The state of two dimensional stress acting on a concrete lamina consists of a direct tensile stress of 6 MPa and shear stress of 4 MPa, which cause cracking of concrete. Then the tensile strength of the concrete in MPa is
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- 4. The state of two dimensional stress acting on a concrete lamina consists of a direct tensile stress of 6 MPa and shear stress of 4 MPa, which cause cracking of concrete. Then what will be the tensile strength of the concrete in MPa.?A reinforced concrete beam of rectangular section has the cross-sectional dimensions shown in the figure below. The concrete of normal density has a compressive strength of 30 MPa and a modulus of rupture of 3.3 MPa. The yield strength of steel is 400 MPa. a) Calculate the stress due to an applied bending moment of 45 kN-m b) Calculate the bending moment at which cracking of concrete will be initiated (cracking moment Mcr) h=600 mm b=300 mm 키 As-2000 mm 4- No 25 d=530 mm> * The state of two-dimensional stresses acting on a concrete lamina consists of a direct tensile stress o. = 1.5 N/mm² and shear stress T = 1.20 N/mm², when cracking of concrete is just impending. The permissible tensile strength of the concrete is
- The flexural strength (fr) of concrete measured from the bending test is always higherthan the direct tensile strength (ft). In reality, the stress in the cracked part of theconcrete is lower than ft. Also, failure may occur on the compressive side. Assuming1) there is no tension softening in the cracked zone, i.e., tensile stress stays at ft; and 2)infinitely large compressive strength, draw the stress distribution over the depth of thebeam at ultimate failure when the curvature of the section is approaching infinity. Withthis stress distribution, show that the upper bound of fr/ft is equal to 3. (Hint: On rotating the section, because the compressive stress can increase well beyondft, the neutral axis will continue to shift towards to the compression side.)At what percent of compressive strength of concrete the stress in beam changes from directly proportional to varying with respect to depth.Problem # 4 A 4 in x 3 in steel block follows an elasto-perfectly plastic stress strain relation with a yielding stress 0,-60 ksi and a modulus of elasticity E=30 x 10° psi. The block is reinforced on the top and bottom by 1 in x 3 in plates that are perfectly elastic having a rupture stress O,-60 ksi and a modulus of elasticity E=10 x10° psi. The stress-strain relation for the two materials are given below. • Determine the largest moment M that can be applied to the cross section before it fails. • Determine the magnitude of the distributed load w (kip/ft) that can be applied to a 12-ft beam made of that cross section. 1 in 4 in. M 1 in 6 ft 6 ft 60 ksi 60 ksi 3 in. Gy E-30x10'ksi E-10x10 ksi 6x 10 stress-strain diagram for 4 in x 3 in stress-strain diagram for 3 in x 1 in plates
- A singly reinforced rectangular section has a width of 250mm and an effective depth of 460mm. The concrete has a compressive strength of 21MPa. The steel has a yield strength of 275 MPa and E = 200,000 MPa. Calculate the ideal flexural strength for the following areas of steel: a. 2580 mm² b. 5160 mm² c. The value at balanced failureBoth of the tension and compression forces are exerted on the concrete specimen when investigating the flexural strength. However, the specimen fails due to the flexural forces. * True False Other:Situation 5: The compound bar, composed of the three segments shown, is initially stress free. Compute the stress in each material if the temperature drops 25°C. Assume that the walls do not yield and use the following data: A (mm²) a (/*C) E (GPa) Bronze segment 2000 19.0 x 10-6 83 Aluminum segrnent 1400 23.0 x 10-6 70 Steel segment 800 11.7 x 10- 200 - 500 mm→- 400 mm- 800 mm Aluminum Steel Bronze 17. Compute the stress in the bronze bar a. 31.6 MPa (T) b. 31.6 MPа (С) 18. Compute the stress in the aluminum bar. а. 45.1 MPа (Т) b. 45.1 MPa (C) 19. Compute the stress in the steel bar. 79.0MPA (T) 79.0MPA (C) 13.6 MPa (T) 13.6 MPa (C) c. d. 54.1 MPа (Т) 54.1 MPa (C) с. d. a. C. 97.0 MPa (T) b. d. 97.0 MPa (C)
- An aluminum pipe must not stretch more than 0.05in when it issubjected to a tensile load. Knowing that E=10.1x10^6psi and that the maximum allowable normal stress is 14ksi, determine (a) the maximum allowable length of the pipe, and (b) the required area of the pipe if the tensile load is 127.5kips.The preliminary design is important because: O A. The designer calculates the required reinforcement amounts in this stage. O B. This stage provides the final dimensions of elements regardless of the design forces. O C. All of the reasons mentioned. D. The size of structural elements determines the stiffness needed for the structural analysis.PROBLEM 6.56 50 mm A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 18 kN and that the modulus of elasticity is 200 GPa for the steel and 73 GPa for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum stress Aluminum 25 mm Steel in the beam. 36 mm