DRCS - Quick Reference (Unit 5)

Column Footing:

In a building, the load path flows from 

• Slabs to Beams, 
• Beams to Columns
• Columns to Column Footings
• Column Footings to the Underlying Soil

Types of Foundations:

• Isolated Footing with Uniform depth
• Isolated Footing with Stepped Footing
• Isolated Footing with Sloped Footing 
• Combined Footings 1) Rectangular 2) Trapezoidal
• Strap Footing (Beam connected columns with slab)

Safe Bearing Capacity of Soil:

• Hard Rocks: (Granite) 2500 to 3500KN/Sqm
• Sand stone, Lime Stone: 1000 to 2500 KN/Sqm
• Soft Rock: 450 to 500 KN/Sqm
• Non-Cohesive Soils: 1) Gravelly Sand: 350 to 450 KN/Sqm 2) Find Sand: 150 to 250 KN/Sqm
• Cohesive Soils: 1)Stiff Clay soft shale: 250 to 350 KN/Sqm 2) Soft Clay: 100 KN/Sqm

General Guidelines:

• Minimum Edge Thickness is 150mm
• Clear Cover is 50mm
• Minimum Steel Reinforcement: same as Slab
• Critical Section for BM is at the face of the column
• Critical Section for Shear: 1) at a distance 'd' from face of column for One-way Shear 2) Punching Shear around the column at a distance of 'd/2' from the face of the column

Area of Footing = Total Load on Soil / Safe Bearing Capacity of Soil

Types of Problems:

1. Design of Square Footing for a given Axial Load

(Design for Working Load Only. Don't consider Factored Load)

Procedure:

1. Given Data
2. Find Total Load to be transferred to soil = Axial Load + Self. wt of footing at 10% of axial load. i.e., Total Axial Load, P = 1.10 x  P 

(if Factored Load is given, Axial Load, P = Pu/1.5)

1. Find Area Required, A = Total Axial Load / SBC
2. Find Side of Column Footing, B =  SQRT of Area
3. Find Projections: l = (B - b)/2 (where b = column width and B = Footing width)
4. Find Pa = Factored Axial Load / Area
5. Find B.M. = Pa² /2
6. Find effective depth, d using Equation: MR = C x Z
7. Find Overall Depth = d + Clear Cover + Half of Dia. of bar
8. Find Ast using Equation: MR = T x Z
9. Find One-way Shear Force: Vu = (Pa x l ) (l-d)
10. Check for Shear: same as Singly Reinforced Beam
11. Check for Two-way Shear: τ_{bd <} τ_permissible
12. _Find τ_bd  = Pa [ L x b – (b+d)(l+d)] / 2 [(b+d)+(l+d)] d
13. τ_permissible = K_s τ_c

τ_c = SQRT of fck

K_s = 0.5 + βc

βc = b/l

1. 
   

1. 17. Draw Plan and Cross Section of Column Footing with Steel Reinforcement Details

2. Design of Square Footing for a given Axial Load

Procedure: same as above except the following:

• Find Size of Rectangular Footings 1) In the same ratio as Column dimentions 2) An equal overhang 'x' kept alround the column (Recommanded)
• Equal Overhang Method: (b+2x) (l+2x) = Area (where b = short side of column, l = long side of column)
• Placement of Reinforcement:

Along long span = same as required

Along short span:

1. Ast for Central Band = 2 Ast / (β + 1)

2. Ast for remaining portion = (Ast - Ast for Central Band)/2

3. Design of Combined Footing for 2 columns carrying Axial Loads

Procedure:

1. Given Data
2. Find Total Load to be transferred to soil =  P1 + P2 + Self. wt of footing at 10% of Total axial load
3. Total Axial Load, P = 1.10 (P1 + P2) (if Factored Load is given, get Axial Load, P = Pu/1.5)
4. Find Area Required, A = Total Axial Load / SBC
5. Assume Width of Footing, B
6. Find Length of Footing, L =  Area / B
7. Find Projections: l = (B - b)/2 (where b = column width and B = Footing width)
8. Find Pa = Factored Axial Load / Area
9. Find B.M. = W² /2
10. Find effective depth, d using Equation: MR = C x Z
11. Find Overall Depth = d + Clear Cover + Half of Dia. of bar
12. Find Ast using Equation: MR = T x Z
13. Find One-way Shear Force: Vu = (Pa x l ) (l-d)
14. Check for Shear: same as Singly Reinforced Beam
15. Check for Two-way Shear: τ_{bd <} τ_permissible
16. _Find τ_bd  = Pa [ L x b – (b+d)(l+d)] / 2 [(b+d)+(l+d)] d
17. τ_permissible = K_s τ_c

τ_c = SQRT fck

K_s = 0.5 + βc

βc = b/l 

1. 
   

