A concrete masonry unit is tested for compressive strength and produces thefollowing results:Failure load = 120,000 lbGross area = 130 in2Gross volume = 960 in3Net volume = 447 in3Is the unit categorized as solid or hollow? Why? What is the compressivestrength? Does the compressive strength satisfy the ASTM requirements forload bearing units shown in Table 8.2? TABLE 8.2 Strength Requirements of Load Bearing and Non-Load-Bearing Concrete Masonry Units (Reprinted, with permission, fromASTM C90, Table 3 and C129, Table 2, copyright ASTM International,100 Barr Harbor Drive, West Conshohocken, PA 19428)Minimum Compressive StrengthBased on Net Area MPa (psi)TypeAverage of Three UnitsIndividual UnitsLoad bearingNon–load-bearing13.1 (1900)11.7 (1700)4.1 (600)3.5 (500)

Calculating whether it is solid or hollow: Hence, the unit categorized as hollow.

Answered: TABLE 8.2 Strength Requirements of Load…

Structural Analysis

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ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A concrete masonry unit is tested for compressive strength and produces the
following results:
Failure load = 120,000 lb
Gross area = 130 in²
Gross volume = 960 in³
Net volume = 447 in³
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?

TABLE 8.2 Strength Requirements of Load Bearing and Non-Load-
Bearing Concrete Masonry Units (Reprinted, with permission, from
ASTM C90, Table 3 and C129, Table 2, copyright ASTM International,
100 Barr Harbor Drive, West Conshohocken, PA 19428)
Minimum Compressive Strength
Based on Net Area MPa (psi)
Type
Average of Three Units
Individual Units
Load bearing
Non–load-bearing
13.1 (1900)
11.7 (1700)
4.1 (600)
3.5 (500)

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