Solution manual Concrete Structures, 2nd Edition, Mehdi Setareh & Robert Darvas (COMPLETE)

Problem 1-1

Hydration is a chemical reaction that starts when water is added to cement. It has three stages:

setting, hardening, and strength development. The hydration process generates heat. It also

continues throughout the life of concrete structures as long as there is free moisture available.




Problem 1-2

Since concrete is a construction material which is very strong in compression but weak in tension,

it is important for the structural designer to know what the compression capacity of concrete is.

The compressive strength of concrete is measured by conducting a “cylinder test”. In this test,

compression force is applied gradually on a standard 6”x12” concrete cylinder. The stress and

strain of the specimen is measured and plotted. The maximum compressive strength is noted as fc’.




Problem 1-3

The air-entraining admixtures are added to concrete to increase the concrete resistance against the

freezing/thawing cycles. As a result, this admixture improves concrete durability.




Problem 1-4

The modulus of elasticity relates the strain to the stress in concrete. It can be determined as a result

of the cylinder test or the use of the ACI approximate equation.




1

,Problem 1-5

The modulus of rupture is the tensile strength of concrete in bending.




Problem 1-6

Deformed bars are usually used as the primary steel reinforcement of structural elements. The bars

have protrusions on the surface to increase their bondage to concrete.

Welded wire reinforcements are thin wires spaced at certain distances in two orthogonal directions

and fabricated in large sheets or long rolls. The welded wire reinforcements are usually used where

large areas need to be reinforced such as floor slabs and walls.




Problem 1-7

 12 24 
110  
weight of beam   12 12   0.22 k/ft
1,000
wT  1.0  0.22  1.22 k/ft

1.22(10)
R  6.1 k
2
wl 2
M max   15.3 ft  k
8




2

, w = 1.22 k/ft



R = 6.1 k R = 6.1 k

2'-0"

6.1k




V (k)




-6.1k




M (ft-k)
M max = 15.3 ft-k


Sketch for Problem 1-7




Problem 1-8

wc  145 pcf

f c  3,500 psi

λ  1.0 (normal weight concrete)

57,000 3,500
E c  57,000 f c   3,372 ksi
1,000

f r  7.5λ f c  7.5(1.0) 3,500  444 psi



3

,Problem 1-9

wc  120 pcf

f c  3,000 psi

P


w = 30 lb/ft



6'-0"




6 6 
w  120    30 lb/ft
 12 12 

wl 2 Pl 30(6)2 P(6)
M max      135  1.5P ft  lb
8 4 8 4

Using Equation (1-3) for sand-lightweight concrete:

f r  f b  7.5λ f c  7.5(0.85) 3,000  349 psi

bd 2 6(6) 2
Sm    36 in 3
6 6
M R  fr Sm

M max  M R

 36 
135  1.5P  (349) 
 12 
135  1.5P  1,047
P  608 lb


then Pmax  608 lb



4

, Problem 1-10

f c  4,000 psi



w = 2.1 k/ft




 8 12 
beam weight  150    100 lb/ft
 12 12 
100
w2  2.1 k/ft
1,000

w 2
M max   0.263 2
8
λ  1.0 (normal weight concrete)

f r  7.5λ f c  7.5(1.0) 4,000  474 psi

bd 2 8(12)2
Sm    192 in 3
6 6
474(192)
M R  fr Sm   7.59 k  ft
12,000
M max  M R

0.263 2  7.59
  5.37'




5