Structural Design [संरचनात्मक डिज़ाइन]

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यह सामग्री किसी भी मौलिकता का दावा नहीं करती है और इसे निर्धारित पाठ्यपुस्तकों के विकल्प के रूप में इस्तेमाल नहीं किया जा सकता है। मैं उन विभिन्न ओपन स्रोतों और NPTEL/SWAYAM पाठ्य सामग्री को मान्यता देना चाहूंगा जिनसे व्याख्यान नोट तैयार किया गया है। जानकारी का स्वामित्व संबंधित लेखकों या संस्थानों के पास है जहाँ ओपन स्रोत की सामग्री तैयार की गई थी। आगे, यह दस्तावेज़ किसी व्यावसायिक उद्देश्यों के लिए उपयोग करने के लिए नहीं है, और ब्लॉगर के मालिक किसी भी मुद्दों के लिए, कानूनी या अन्यथा, जिम्मेदार नहीं हैं जो इसके उपयोग से उत्पन्न होते हैं।

 Course Name: Structural Design [संरचनात्मक डिज़ाइन]

Course Code: MZSCEA-SD-06

Content Creator: Dr. Mohd. Zameeruddin

[1] While designing a beam, what points are considered from the following?

A. Bending Moment

B. Shear Force

C. Neither Shear force nor Bending moment

D. Both Shear force and Bending moment

Answer: D – Both Shear Force and Bending Moment

[2] What is the ratio of the strength of material to the permissible stress is called?

A. Factor of Safety

B. Shear factor

C. Creep factor

D. Strength Modulus

Answer: A – Factor of Safety

[3] The concrete in the ______ zone should not be taken into account while determining the neutral axis.

A. Compression

B. Neutral Zone

C. Centroidal Zone

D. Tension Zone

Answer: D-Tension Zone

[4] Which of the following is the function of the transverse reinforcement in a column?

A. To prevent longitudinal buckling of longitudinal reinforcement

B. To prevent certain brittle failure

C. To impart certain ductility to the column

D. To reduce effect of creep

Answer: A-To prevent longitudinal buckling of longitudinal reinforcement

[5] A column is 5m long (effective) and 500 mm in diameter. What kind of column is this?

A. long

B. Slender

C. Short

D. Thick

Answer: C-Short

[6] What is the horizontal distance or spacing between the reinforcement bars of an 80 mm thick slab?

A. 340 mm

B. 440 mm

C. 240 mm

D. 540 mm

Answer: C-240 mm

Explanation: According to the codal provision, the spacing = 3*thickness of the slab or 300mm whichever is less. Hence the spacing s = 3*80mm=240mm.

[7] Which one is not the function of transverse reinforcement?

A. It distributes the load more evenly and uniformly on the slab

B. It prevents the shrinkage or temperature effect

C. It keeps the main reinforcement in position

D. It works as the main reinforcement

Answer: D-It works as the main reinforcement

[8] What is the objective of providing foundation to a structure?

A. To provide a base to the structure

B. To stabilize the soil below the structure

C. To distribute the load to the soil

D. For the compaction of the soil below the structure

Answer: C-To distribute the load to the soil

[9] Which of the following is not an assumption for working stress design method?

a. Tensile strength of concrete is considered

b. Bond between steel and concrete is perfect within the elastic limit of steel

c. Concrete is elastic

d. A section which is plane before bending remains plane after bending

Answer: a-Tensile strength of concrete is considered

[10] The stress strain curve of concrete as per IS-456 is

A. Perfect straight line up to failure

B. Straight line up to 0.002 strain value and then parabolic up to failure

C. Parabolic up to 0.002 strain value and then uniform up to failure

D. Linear up to 0.002 strain value and then uniform up to failure

Answer: C- Parabolic up to 0.002 strain value and then uniform up to failure

[11] Which of the following is not a limit state of serviceability?

