Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

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1 Serial : IG1_CE_G_Concrete Structures_ Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: info@madeeasy.in Ph: CLASS TEST CIVIL ENGINEERING Subject : Concrete Structures Date of test : 10/08/018 Answer Key 1. (a) 7. (b) 13. (a) 19. (c) 5. (a). (a) 8. (c) 14. (c) 0. (d) 6. (b) 3. (c) 9. (a) 15. (d) 1. (b) 7. (c) 4. (d) 10. (c) 16. (a). (a) 8. (c) 5. (a) 11. (b) 17. (b) 3. (c) 9. (b) 6. (b) 1. (c) 18. (b) 4. (b) 30. (c)

2 CT-018 CE Concrete Structures 7 Detailed Explanations 4. (d) Modulus of elasticity of concrete is dependent on its compressive strength. Lower w/c ratio,larger curing period, higher vibration will produce concrete of higher compressive strength. With increase in age, compressive strength increases and therefore modulus of elasticity of concrete increase with increase in E age. Also, Modulus of Elasticity, E c, where E 1 +θ c is intial Modulus of elasticity and θ is creep coefficient. θ decreases with age, so E increase with age. 5. (a) For isolated T-beam, b f l l b + b w where, l 0 6m, b 1000 mm, b w 300 mm b f mm 6. (b) Strain distribution is linear and stress distribution is non-linear. 7. (b) Central dip, h 8. (c) Let us design a square section of size B B A B wl 8P 40 ( 10) m 00mm A s 1 B 0.01 B 100 Now for a short column A A s B 0.01 B 0.99 B P σ cc + σ sc A s B B B mm Hence, column is a long column L eff B >

3 8 Civil Engineering Now, C R B 4000 P ( B B ) B B B B Area ( 49.5 ) mm 9. (a) For deformed bars value of bond stress is increased by 60%. For bars in compression the value of bond stress is increased by 5% Development length ( 0.87f y ) φ τ bd ( ) mm 11. (b) b f t f b w 1000 mm 100 mm 50 mm d 500 mm π mm Assuming N.A. to lie within the flange and equating the forces of compression and tension, we get 0.36 f ck b f x u 0.87 f y x u 0.87f A y 0.36 f ck st b f mm x u < t f (ok) x u max 0.53 d mm x u < x u max, section is under reinforced ultimate moment of resistance, M u 0.87 f y (d 0.4 x u ) ( ) KNm 1. (c) Area of concrete A g A s

4 CT-018 CE Concrete Structures 9 π π mm Ultimate load, P u 1.05(0.4 f ck ) 1890 KN Safe load KN 13. (a) σ max P + Pe A Z N/mm 14. (c) According to IS: Span Effective depth (A) value 10 Span in metres (for span > 10 m) Effective depth Span 10 (A) value Span in metres (A) value for simply supported beam is 0. effective depth > 1,15 mm 15. (d) P u KN A mm P u 0.4 f ck f y A sc (76000 A sc ) A sc A sc 5710 mm 16. (a) Applying the yield line theory with, R 1, µ i x i y 1 l x l y 4 m µ (3 + iy) K 1 R (1 i ) + x K 4(1 + i x ) 4 µ (1 + i ) y

5 10 Civil Engineering α 1 α K1.35 K 1 K 1 3K M + uy wu ly (3 α 1) 6( iy +α1) w 6(1 ) wl uly α wu ly + ix (Greater) 17. (b) Loss of prestress, stress in concrete at level of steel M + + uy Mux Muy Mu σ m θ σ c x knm/m Loss of prestress, σ c P Pey A + I m N/mm E E s c σ N/mm Prestress P A π s N/mm Loss of prestress % (c) Loss of prestress due to anchorage slip L Es L N/mm % loss of prestress %

6 CT-018 CE Concrete Structures (d) Initial prestressing force, P kn 0.85 e 75 mm. Area of beam, A mm Top fiber stress P 6e 1 A d 3.9 N/mm Bottom fiber stress P 6e 1+ A d 19.6 N/mm 1. (b) For HYSD bars in compression, actual bond stress φσs L d 4 τ bd τ bd N/mm. (a) 1 ( ) L d mm 4.8 According to IS 456 : 000, the development length should be increased by 0% for three bars in contact. L d is to be increases by 0% for three bundled bars. L d mm b x u C 0.67 f ck B 0.45 f ck 0.36 fckbxu A 0.87 f y 0.87 f y Assumed IS : AB AC AB AC 4 7 [from assumed stress diagram] AB 4 7 x u But BC AC AB x u 4 7 x u 3 7 x u Total compressive force 0.67 f ck 3 7 x u b x 7 u b f ck b x u f ck

7 1 Civil Engineering 3. (c) Total tensile force 0.87 f y For actual depth of neutral axis, Total compressive force Total tensile force f ck b x u 0.87 f y x u π (16) x u mm 1000 mm mm mm 570 mm Given, X u > 100 mm and section is under reinforce 0.36 f ck (35)X u f ck (675) f y (This expression is valid only when 3 X 7 u > D f, > OK) X u X u mm Take, X u 36.3 mm Check, X u(l) X u < X u(l) (OK) 4. (b) Ultimate design shear force, V u 10 kn Ultimate shear capacity of concrete, V uc τ c bd kn Ultimate shear to be resisted by shear reinforcement, V us V u V uc kn For vertical stirrups, spacing is given as S v 0.87 f A d V y us sv π (8) mm

8 CT-018 CE Concrete Structures (a) V s 75 kn, τ c 0.48 N/ mm V c kn Vertical stirrups to be designed for ( ) 9.4 kn S v (spacing) Area of stirrups σ effective depth design value for vertical stirrups y π mm 6. (b) Maximum spacing allowed for vertical stirrups (1) 0.75 d 85 mm, () 300 mm S V 85 mm C min < 0.05 D therefore we can apply, P u 0.4 f ck f y A sc π Asc A 4 sc 7. (c) on solving, A sc mm x u,lim 0.48 d mm Moment of resistance 0.36 f ck b x u,lim (d 0.4 x u,lim ) ( ) kn-m 8. (c) Bending tensile strength, σ bt. N/mm Characteristic compressive strength, f ck 0 N/mm Maximum stress in concrete, σ cbc 0.45 f ck N/mm P P L/3 L/3 L/3 PL/3 PL/3 BMD Section modulus, Z bd 00 (300) mm 3 Maximum moment, M max Z σ bt kn-m

9 14 Civil Engineering PL M max 3 P P 6.6 kn 9. (b) Effective width of flange, b f l l b b w mm This is greater than b, which is not possible. So, effective width of flange is 1000 mm. 30. (c) Axial load 1400 kn 10% of column load 140 kn Total load 1540 kn Actual area of footing required m Net earth pressure acting upward due to factored load is Punching shear w kN/m 15.4 Critical section 400 d The critical section is taken at a distance d Two way shear force, V u Punching shear stress away from the face of column as shown in above figure ( ) kN Factored shear force Perimeter effective depth of the the critical section N/mm 4( ) 600

Pre-stressed concrete = Pre-compression concrete Pre-compression stresses is applied at the place when tensile stress occur Concrete weak in tension

Pre-stressed concrete = Pre-compression concrete Pre-compression stresses is applied at the place when tensile stress occur Concrete weak in tension but strong in compression Steel tendon is first stressed