Qui uiz e A normal weight standard 6 in x 12 in concrete cyl

Qui uiz e A normal weight standard 6 in x 12 in concrete cylinder specimen tested at 28 days yielded the following results: . It failed at an axial compressive force of 105,000 lb It failed at a transverse force of 45,000 lb in a split tensile test. Calculate ultimate compressive strengh fe and splitting Tensile Strength, ft

Solution

Solution:-

Given

Size of specimen = 6 inch x 12 inch

Axial compressive force (P) = 105000 lb

Transverse force = 45000 lb

Ultimate compressive strength (f_c’) = P/A

                                                       = P/( r² )           

                                                       = 105000/ ( * 3²)

                                                       = 3713.6 lb/inch²       Answer

Splitting tensile strength (f_ct’) = 2 * transverse force/( D L )

                                            = 2 * 45000 /( * 6 *12)

                                            = 397.8 lb /inch² Answer