1) The gross area compressive strength is calculated by dividing the load at failure by the gross cross-sectional area of the unit. (T or F)
2)A hollow concrete masonry unit has actual gross dimensions of
7-5/8” x 7-5/8” x 15-5/8”. If the unit fails under a load of 119.14
kips, determine the gross area compressive strength.
3)
A concrete masonry unit tested as follows:
Saturated mass - 10,089 g Absorption ______%
Oven dry mass - 8,893 g Moisture ______%
Field mass - 9,354 g
Determine the absorption and moisture content percentages.
1) The gross area compressive strength is calculated by dividing the load at failure by the gross cross-sectional area o...
Hollow concrete masonry blocks typically have a net cross sectional area which is at least 25% less than the gross area. (T or F) A concrete masonry unit has actual gross dimensions of 7-5/8” x 7-5/8” x 7-5/8”. The unit is tested in a compression machine with the following results. Max. failure load = 98 kips. Net volume = 348.1 in3 T or F 4. The unit is categorized as solid. _____ 5. The gross area...
Problem #3 A concrete masonry unit is tested for compressive strength and produces the following results: Failure load- 726 kN Gross area 0.081 m2 Gross volume 0.015 m Net volume = 0.007 m? Is the unit categorized as solid or hollow? Why? What is the compressive strength? Does the compressive strength satisfy the ASTM requirements for load bearing units shown in Table 8.2?
1. The fracture stress of your sample (in units of N m−2) is given by dividing the load (in N) at the fracture point by the cross-sectional area of the original sample (in m2). The typical thicknesses of plastic bags are 8 × 10−6 m for flimsy bags and 7 × 10−5 m for thicker bags. Select one of these values and use it to obtain an estimate for the fracture stress of your bag. Ensure you state the thickness...