A. Deflection

B. Cracking

C. Torsion

D. Durability

Answer: C-Torsion

[12] Limiting moment of Resistance of R.C. beam for Fe415 grade steel is

a. M_u lim= 0.138 F_ck bd²

b. M_u lim= 0.148 F_ck bd²

c. M_u lim= 0.133 F_ck bd²

d. M_u lim= 0.130 F_ck bd²

Answer: a. M_u lim= 0.138 F_ck bd²

A)  Differentiate between Under-Reinforced and Over –Reinforced Section

Under- Reinforced Section	Over –Reinforced Section
A concrete element is considered under-reinforced when it has less steel reinforcement than that required by balanced section. This means that the concrete stresses do not reach its maximum allowable value while steel reaches its maximum permissible value.	A concrete element is considered over-reinforced when it has more steel reinforcement than required for a balanced section. In this case, stress in concrete reaches its maximum allowable value earlier than that in steel.
The neutral axis depth will be smaller than that in balance section as position of neutral axis shift upwards.	The neutral axis depth will be greater than that in balance section as position of neutral axis shift towards steel.
Moment of resistance is governed by allowable tensile stress in steel	Moment of resistance is governed by compressive stress in concrete
Under-reinforced sections typically exhibit more predictable behaviour. When they reach their load capacity, they tend to fail gradually, allowing for warning signs (such as cracking) before ultimate failure.	Over-reinforced sections can lead to sudden failure with little or no warning. The concrete may crack and fail before the steel has fully yielded, which can be dangerous because it lacks the ductility that allows for warning signs.
Under-reinforcement is often preferred in structural design because it provides safety through ductility. It ensures that the concrete will crack and fail before the steel reinforcement can yield, giving engineers and occupants time to react.	Over-reinforcement is generally not desirable in structural design because it can compromise safety. It can lead to brittle failure modes, where the structure collapses suddenly without adequate warning.

Find the moment of resistance of RC beam 250 mm wide and 500 mm effective depth. The area of tensile steel is 1200 mm². Also calculate the actual stresses in concrete and steel at the maximum moment of resistance, if permissible stresses in steel and concrete are 140 N/mm² and 5 N/mm² respectively. m= 18.66

Solution:

Step 1: Given Data

    Width of beam = 250 mm

    Effective depth = 500 mm

    Area of tension stee (A_st) = 1200 mm²

    Permissible stress in steel = 140 MPa

    Permissible Stress in Concrete = 5 MPa

    Modular ratio = 18.66

    Evaluate Maximum Stress in Concrete and Steel

Step 2: Determination of Neutral Axis Depth

    bx²/2 = m Ast (d-x)

    250 x²/2 = 18.67 x 1200 x (500 -x)

    125 x² + 22404 x -11202000 =0 

    x² + 179.23 x -89616 =0 

    x = 222.87 mm

Step 3: Depth of Critical Neutral Axis

    k = [1/ (1+ (σ_st /m* σ_cbc)] = [1/ (1+(140/18.66*5)] = 0.40

    x_u,max = k*d = 0.4*500 = 200 mm

    x > x_umax, hence the section is over-reinforced

    j = (1-0.4/3) = 0.87

    z = d -x/3= 500-(222.87/3) = 425.71 mm

 Step 4: Maximum Bending Moment

    M =b*x*1/2* σ_cbc*(d-x/3)

    M = 250*222.87*0.5*5*(500-222.87/3) = 59.298 kNm

Step 5: Maximum Stress in Concrete

    Moment of resistance      = b*x*(σ_cbc/2)*z

                    59.28 x 10⁶ = 250*222.87*0.5* σ_cbc*427.71

                                  σ_cbc = 4.975 Mpa

Step 6: Maximum Stress in Steel

    Moment of resistance      = A_st* σ_st*z

                    59.28 x 10⁶  = 1200* σ_st*(500-222.87/3)

                                 σ_st     = 116.041 MPa

Difference Between Limit State Method and Working Stress Method

Limit State Method [LSM]	Working Stress Method [WSM]
LSM considers ultimate load failure and ensures usability	WSM limits stress within elastic range, making it overly conservative
LSM allows efficient material use, reducing costs.	WSM requires larger sections, increasing construction costs
LSM accounts for realistic conditions like overloading and serviceability. Factor of safety for loads and stresses are considered.	WSM assumes all loads are within safe limits, ignoring possible failures. Factor of safety is considered for stresses only.
LSM ensures that structures can withstand loads while maintaining serviceability, making it more suitable for modern construction.	Working Stress Method focuses only on elastic stresses

Derive the stress block parameters for flexure

Consider a beam section with overall dimension as width (b) and overall depth (D) of the section. Let the effective depth be ‘d’ and effective cover (d’) and neutral axis depth (x).