1. 17. Draw Plan and Cross Section of Combined Column Footing with Steel Reinforcement Details

DRCS - Quick Reference (Unit 6)

General Design Approach:  Top to Bottom

• Design of Slab (Unit 4)
• Design of Roof Beam (Unit 2 & 3)
• Design of Columns (Unit 6)
• Design of Column Footings (Unit 5)

Columns:

Compression members are called as 'Strut' or 'Pillar' or 'Columns'

Types of Columns: (based on slenderness ration)

• Short Columns - Slenderness Ratio <= 12
• Long Columns  - Slenderness Ratio > 12 

Slenderness Ratio = eff. length of Column / least lateral dimension

Columns are subjected to 

1. Axial Load
2. Axial Load + Uni-axial Bending 
3. Axial Load + Bi-axial Bending 

Load carrying capacity of Column: Pu = 0.4 fck Ac + 0.67 fy Asc 

(for columns subjected to minimum eccentricity of 0.05 times of least lateral dimension of column)

Ac = Cross Section Area of Concrete

Asc = Area of Compression Steel Reinforcement

Design Guidelines:

• Minimum Steel Reinforcement is 0.8%
• Maximum Steel Reinforcement is 6%
• No. of bars: 1) 4 for Square Column 2) 6 for Circular Columns
• Minimum dia. of steel bar is 12mm
• Minimum Clear cover is 40mm
• Spacing of longitudinal bars around the circumference of the column shall be not less than 300mm

Type of Problems:

1. Design Column of given size subjected to Axial Load

Procedure:

1. Given Data
2. Assume % of Steel Reinforcement (Minimum is 0.8)
3. Find Steel Required by equating, Pu = 0.4 fck Ac + 0.67 fy Asc 
4. Assume dia. of longitudinal steel bar
5. Provide No. of bars (minimum 4 for Square Column)
6. Find Dia. of Later Tie: greater of 1) 1/4 of dia. of main bar 2) 6mm
7. Find Pitch or Spacing of Lateral Tie: least of 1) Least Lateral Dimension of Column 2) 16 times of smallest longitudinal bar 3) 300mm
8. Draw Longitudinal and Cross Section of Column with Steel Reinforcement Details

2. Design Column of given size subjected to Axial Load + Uni-axial Bending

Procedure:

1. Given Data
2. Find d'/D
3. Find Pu / fck b D
4. Find Mu / fck b D²
5. Referring Charts 27 to 62 based on Grade of Steel, Shape of Column, d'/D, Reinforcement Placement (Opposite Faces or Equally All faces): Find p/fck value 
6. Find % of Steel, p = fck x value obtained for chart
7. Check for Minimum and Maximum Reinforcement
8. Provide No. of bars
9. Find Dia. of Later Tie: greater of 1) 1/4 of dia. of main bar 2) 6mm
10. Find Pitch or Spacing of Lateral Tie: least of 1) Least Lateral Dimension of Column 2) 16 times of smallest longitudinal bar 3) 300mm
11. Draw Longitudinal and Cross Section of Column with Steel Reinforcement Details

3. Design Column of given size subjected to Axial Load + Bi-axial Bending

Procedure:

1. Given Data
2. Find d'/D\
3. 1st Trail: Assume % of Steel
4. Find Load Carrying Capacity, Puz
5. Find Pu/Puz where Pu = Load coming on the column
6. Find α_{n (based Pu/Puz values)}
7. Find Pu / fck b D
8. Find p/fck = % of steel / fck
9. Find Mux1 / fck b D² from chart
10. Similary Find Muy1 / fck b D² from chart
11. Apply Check: (M_ux/M_ux1) ^αn   + (M_uy/M_uy1) ^αn   <= 1.0
12. If value is too less than 1, Assume less % of steel than before and do steps 3 to 11 again
13. Once Check is OK, find Longitudinal Steel Reinforcement
14. Provide No. of bars
15. Find Dia. of Later Tie: greater of 1) 1/4 of dia. of main bar 2) 6mm
16. Find Pitch or Spacing of Lateral Tie: least of 1) Least Lateral Dimension of Column 2) 16 times of smallest longitudinal bar 3) 300mm
17. Draw Longitudinal and Cross Section of Column with Steel Reinforcement Details

