4

𝐶 =𝑇 0.85𝑓𝑐 ′𝑎𝑏 = 𝐴𝑠 𝑓𝑦 0.85(20.7)𝑎(400 ) = 1472.622(415 ) 𝑎 = 86.834 𝑚𝑚 𝑎 𝑀𝑢 = ∅𝐴𝑠 𝑓𝑦 (𝑑 − ) 2

𝑀𝑢 = 0.9(1472.622)( 415) (450 − 𝑀𝑢 = 223.63 𝑘𝑁. 𝑚

𝑀𝑢 =

86.834 ) 2

𝑊𝑢 𝐿2 8

223.63 =

𝑊𝑢 82 8

𝑊𝑢 = 𝟐𝟕. 𝟗𝟓𝟒 𝒌𝑵⁄𝒎

b) Value of WU based on the shear capacity of the stirrups from A-B. 𝜋 𝐴𝑣 = (102 )(2) 4 Av = 157 mm2 𝐴𝑣 𝑓𝑦𝑡 𝑑 𝑉𝑠 (157)(415)(450) 100 = 𝑉𝑠 𝑉𝑠 = 293197.5 𝑘𝑁 𝑆=

𝑉𝑐 = 0.17𝜆√𝑓′𝑐 𝑏𝑤 𝑑 𝑉𝑐 = 0.17(1)√20.7(400)( 450) 𝑉𝑐 = 139221.59 𝑁 𝑉𝑠 =

𝑉𝑢 − 𝑉𝑐 ∅

𝑉𝑢 − 139.221 0.75 𝑉𝑢 = 324.314 𝑘𝑁

293.198 = 2𝑅 = 8𝑊𝑢 𝑅 = 4𝑊𝑢

𝑉𝑢 = 𝑅 − 𝑊𝑢 (0.45 ) 324.314...