Let     ε_c = maximum strain in concrete

          Ε_s = maximum strain at the centroid of the steel

          σ_cbc = Maximum compressive stress in concrete, in bending

          σ_st = Stress in steel

            m = modular ratio

Fig. 1: Stress-strain distribution of singly reinforced concrete beam

From strain distribution

(ε_c/ε_s) = (x/d-x)

(d-x)/x = ε_s/ε_c

(d/x-1) = σ_st E_c/E_s σ_cbc

(d/x-1) = σ_st /m σ_cbc

d/x = 1+σ_st /m σ_cbc

x = 1/[1+σ_st /m σ_cbc]d

x = k d

k = 1/[1+σ_st /m σ_cbc]

From Stress distribution diagram

Total compressive force C =1/2*b*x* σ_cbc

Total tensile force T = σ_st * A_st

Lever Arm (z) = d – x/3 = d-kd/3 =d(1-k/3) = j. d   where j = 1-k/3

Moment of resistance      = C*z

                                        = 1/2*b*x* σ_cbc*j.d = 1/2*b*kd* σ_cbc*j.d = Q*b*d²

                                                            Q = 1/2*k* σ_cbc*j

The parameters k, j and Q are said to be stress block parameter

Singly reinforced beam 250 mm × 500 mm in section is reinforced with 4 bars of 16 mm diameter with an effective cover of 25 mm. Effective span of the beam is 5 m. Assuming M20 concrete and Fe 415 steel, determine the central concentrated load P that can be carried by the beam in addition to its self-weight. Use LSM.

Solution:

Step 1: Given Data                                                                             

Width of beam = 250 mm

Effective cover = 25 mm

Effective depth = 500 - 25 = 475 mm

Length of beam = 5 m = 5000 mm

Area of tension stee (Ast) = 4 bars of 16mm dia. =804.25 mm2

Steel 415

M 20 Concrete

Step 2: Determination of Neutral Axis Depth                                             

x_u/d = (0.87f_yA_st)/(0.36f_ckbd)

x_u/d = (0.87*415*804.25)/ (0.36*20*250*500)

x_u/d = 290374.46 / 900000 = 0.322

x_u = 153.25 mm

(x_u,_lim/d) = [700/(1100+0.87*f_y] = [700/ (1100+0.87*415)] = 0.48

x_u,_lim = 0.48*475 = 240 mm

hence x_u < x_u,lim its under reinforced

Step 3: Ultimate moment of resistance

Mu     = 0.87f_yA_std[1-(A_stf_y/bdf_ck)]

          = 0.87*415*804.25*475 [1-(804.25*415/250*475*20)]

          = 118.617 kNm

Step 4: Bending moment over the span of beam

            (Assuming simply supported end)

M = W*L/4

118.617 x 10⁶ = W*5000/4

W = 94.893 kN

Self-weight of the beam = 0.25*0.5*25 = 3.125 kN/m

Equivalent Point load = 15.625

Additional load that beam can carry = 94.893-15.625 = 79.268 kN

Design a simple supported singly reinforced beam of 230 mm wide which carries a factored load of 30 kN/m including its self-weight. The span of beam is 6 m. Design the beam for flexure and shear by limit state method if materials are M 25 concrete and HYSD bars of Fe 415.