Note: (from IS Code book)

_{for Pu/Puz <= 0.2,} α_{n is 1.0}

_{for Pu/Puz >= 0.8,} α_{n is 2.0}

DRCS - Quick Reference (Unit 4)

Slabs:

A slab is a plate like structural element transferring the load on to the supporting beams some times directly on to Columns, basically by bending action.

For design purpose, a slab is considered to be a beam of unit width

Types of Slabs:

By Shape: 1) Rectangular 2) Non-Rectangular
By Support Condition: 1) Cantilever 2) Simply Supported 3) Continuous
By Span Ratio: 1) One-way Slab 2) Two-way Slab

One-way Slab: Ly/Lx >2
Two-way Slab: Ly/Lx <=2

Ly = Effective long span
Lx = Effective short span

Types of Problems:

1. Design of Simply Supported One-way Slab (Slab over load bearing walls)

Procedure:

1. Given Data
2. Assume Slab Thickness @ 40mm per metre of short span
3. Assume Clear Cover = 15mm
4. Assume dia. of main steel bar is 10mm
5. Find Effective Span = Clear Span + 2 x Half of Support Width
6. Find Load on Slab 1) Self weight of slab 2) Floor Finish Load at 1 KN/Sqm 3) Live Load
7. Find Max. BM = Wl²/8 for UDL and S.F. = Wl/2 for UDL
8. Check for Depth: MR = C x Z
9. Find Ast: MR = T x Z
10. Spacing of Main Steel Reinforcement, S = 1000 x (ast / Ast)
11. Check for Maximum spacing of main steel bars is not to exceed least of  1) 3d  2) 300mm
12. Provide Distribution Steel at 0.12% = (0.12/100) * bd  (where b = 1000mm)
13. Check for Maximum spacing of Distribution Steel bars is not to exceed least of 1) 5d 2) 450mm 
14. Check for Shear same as Singly Reinforced Beam
15. Check for Bond same as Singly Reinforced Beam
16. Draw Cross Section and Plan of Slab with Steel Reinforcement Details

2. Design of Continuous One-way Slab (Slab over load bearing walls)

Procedure: Same as Simply Supported Slab except the following:

• Assume thickness of slab around 1/30 of span subjected to a minimum of 100mm
• Support Moment, Mu -ve = W_dl²/12 + W_l²/9
• Mid Span Moment, Mu +ve = W_dl²/16 + W_l²/12

W_d = Dead Load

W_l = Live Load

3. Design of Cantilever Slab

Procedure: Same as Simply Supported Slab except the following:

• Assume thickness of slab around 1/7 of span subjected to a minimum of 100mm
• Support Moment, Mu -ve = Wl²/2 for UDL

Note:

1. Assumption of Live Load for Roof Slab (top floor) = 1.5 KN/Sqm
2. Assumption of Live Load for Floor Slab (Residential Building) = 3.0 KN/Sqm
3. Assumption of Live Load for Floor Slab (Commercial Building) = 4.0 KN/Sqm
4. Bent-up bars should be bent at an angle of 45 degree at a distance of L/7 from inner edge of support

4. Design of Simply Supported Two-way Slab (Slab over load bearing walls)

Methods:

Using Rankine Grashoff's Theory

M_x = α_x w l_x²

M_y = α_y w l_x²

^Where 

α_{x =}  r⁴ / 8 (1+r⁴)

α_{y =} r² / 8 (1+r⁴)

r = ly / lx

IS code Method:

Use IS 456:2000 (table 26) for difference ly/lx

Procedure:

1. Given Data
2. Assume Slab Thickness @ 40mm per metre of short span
3. Assume Clear Cover = 15mm
4. Assume dia. of main steel bar is 10mm
5. Find Effective Span = Clear Span + 2 x Half of Support Width
6. Find Load on Slab 1) Self weight of slab 2) Floor Finish Load at 1 KN/Sqm 3) Live Load
7. Find Max. BM in both directions: Mx and My (Use Any one of methods given above)
8. Check for Depth: MR = C x Z
9. Find Main (along short span )As:: Mx = T1 x Z1
10. Spacing of Main Steel Reinforcement, S = 1000 x (ast / Ast)
11. Check for Maximum spacing of main steel bars is not to exceed least of  1) 3d  2) 300mm
12. Find Distribution Steel (along long span) Ast: My = T2 x Z2
13. Spacing of Main Steel Reinforcement, S = 1000 x (ast / Ast)
14. Check for Maximum spacing of main steel bars is not to exceed least of  1) 3d  2) 300mm
15. Check for Shear same as Singly Reinforced Beam
16. Check for Bond same as Singly Reinforced Beam
17. Draw Cross Section and Plan of Slab with Steel Reinforcement Details

5. Design of Simply Supported Two-way Slab (with Provision for Torsion ) - Using IS code Method

Procedure:  Same as above except the following:

1. As per clause D-1.8 of Code, torsion reinforcement must be provided on all simply supported corners of the slab.
2. At each of the four corners this provided in 4 layers in a Square portion of lx/5
3. Ast for Torsion = (3/4 of Ast required for Short Span ) x (lx/5)

DRCS - Quick Reference (Unit 3)

Unit 3:

Shear in beams:

Forms of Shear Reinforcement:

1. Vertical Stirrups
2. Inclined Stirrups
3. Bent-up bars

Max. shear generally occur at a distance 'd' from the support. (d = effective depth of beam)

Types of Problems:

1. Design Shear Reinforcement (Vertical Stirrups only)

Procedure:

1. Given Data
2. Find Max. Shear Force, Vu (wl/2 for Simply Supported beam, wl for cantilever beam under udl)
3. Find Shear Stress Developed, Tv (tow v) = Vu / b d
4. Find Shear Strength of Concrete, Vc = Tc  bd 
5. Find Tc based on 1) % of Steel Reinforcement used 2) Grade of Concrete (Use IS Code book - Table 19)
6. Find % of Steel = (Ast x 100) / b d
7. If Vu > Vc, then Shear Reinforcement has to be provided
8. Find Shear Resistance offered by Vertical Stirrups, Vus = Vu - Vc
9. Assume 2 legged stirrups of 8mm dia bar
10. Find Spacing of Stirrups, Sv = (0.87fy Ast d) / Vus
11. Check for minimum spacing from the least of 1) 0.75d  2) 300mm 3) using Eqn.: Asv / b. Sv = 0.4 / 0.87fy
12. Draw cross section of beam and show shear reinforcement details

2. Design Shear Reinforcement (Vertical Stirrups + Bent-up Bars)

Procedure:

1. Given Data
2. Find Max. Shear Force, Vu (wl/2 for Simply Supported beam, wl for cantilever beam under udl)
3. Find Shear Stress Developed, Tv (tow v) = Vu / b d
4. Find Shear Strength of Concrete, Vc = Tc  bd 
5. Find based Tc based on 1) % of Steel Reinforcement used 2) Grade of Concrete (Use IS Code book - Table 19)
6. Find % of Steel = (Ast x 100) / b d
7. If Vu > Vc, then Shear Reinforcement has to be provided
8. Find Shear Resistance Required, Vus1 = Vu - Vc
9. Vus1 = Vus + Vub
10. Shear Resistance offered by Bent-up bars, Vub = 0.87 fy Asb (Sin α) (Limited to 1/2 of Vus1)
11. Assume α is 45⁰
12. Find Shear Resistance offered by Vertical Stirrups, Vus = Vus1 - Vub
13. Assume 2 legged stirrups of 8mm dia bar
14. Find Spacing of Stirrups, Sv = (0.87fy Ast d) / Vus
15. Check for minimum spacing from the least of 1) 0.75d  2) 300mm 3) using Eqn.: Asv / b. Sv = 0.4 / 0.87fy
16. Draw cross section of beam and show shear reinforcement details

Torsion: (Clause 41.1 of IS: 456-2000)

A beam subjected to torsion develops spiral cracks in structure

Types of Torsion:

1. Equilibrium Torsion - Ex. Sunshade, Cantilever Balcony
2. Compatibility Torsion - Ex. Ring Beams, Curved Beams

Equilibrium Torsion:

Torsion Resistance can be provided by providing additional Longitudinal Reinforcement and Shear Reinforcement

Types of Problem:

1. Design beam subjected to a Torsion

Procedure:

1. Given Data
2. Assume Clear Cover and find effective Cover (use 8mm dia for vertical stirrups)
3. Design Longitudinal Reinforcement, Mel = Mu + Mt
4. Find Mt = Tu [(1 + D/b) / 1.7] where Tu = Torsion Moment
5. Find Steel Reinforcement same as Singly Reinforced Beam
6. Design Max. Shear Force due to Torsion, Ve = Vu + (1.6 Tu / b)
7. Find Shear Reinforcement. Assume 8mm dia 2 legged stirrups
8. Spacing of Stirrups, Sv = Asv / [ Tu / (b1 d1 0.87fy) + Vu / (2.5 d1 0.87 fy)]
9. Find b1 = c/c of corner rods in the direction of width
10. Find d1 = c/c of corner rods in the direction of depth
11. Check for minimum spacing 1) Sv = Asv 0.87fy / (Tve - Tc) b 2) Sv = x1 3) Sv = (x1 + y1)/4  4) Sv = 300mm
12. x1 = c/c of stirrup width
13. y1 = c/c of stirrup depth
14. Provide Side Face Reinforcement of 12mm dia. bars 2 nos at the sides in the middle, if depth exceeds 450mm
15. Draw Cross Section of beam with Steel reinforcement details

Bond:

Bond is the binding force between steel and concrete that makes them act in an integrated manner.

Types:

• Anchorage Bond - Ex. Hooks, U - Bends
• Development Bond - reinforcement bars embedded into the concrete by the required length

Development Length of Bar, Ld = φ 0.87 fy / 4 τbd

Find τbd from IS Code for given grade of Concrete

Note:

• τbd for Tension can be increased by 60% for HYSD bars like Fe415 and Fe500 = 1.60 τbd
• τbd for Compression can be increased by 25% for plain bars like Fe250 = 1.25 τbd
• τbd for Compression can be increased by 60% for plain bars like Fe250 = = 1.60 x 1.25 τbd

Check for Bond:

In order to ensure bond between the positive moment reinforcement and the surrounding concrete, the dia. of tension reinforcement is limited to 

Ld < = (M1 / V) + Lo

Anchorage for Reinforcing bars:

• hooks
• bends - 1) 4 times dia. of bar for 45 degrees bend 2) 16 times of dia. of bar for U-bend

DRCS - Quick Reference (Unit 1 & 2)

Unit I:

Grade of Concrete:
Grade of Concrete is a measure of its strength and is designated by writing the value of the 28days cube strength in KN/Sqm after an alphabet 'M'

Various Grades:
M15, M20, M25, M30, M35,....

As per IS:256-2000, minimum grade of M20 should be used for design

Grades of Steel:
Fe250 - Mild Steel - will have characterestic strength of 250N/Sqm
Fe415 & Fe500 - High Yield Strength Deformed (HYSD) bars

Most commonly used grade is Fe415

Unit II

Flexural Members or Beams:
1. Singly Reinforced Beams - simply supported / fixed/ cantilever beams where depth is not a constraint.
2. Doubly Reinforced Beams - designed when depth of beam is limited. Usually at the supports of beam.
3. T Beams & L Beams- beams can be designed as T-beam or L-beam by taking the advantage of compressible strength of slab. Usually at the mid span of simply supported/fixed beams

Methods of Design:
1. Working Stress Method
2. Limit State Design Method

In Limit State Design Method:

Based on depth of NA, beams are

1. Balanced beams (if Xu = Xumax)

 MR = C x Z or T x Z

2. Under Reinforced Beams (if Xu< Xumax)

MR = T x Z

3. Over Reinforced Beams (if Xu > Xumax)

MR = C x Z

Where

C = 0.36 fck b Xu
T = 0.87 fy Ast
Z = d-0.42Xu

Xumax Equation: Xumax / 0.0035 = d-xumax / (0.87fy/Es + 0.002)

