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• Bryan Ewe

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YIELD LINE ANALYSIS Energy Method 1. Determine the ultimate moment of a square isotropic slab simply supported on three sides and subjected to a uniform load q per unit area. Solution: l

Part-1 θ1

x

m

m Part-3

Part-2

l

a

l/2

b 1

l/2

c

m θ2

θ3

l/2

l/2

I =¿ ∑ ¿ Internal work done

∑E

=External work done

∑ I =∑ E For part 1,

Tan 1 

1 x

, For small deflection Tan θ is equal θ (Small-angle approximation)



θ2

θ1

1 = x

=

θ3

=

1 l 2

=

2 l

Page 1 of 25

∑I

Internal work done,

=

I 1 + I 2+ I 3

I=( m+m' ) lθ l

= Projected length of yield line on axis of rotation of that region

(m) θ= Angle of rotation of slab panel (rad) I1

=

I 3 =I 2

1 x

=

ml x

xlx

2 l

=

m1 x l x

m2

=

Internal work done,

∑I

2m 1

=

ml x

=

E1 + E2 + E3

+

4m 1

l = m[ x + 4] External work done,

∑E

It is equal to resultant on each part multiplied by the vertical displacement of its point of application. External work done (

E1

) for part-1

1 1 E1=q x x x x l x 2 3

=

qxl 6

External work done (

E3∨E 2

) for part-3 or part-2

E3=E 2=¿ One Triangle +One Rectangle 1 l 1 l 1 E3=E 2=q x x x x x +q x x ( l−x ) x 2 2 3 2 2 qxl ql(l−x ) = 12 + 4 Page 2 of 25

External work done,

=

=

=

qxl 6

qxl + 6

q l2 2

∑E

+

q l2 2

=

-

qxl 6

2[

+

qxl ql ( l−x ) + ] 12 4

qxl 2

qxl 6

-

1 x ql 2 ( − ) 2 6l

∑ I =∑ E l m[ + 4] = x

1 x ql 2 ( − ) 2 6l q l 2( 3l −x) = 6l( l + 4) x

m

2

=

q l (3l −x) x 6 l (l +4 x )

=

ql(3l −x) x 6(l+ 4 x)

The maximum value of m is obtained when

dm =0, dx

( 6 l+ 24 x )( 3 l−2 x )=( 3lx −x 2)(24) 18 l2 −12lx+72 lx−48 x 2=72lx−24 x 2 2

2

18 l −12lx−24 x =0 x=

−(−12 l) ± √ (−12l)2−4 (−24)(18 l 2) 2(−24)

= −1.15 l∨0.65l , Value of

x

can’t be negative

Page 3 of 25

m=

ql (3 l−0.65 l)(0.65l) 6 {l+4 ( 0.65 l ) }

=

ql 2 14.1

2. Consider the isotropic slab with an opening which is simply supported along the edge AB, built-in along edge BC and CD and supported on a column at E as shown in figure below. The edges DE, EF and AF are free and the effect of column size can be ignored. The slab is subjected to a uniformly distributed load q (KN/m 2) throughout and a live load intensity p=0.9 qb (KN/m) acting on the free edge AF. Assume that m’=2m (KNm/m). Where m and m’ are the sagging (positive) and hogging (negative) ultimate moments of resistance per unit length respectively. If a yield line pattern during collapse is proposed as shown in figure, apply the virtual work method to determine the optimum value β and hence the ultimate moment m of the slab for b=2.5m and q=12 KN/m2 . B

K

C

J

UDL=q

Part-2 3b

p=0.9 bq Part-1 P

b

A

βb

Part-3 I F G Part-4 E

1.5b

m'=2 m

b=2.5 m

Part-5 H b

D

q=12 KN /m

2

4b

Solution: EH=b

Page 4 of 25

KJ=1.5b-βb=b(1.5-β) JC=4b BK=βb HD=3b Assume max displacement of 1 unit along IF

1  Part-1

1 b

tan  2  Part-2

2 

1 3b

1 3b

3  Similarly

4  5 

1 1  HD 3b

1 1 1   EH EF b

m’=2m I  (m  m' )l

Internal work done m = Sagging ultimate moment (+ve) m’= Hogging ultimate moment (-ve) i.e on fixed support only. Internal work done,

∑I

=

I 1 + I 2+ I 3

+

I 4 + I5

I =( m+m' ) lθ I 1 =m x 3 b x

1 βb

Page 5 of 25

3m β

=

m (¿¿ ' ) (5.5 b ) x

1 1 +m x ( 4 + β ) b x 3b 3b I 2 =¿

=

11 m 4 m βm + + 3 3 3

=

15 m+ βm 3 m+m 1 3b

(¿¿ ' ) ( 4 b ) x I 3=¿

=

12 m 3

m+m 1 b I 4=I 5=¿

(¿¿ ' ) x b x

=

3m

Internal work done, =

∑I

=

I 1 + I 2+ I 3

+

I 4 + I5

3m 15 m+βm 12 m 3m + 3m + + β 3 3 +

3 15+ β = β + +10] 3 m¿ External work done,

∑E

=

E1 + E2 + E3

+

E4 + E 5

1 1 1 E1=q x x βb x 3 b x +0.9 qb x βb x +0.9 qb(1.5−β) b 2 3 2

Page 6 of 25

2

=

3 βq b +0.45 βq b 2+1.35 q b2−0.9 βq b 2 6

=

0.05 βq b +1.35 q b

2

2

1 1 1 1 E2=0.5 βq b 2+ q x ( 1.5−β ) b x 3 b x +q x b x 3 bx + q x x 3 b x 3 b x 2 2 2 3 =