Solution:

Step 1: Given Data                                                                   

Width of beam = 230 mm

Length of beam = 6 m = 6000 mm

Steel 415

M 25 Concrete

Ru = 3.45

Factored load = 30 kN/m

Step 2: Factored Moment

M_u = wL²/8 = 30 x (6)² /8 = 135 kNm

Step 3: Design of Beam

M_u,lim = 0.138f_ckbd² = R_ubd²

Effective depth = SQRT (135 x 10⁶/3.45*230) = 412.47 mm, Say 415 mm

Overall depth = 415 + 35 = 450 mm

Adopt size of the beam to be 230 mm x 450 mm

Step 4: Design of reinforcement

M_u/bd² = [135 x 10⁶/(230 x 415²)] = 3.408

From table 1 SP 16

 P_t = 1.717;   A_st = P_t (b x d/100) = 1.717*(230x415/100) = 1632.19 mm²

Alternative

A_st = (0.5*f_ck/f_y) *[1-SQRT(1-(4.6M_u/f_ckbd²)] (b*d)

A_st = (0.5*25/415) *[1-SQRT(1-(4.6*135x10⁶/25*230*415²)] (230*450)

A_st = 1118.375 mm2, hence provide 4 nos-20 mm diameter bars

Step 5: Check for shear

Maximum factored shear = wl/2 = 30*6/2 = 90 kN

Nominal shear stress = 90 x 10³/ (230 * 415) = 0.943 MPa

Actual P_t = (4*314.56*100)/(230*415) = 1.31

Permissible shear stress;

For 1.25 ----- 0.70

For 1.50 ----- 0.74

For 1.31 = 0.70 + [(0.74-0.70)/(1.50-1.25)]*(1.31-1.25) = 0.7096 MPa

Hence safe,

Provide minimum shear reinforcement 8mm diameter 2 legged at

Sv = (A_sv*f_y)/(0.4*b) = (2 *50.27*415)/(0.4*230) = 453.53 mm, say 450 mm

A T-beam of effective flange width of 1800 mm, thickness of slab 100 mm, width of rib 230 mm and effective depth of 500 mm is reinforced with 4 Nos. 25 mm diameter bars. Calculate the factored moment of resistance if M 20 and Fe 500 is used. Use LSM.

Solution:

Step 1: Given Data                                                                                        

Width of flange = 1800 mm

Width of beam = 230 mm

Depth of flange = 100 mm

Effective depth of rib = 500 mm

Area of steel = 4-25 mm diameter = 4*490.87 = 1963.50 mm²

Steel 500 and M 20 Concrete

Step 2: Determination of Neutral Axis Depth                                       

Depth of flange/effective depth of rib = 100/500 = 0.20

x_u = (0.87*f_y*A_st)/ (0.36*f_ck*b_f)

x_u = (0.87*500*1963.50)/ (0.36*20*1800)

x_u = 65.90 mm < 100 mm

x_{u, max}/d = [700/(1100+0.87*f_y)]

x_{u, max}/d = [700/(1100+0.87*500)] = 0.46

Step 3: Determination of Moment of resistance

M_ur = 0.36 (x_u,max/d)*[1-0.42(x_u,max/d)]f_ck*b_w*d² + 0.45*(b_f-b_w)*D_f*(d-D_f/2)

M_ur

= 0.36 (0.46) *[1-0.42(0.46)]20*230*500² + 0.45*(1800-230) *100*(500-100/2)

M_ur = 185.43 x 10⁶ kNm

A rectangular beam is restricted to a depth of 350 mm due to architectural constraints. Design the beam section to resist a factored moment of 180 kNm using, M 25 concrete, Fe 500 steel, Width = 300 mm, Effective cover = 40 mm (Both side) Find the required areas of tension and compression reinforcement.

Strain	0.00174	0.00194	0.00226	0.00277	0.00312	0.00417
Stress (N/mm2)	347.8	369.6	391.3	413.0	423.9	434.8

Solution:

Step 1: Given Data                                                                                        

Restricted depth of beam = 350 mm

Width of beam = 300 mm

Effective cover (for both tension and compression side) = 40 mm

Effective depth of beam for tension side = 350-40 = 310 mm

Steel 500 and M 25 Concrete

Factored bending moment = 180 kNm

Step 2: Calculation of steel Reinforcement for tension side

x_{u, max}/d = [700/(1100+0.87*f_y)]

x_{u, max}/d = [700/ (1100+0.87*500)] = 0.46

M_u,lim = 0.36 (x_{u, max}/d) [1-0.42(x_{u, max}/d)] f_ck bd²

          = 0.36*(0.46) [1-0.42(0.46)] *25*300*310² 

          = 96.30 kNm

Ast required

M_u = 0.87*f_y*A_st*d[1-(A_st*f_y/b*d*f_ck)]