Fe250 -  0.53d
Fe415 -  0.48d
Fe500 -  0.46d

d = effective depth = Overall depth of beam - Clear Cover - Half of Dia. of Steel bar

Clear Cover = 25mm for beams

Xu:

Find Xu by Equating C = T

Singly Reinforced Beams:

Type of Problems:

1. Find Ultimate Moment of Resistance of a given Beam

Procedure:

1. Write Given Data
2. Find Xu by equating C=T
3. Find Xumax for given grade of Steel
4. Find section of beam (balanced / under reinforced / over reinforced)
5. Find MR based on section
6. Draw Cross Section of beam and show steel reinforcement details

2. Design Beam for a given Bending Moment

Procedure

1. Write Given Data
2. Find Load, Factored Load, and Factored BM
3. Find Xumax for a given grade of steel (assume balanced section)
4. Find 'd' by equating BM =C x Z
5. Find Overall depth of beam
6. Find Steel Reinforcement by equating BM = T x Z
7. Find No. of bars. (Assume 12mm/16mm/20mm/25mm dia based on Ast required)
8. Draw Cross Section of beam and show steel reinforcement details

Check: D is in between 1/10 to 1/15 of the span

Doubly Reinforced Beams:

Type of Problems:

1. Find Ultimate Moment of Resistance of a given Beam

Procedure:

1. Write Given Data
2. Find Mu = Mu limited + Mu2
3. Find Mu Limited = C1 x Z1 or T1 x Z1 (assume balanced section like singly reinforced beam)
4. Find Mu2 = C2 x Z2 or T2 x Z2
5. Find C1 = 0.36fck b Xu
6. Find Z1 = d-0.42Xu
7. Find T1 = 0.87fy Ast
8. Find C2 = fsc Asc - fcc Ac
9. Find T2 = d-d'
10. Find d' = d - top clear cover - half of dia of top steel bar
11. Draw Cross Section of beam and show steel reinforcement details

2. Design Beam for a given Bending Moment

Procedure

1. Write Given Data
2. Find Load, Factored Load, and Factored BM
3. Find Mu limited = C x Z
4. If Mu > Mu limited, then design as Doubly reinforced beam
5. Mu = Mu limited + Mu2
6. Find Mu2 = Mu - Mu limited
7. Find Ast1 by equating Mu limited = T1 x Z1
8. Find Ast2 by equating Mu2 = T1 x Z2
9. Total Ast = Ast1 + Ast2
10. Find Asc by equating Mu2 = T2 x Z2
11. Draw Cross Section of beam and show steel reinforcement details

T - Beams & L - Beams:

Type of Problems:

1. Find Ultimate Moment of Resistance of given T-beam or L-Beam

Procedure:

1. Write Given Data
2. Assume Xu lies in Slab and find Xu by Equating C = T
3. Find C = 0.36 fck bf
4. Find T = 0.87 fy Ast
5. bf for T - beam = L/6 + bw + 6 Df (for multiple spans)
6. bf for L - beam = L/12 + bw + 3 Df  (for multiple spans)
7. Find MR based on Xu lies in 1) xu < Df    2)  3/7 Xu > Df or Df/d < 02  3) 3/7 Xu< Df or Df/d > 0.2
8. Draw Cross Section of beam and show steel reinforcement details

2. Design T - Beam or L - Beam:

Procedure:

1. Write Given Data
2. Find Load, Factored Load, and Factored BM
3. Assume Xu lies in Slab
4. Assume Depth of Beam (1/2 to 1/15 of span)
5. Find Steel Reinforcement by equating BM = T x Z
6. Find No. of bars. (Assume 12mm/16mm/20mm/25mm dia based on Ast required)
7. Draw Cross Section of beam and show steel reinforcement details

Note: bf for Isolated T-beam and L-beam can be found from IS Code

Tips to get Maximum Marks:

1. Given Data
2. Assumptions
3. Design Procedure
4. Write references taken from IS Code book
5. Formula or Equation with terminology Ex. MR = Momnet of Resistance, d = effective depth
6. Draw Cross Sections of structures with Steel Reinforcement Details
7. Units

Surveying Lab External - Exam Pattern

Sample Questions:

A) Experiment (30 Marks : 10 for Theory and 20 for Practicals) 

