Provided solution

SINGLY REINFORCED BEAM PROBLEMS

PROBLEM #1
GIVEN :
b = 250mm
h = 350mm
f ' c = 21 MPa
fy = 276 MPa
s = 10 mm
cc = 50mm
B = 16 mm
As = 192
SOLVING THE ACTUAL MOMENT
@123
M1L1 + 2M2 ( L1 + L2 ) + M3 ( L2 ) + A + B = 0
( 30 )( 3 )
3

0 ( 0 ) + 2 ( M2 )( 0 + 3 ) + M3 ( 3 ) + 0 + =0
4
6M2 + 3M3 = −202.5
@ 234
M2 ( 3 ) + 2M3 ( 3 + 3 ) + M4 ( 0 ) + 202.5 + 0 = 0
6M2 + 12M3 = −202.5
M2 =
M3 =

SOLUTION @ POSITIVE MOMENT :
16
d = 350 − 50 − 10 −
2
= 282 mm
B1 = since f ' c = 21 and it is less than to 28  B1 = 0.85
*ASSUMPTION
fs = fy
C=T
0.85 ( 21)( a )( 250 ) = 192 ( 276 )
a = 37.306 mm
a = B1c ; c = 43.889 mm
*STRESS CHECK !
d−c
fs = 600  
 c 
 282 − 43.889 
= 600  
 43.889 
= 3255.180  1000MPa  TENSION CONTROLLED
SOLVE MU :
 37.306 
M T = 0 ; 0.90(0.85 ( 21)( 37.306 )( 250 )  282 −
 2 
) ; Mu = 39.457 KN.M

,PROBLEM #1
GIVEN :
b = 250mm
h = 350mm
f ' c = 21 MPa
fy = 276 MPa
s = 10 mm
cc = 50mm
B = 16 mm
As = 512 

SOLUTION @ NEGATIVE MOMENT :
16
dt = 350 − 50 − 10 −
2
= 282 mm
25
d = 350 − 50 − 10 − 16 −
2
= 261.5
B1 = since f ' c = 21 and it is less than to 28  B1 = 0.85
*ASSUMPTION
fs = fy
C=T
0.85 ( 21)( a )( 250 ) = 512 ( 276 )
a = 99.483 mm
a = B1c ; c = 117.038 mm
*STRESS CHECK !
d−c
fs = 600  
 c 
 282 − 117.038 
= 600  
 117.038 
= 276  845  1000MPa  TRANSITION CONTROLLED
0.90 1000
X 845
0.65 276

0.90 − 0.65 X − 0.65
= ; X = 0.846
1000 − 276 845 − 276
SOLVE MU :
 99.483 
M T = 0 ; 0.846(0.85 ( 21)( 99.483 )( 250 )  261.5 −
 2 
) ; Mu = 79.531 KN.M


ALTERNATE SOLUTION MOMENT @COMPRESSION
 37.306 
M C = 0 ;0.846(512 ( 276 )  261.5 −
 2 
) ; MU = 79.531 KN.M

,PROBLEM # 2
GIVEN :
b = 250mm
h = 350mm
L = 2m
KN
WDL = 10
m
KN
WLL = 9
m
PDL = 15 KN
PLL = 20 KN
f ' c = 35MPa
fy = 415 MPa
s = 10 mm
cc = 40 mm
B = 28 mm
As = 980
WU = 1.2 (10 ) + 1.6 ( 9 ) = 26.4 CORRECTION ASSUMPTION
PU = 1.2 (15 ) + 1.6 ( 20 ) = 50 Fs  Fy
MOMENT ACTUAL : C=T
  264.8 − c  
WL2 0.85 ( 35 )( 0.80c )( 250 ) = 980  600  
+ PL   c 
2
c = 173.522
26.4 ( 2 )
2

MA = + 50 ( 2 ) a = 138.818
2
= 152.8 KN.M
d−c
fs = 600  
 c 
SOLUTION @ NEGATIVE MOMENT :  286 − 173.522 
= 600  
28  173.522 
dt = 350 − 40 − 10 −
2 = 388.924  415  COMPRESSION CONTROLLED
= 286 mm SOLVE MU :
2  138.818 
( 28 + 25 ) = 21.2 M = 0 ; 0.8(0.85 ( 35 )(138.818 )( 250 )  264.8 − ) ;
5 T
 2 
d = 286 − 21.2 Mu = 131.127 KN.M − INADEQUATE
= 264.8
0.05 ( 35 − 28 )
B1 = 0.85 −
7
= 0.8

*ASSUMPTION
fs = fy
C=T
0.85 ( 35 )( a )( 250 ) = 980 ( 415 )
a = 171.790 mm
a = B1c ; c = 214.738mm

, PROBLEM # 3
GIVEN :
b = 400mm
h = 700mm
P = 100
L = 8m
f ' c = 28MPa
fy = 340 MPa
s = 10 mm
cc = 60 mm
B = 32 mm
As = 1536

MOMENT ACTUAL :
WL2
+ PL
2
26.4 ( 2 )
2

MA = + 50 ( 2 )
2
= 152.8 KN.M
@QUESTION C.
400  2 ( 60 ) + 2 (10 ) + 3 ( 32 ) + 2 ( 32 ) = 364 mm
SOLUTION:
32
dt = 700 − 60 − 10 −
2
= 614 mm
2
( 32 + 25 ) = 19
5
d = 614 − 19
= 595
B1 = 0.85


*ASSUMPTION
fs = fy
C=T
0.85 ( 28 )( a )( 400 ) = 1536 ( 340 )
a = 172.339 mm
a = B1c ; c = 202.752 mm


*STRESS CHECK !
 614 − 202.751 
fs = 600  
 202.751 
 614 − 202.751 
= 600  
 202.751 