96.30 x10⁶ = 0.87*500*A_st*310[1-(A_st*500/300*310*25)]

96.30 x10⁶ = 0.87*500*A_st*310[1-(A_st*0.000215)]

29A_st² – 134850A_st + 96.30 x10⁶ = 0

A_st = 881.07 mm² or 3768.93 mm²

Adopt A_st = 881.07 mm²

Step 3: Calculation of steel Reinforcement for compression side

M_u1 = M_u – M_{u, lim}

M_u1 = 180 – 96.30 = 83.70 kNm

Ԑ_sc = 0.0035*(x_u,max-d’)/ x_u,max  = 0.0035*(0.46*310-40)/ (0.46*310) = 0.00251

For Ԑ_sc = 0.00226    f_sc = 391.3

For Ԑ_sc = 0.00277    f_sc = 413.0

f_sc = 391.3 + (413.0-391.30) * (0.00251-0.00226) / (0.00277-0.00226)

    = 401.93 MPa   

M_u1=f_sc*A_sc*(d-d’)

83.70 x 10⁶=401.93*A_sc*(310 - 40)

A_sc = 771.37 mm²

Corresponding tension steel is

A_st1 = (A_sc* f_sc)/(0.87f_y)

A_st1 = (771.37* 401.93)/(0.87*415) = 858.70 mm²

Total tension steel = A_st + A_st1

                             = 881.07 + 858.70 = 1739.78 mm²

Design a two-way slab for a panel with all four edges continuous. The size of room is 3.5 m x 4.2 m. The L.L. on the slab may be taken as 2 kN/m² and F.F. 1 kN/m². Use M 20, and Fe 500, Use LSM.

Solution:

Step 1: Given Data                                                                                       

Two-way Slab

Lx = 3.5 m

Ly = 4.2 m

Live Load = 2 kN/m²

Finish Load = 1 kN/m²

Steel 500 and M 20 Concrete

Assume effective depth of slab be Lx/26 = 3500/26 = 134.61 say 135 mm

Overall depth = 135 + 10/2 + 15 = 155 mm

Effective length of slab in x direction = 3500 + 135 = 3635 mm

Effective length of slab in x direction = 4200 + 135 = 4335 mm

Self-weight of the slab = 0.155*25 = 3.875 kN/m²

Total load on Slab = Dead Load + Live Load + Finish Load

                              = 3.875 + 2 + 1 = 6.875 kN/m²

Factored load = 1.5*6.875 kN/m² = 10.31 kN/m²; say 10.50 kN/m²

Ratio of Ly/Lx = 4.2/3.5 = 1.20

Short span coefficient (α_x) read table 26 of IS 456-2000

αx = 0.043 (-ve)	αy = 0.032 (-ve)
αx = 0.032 (+ve)	αy = 0.024 (+ve)

Step 2: Calculation of design moments

M_x = α_xwl²_x = 0.043*10.50*(3.635)² = 5.965 kNm

M_x = α_ywl²_x = 0.032*10.50*(3.635)² = 4.439 kNm

M_y = α_xwl²_y = 0.032*10.50*(4.335)² = 6.314 kNm

M_y = α_ywl²_y = 0.024*10.50*(4.335)² = 4.735 kNm

Step 3: Calculation of effective depth

M_u,lim = 0.138f_ckbd² = R_ubd²

Effective depth = SQRT (6.314 x 10⁶/2.76*1000) = 47.83 mm, Say 50 mm

Which is smaller than the assumed value, hence ok

Step 4: Calculation of reinforcement

a)    Along the shorter span

Middle strip = 0.75 Lx =0.75*3.635 = 2.726 mm

M_u = 0.87*f_y*A_st*d[1-(A_st*f_y/b*d*f_ck)]

5.965 x10⁶ = 0.87*500*A_st*135[1-(A_st*500/1000*135*20)]