• Find Area using Chain Surveying (Traversing / Triangulation Method)
• Find Area using Compass Surveying (Traversing)
• Find Area using Plane Table Surveying (Radiation / Intersection Method)
• Find Distance between two in-accessible points (Chain Surveying / Compass Surveying)
• Find Level Difference between two given points (Fly Leveling)

B) Short Notes (10 Marks)

• Objective & Uses of Surveying
• Principle of Surveying
• Direct Ranging vs Indirect Ranging
• Explain:
• True Meridian
• Magnetic Meridian
• Arbitrary Meridian
• Magnetic Bearing (WCB / RB)
• Bearing (Fore Bearing / Back Bearing)
• Dip and Declination
• Local Attraction and Correction
• Bench Marks (GTS Bench-marks / Permanent Bench-marks / Arbitrary Bench-marks/ Temporary Bench-marks)
• Contour Interval vs Horizontal Equivalent

C) Viva Voce (10 Marks)

External Lab Exam Schedule

BHARAT INSTITUTE OF ENGINEERING AND TECHNOLOGY Civil Engineering Department External Lab Exam Schedule
S. NO	Year - Lab	Class	Internal Examiner	Date
1	Survey Lab	II A	P. Rambabu	9-Nov-12
2	Strength of Materials Lab		K. Vijaya Kumar	15-Nov-12
3	Strength of Materials Lab	II B	C. S. Rao	10-Nov-12
4	Survey Lab		N. Ramachandra Rao	15-Nov-12
5	Engineering Geology Lab	III	Arati Reddy	9-Nov-12
6	FM & HM Lab		C. S. Rao	12-Nov-12
7	Concrete Technology &Transporation Engineering Lab	IV	D Lakshmi Lavanya	10-Nov-12
8	Environmental Engineering Lab		P Sudha Rani	14-Nov-12

RCC Model Questions for Mid Exam 2

Unit 5: Column Footings:

1.    Design an isolated square footing for the column of 300mmx300mm size carrying capacity of 1000KN taking safe bearing capacity of the soil as 300KN/m². Use M20 concrete and Fe 415 steel.

2.    Design an isolated Rectangular footing for the column of 300mmx450mm size carrying capacity of 1000KN taking safe bearing capacity of the soil as 300KN/m². Use M20 concrete and Fe 415 steel.

3.    Design a combined Rectangular footing for column of 300mm x 400mm size carrying capacity of 1200KN and 1800KN. Spacing between Columns is 3m. Safe bearing capacity of the soil as 250KN/m². Width is limited to 1.5m. Use M20 concrete and Fe 415 steel.

4.    Design a combined Trapezoidal footing for column of 300mm x 400mm size carrying capacity of 1200KN and 1800KN. Spacing between Columns is 3m. Safe bearing capacity of the soil as 250KN/m². Width is limited to 1.5m one end and 2.5m at other end. Use M20 concrete and Fe 415

Unit 6: Columns

5.    Design a Circular column to carry a factored axial load of1200 KN. Use M20 concrete and Fe-415 steel. Show the details of reinforcement.

6.    Design a Rectangular column of 300mm x 450mm to carry a factored axial load of1500 KN. Use M20 concrete and Fe-415 steel. Show the details of reinforcement.

7.    Design a Square column 300mmx300mm size to carry a factored axial load of1200 KN and 100KN-M moment. Use M20 concrete and Fe-415 steel. Show the details of reinforcement.

Unit 7: Serviceability

8.    Deflection limits in IS: 456-2000

9.    General aspects of serviceability

10.  Calculation of deflection

11.  Explain Cracking in structural concrete members

Unit 8: Design of Staircase and Canopy

12.  Design one flight of dog legged stair case supported on beams of 230m wide. No. of steps are 12. Tread is 300mm. Rise is 150mm. Live load 5 KN/m^2. Use M20 & Fe415. Beams are provided at top and bottom of risers.

13.  Design one flight of dog legged stair case supported on beams of 230m wide. No. of steps are 12. Tread is 300mm. Rise is 150mm. Live load 5 KN/m^2. Use M20 & Fe415. Beams are provided at the edge of landings

Landing width is 1.20m

14.  Design a Cantilever Portico of span 4m and 6m long. Use M20 & Fe415