For concrete of grade M50, the value of flexural tensile strength will be nearly 

This question was previously asked in
UKPSC JE Civil 8 May 2022 Official Paper-I
View all UKPSC JE Papers >
  1. 5 N/mm2
  2. 10 N/mm2
  3. 25 N/mm2
  4. 50 N/mm2

Answer (Detailed Solution Below)

Option 1 : 5 N/mm2
Free
UKPSC JE CE Full Test 1 (Paper I)
1.8 K Users
180 Questions 360 Marks 180 Mins

Detailed Solution

Download Solution PDF

Concept:

As per IS 456 : 2000, clause 6.2.2:

Tensile Strength of Concrete in Flexure: The flexural and splitting tensile strengths shall be obtained as described in IS 516 and IS 5816 respectively. When the designer wishes to use an estimate of the tensile strength from the compressive strength, the following formula may be used:

Flexural strength (fcr) = \(0.7\sqrt{f_{ck}}\) N/mm2

where fck is the characteristic cube compressive strength of concrete in N/mm2.

Calculation:

Given

Grade of concrete is M50

Flexural tensile strength (fcr) = 0.7\(\sqrt{f_{ck}}=0.7\sqrt{50}=4.949\) N/mm2

Flexural tensile strength (fcr) = 4.949 ≈ 5 N/mm2

Hence option (1) is the correct answer.

Latest UKPSC JE Updates

Last updated on Mar 26, 2025

-> UKPSC JE Notification for 2025 will be out soon!

-> The total number of vacancies along with application dates will be mentioned in the notification.

-> The exam will be conducted to recruit Junior Engineers through Uttarakhand Public Service Commission. 

-> Candidates with an engineering diploma in the concerned field are eligible. 

-> The selection process includes a written exam and document verification. Prepare for the exam with UKPSC JE Previous Year Papers.

Get Free Access Now
Hot Links: teen patti gold real cash teen patti wealth teen patti joy official teen patti game - 3patti poker