10.875A_st²- 58725A_st + 5.965x10⁶=0

A_st =103.561 mm² and 5296.438 mm²

Using 8mm bars @ (As/Ast) *1000 = (50.27/103.561) *1000 = 485.41 mm

Spacing of main steel must not be more than (i) 3*d = 3*135 = 405 mm and (ii) 300 mm whichever is less. Hence Using 8mm bars @ 300 mm c/c

 b)   Along the Longer span

Middle strip = 0.75 Lx =0.75*4.335 = 3.251 mm

M_u = 0.87*f_y*A_st*d[1-(A_st*f_y/b*d*f_ck)]

6.314 x10⁶ = 0.87*500*A_st*135[1-(A_st*500/1000*135*20)]

10.875A_st²- 58725A_st + 6.314 x10⁶=0

A_st =109.74 mm² and 5290.25 mm²

Using 8mm bars @ (As/Ast) *1000 = (50.27/109.74) *1000 = 458.20 mm

Minimum area of reinforcement provided

= 0.12%*b*D = (0.12/100) *1000*150 = 210 mm²

Hence provide A_{st, min} as main reinforcement in both x and y direction

Using 8mm bars @ (As/Ast) *1000 = (50.27/210) *1000 = 239.8 mm

Spacing of main steel must not be more than (i) 3*d = 3*135 = 405 mm and (ii) 300 mm whichever is less. Hence Using 8mm bars @ 300 mm c/c

Step 5: Calculation of torsional reinforcement

At corners where slab is discontinuous over both the edges

Area of reinforcement in each layer = ¾ *Ast_x = ¾*210 = 157.5 mm²

Distance over which torsional reinforcement to be provided

= 1/5*Lx =1/5*3365 = 673 mm

Provide 6mm @ (28.27/157.5) *1000 = 170.49 mm, Say 180 mm at four corners in four layers

Step 6: Check for deflection

P_t corresponding to maximum mid span moment

P_t = (100 x 210)/(1000 x 135) = 0.155 %

fs = 0.58f_y(A_st,_reqd/A_{st, pro}) = 0.58*500*(210/210) = 290 MPa

(L/d) _max = Basic value x K₁

(L/d) _max = 26 x 2 = 52 mm

(L/d) _provided = (3635/135) = 26.9 mm, hence safe

Step 7: Check for deflection

L_d < = (1.3M₁/V_u) + L₀

M₁ = 0.87f_y A_st (d-0.42 x_u,max)

M₁ = 0.87*500*210*(135-0.42*0.46*150) = 9.68 kNm

V_u = wL_x/2 = 10.5*3.635/2 = 19.08 kN

L₀ = bw/2 + 8* diameter of bar = 300/2 +8*8 = 214 mm

L_d = 1.3*(9.68*10⁶/19.08*10³) + 214 = 659.53+214 =873.53 mm

Alternate

L_d = 0.87f_yϕ/4τ_bd = 0.87*500*8/1.6*1.2*4 = 376 mm, hence safe 

Design a square column for a working load of 1000 kN. The unsupported length of column is 4.2 m. The ends of column are effectively held in position but not restrained in direction. Use M 25 and Fe 500 Adopt LSM.

Solution:

Step 1: Given Data                                                                                       

Working Load = 1000 kN

Factored Load = 1.5*1000 = 1500 kN

Unsupported length = 4200 mm

Effective length = 4200 mm

Steel 500 and M 25 Concrete

Step 2: Calculation of Area

P_u/A_g = 0.4f_ck+(p_t/100) *(0.67f_y-0.4f_ck)

Assuming1% reinforcement

1500*1000/A_g = 0.4*25+(1/100) * (0.67*500-0.4*25)

                   A_g = 113207.51 mm²

         A_g = b*b

          b = 336.46 mm, say 340 mm

          B = 340 + 40 + 10 = 390 mm, say 400 mm

A_s = (1/100) *400*400 = 1600 mm²

Step 3: Check for minimum eccentricity

e_min = (L/500) + (B/30)

e_min = (4200/500) + (400/30) = 8.40 + 13.33 = 21.73 mm

e_min/B = 21.73/400 = 0.054 = 0.05