0.5 βq b2 +2.25 q b 2−1.5 βq b2 +1.5 q b2 +1.5 q b 2

=

5.25 q b2− βq b2

E3=1.5 q b2 +q x 3 b x b x 2

1 2

2

= 1 .5 q b +1 .5 q b

b

2b/3

2 = 3 qb

b/3

CG

a a/3

2a/3

Axis of rotation pass through column

Page 7 of 25

C

1/3

J

2/3

Part-2 CG Part-3 CG

F G Part-4 E

Part-5 H

D

2/3

1/3

1 2 E4 =E5=q x x b x b x 2 3 2 = 0.33 q b

External work done, = qb

∑E

2

0.05 βq b +1.35 q b

2

=

E1 + E2 + E3 2

+ 2

+ 5.25 q b − βq b

E4 + E 5

+3 q b

2

+0.33 q b

2

+0.33

2

2 2 = 10.27 q b −0.95 βq b

=

q b 2 [10.27−0.95 β ]

∑ I =∑ E 3 15+ β β + +10] = 3 m¿ m=

q b 2 [10.27−0.95 β ]

q b2 [10.27−0.95 β] 3 15+β + +10 β 3

Page 8 of 25

The maximum value of m is obtained when 3 15+ β ( + +10)(−0.95) = β 3

(10.27−0.95 β )(

dm =0, dβ

−3 1 + ) 2 β 3

−2.85 −30.81 2.85 −4.75−.32 β−9.5= + 3.42+ −0.32 β 2 β β β 30.81 5.7 − −17.67=0 β β2 30.81−5.7 β−17.67 β 2=0 β=

5.7 ± √ (−5.7)2−4(30.81)(−17.67) 2(−17.67) β

= -1.49 or 1.17, Value of

can’t be negative

12(2.5)2 [10.27−0.95(1.17)] m= 3 15+1.17 + +10 1.17 3 = 38.26 KN-m/m

 3. Determine the ultimate moment and find out the optimum value of for the given pattern of yield line as shown in Fig. 2

B

A

m' =2 m UDL=q

b

Part-3 Part-4 Part-2

M

F

N

0.75b

βb

Solution:

βb Fig. 2

Part-1

Part-5

G

H b

E

C

D b

b

Page 9 of 25

Assume maximum vertical displacement=1 along MN Then find deflections at F and E ( Displacement at E and F ,

θ1=θ 5=

2  4 

3 

δ F ∧δ E

δ E =δ F =

) by similar triangles method

b 1 = b+ βb 1+ β

δF 0.75b

1 1  b   b (1   )b

1 b

Internal work done,

∑I

=

I 1 + I 2+ I 3

+

I 4 + I5

I =( m+m' ) lθ I 1 =I 5=m x b x

=

mxbx

δF 0.75b

1 0.75 b(1+ β )

=

m 0.75(1+ β)

=

1.33 m 1+ β m+m 1 b ( 1+ β ) I 2=I 4=¿

(¿¿ ' ) x 1.75 b x

=

5.25 m 1+ β

Page 10 of 25

I 3 =m x 2(b+βb) x

1 b

2 (1+ β ) m

=

Internal work done, 2

=

= 2

[

[

∑I

=

I 1 + I 2+ I 3

+

I 4 + I5

]

1.33 m 5.25 m + +2 ( 1+β ) m 1+ β 1+ β

2 1.33 m+5.25 m 2(1+ β) m + 1+ β (1+ β )

]

2 6.58 m 2(1+ 2 β + β ) m 2 + = 1+ β (1+ β )

[

]

13.16 m+2 m+ 4 mβ +2 m β 2 = 1+ β

( 15.16+4 β +2 β 2 ) m

=

1+ β

External work done,

∑E

=

E1 + E2 + E3

+

E4 + E 5

1 1 1 E1=E 5=q x x 0.75 b x b x x 2 1+ β 3 0.125 q b2 1+ β

=

1 1 1 1 1 E2=E 4=q x x 0.75 b x b x x +q x x b x (1+ β) b x 2 1+ β 3 2 3 2

0.125 q b + 0.17 q b2 +0.17 βq b 2 1+ β

=

[

]

1 1 1 E3=2 q x x b x (1+ β ) b x +q x (1−2 β ) b xbx 2 3 2

Page 11 of 25

( 1+ β ) q b2 (1−2 β ) q b 2 + = 3 2 2

2

2

2

= 0.33 q b + 0.33 βq b +0.5 q b − βq b 2 2 = 0.83 q b −0.67 βq b

External work done,

∑E

=

E1 + E2 + E3

+

E4 + E 5

=

2( 0.125 q b2 ) 0.125 q b2 +2 + 0.17 q b 2+0.17 βq b2 +0.83 q b 2−0.67 βq b 2 1+ β 1+ β

=

0.5 q b2 +1.17 q b2−0.33 βq b 2 1+ β

[

]

q b2 = 1+ β [0.5+1.17 ( 1+ β ) −0.33 β ( 1+ β ) ] 2

qb 2 = 1+ β [1.67 +0.84 β−0.33 β ]

∑ I =∑ E ( 15.16+4 β +2 β 2 ) m 1+ β

Therefore,

m=

2

=

qb [1.67+0.84 β −0.33 β2 ] 1+ β

q b2 [1.67+0.84 β−0.33 β 2] ( 15.16+4 β +2 β 2 )

The maximum value of m is obtained when

dm =0, dβ

( 15.16+ 4 β+ 2 β 2 ) ( 0.84−0.66 β )=[1.67+ 0.84 β−0.33 β 2](4 +4 β ) 12.73−10.01 β+ 3.36 β −2.64 β 2+1.68 β 2−1.32 β 3=6.68+6.68 β +3.36 β+3.36 β 2−1.32 β 2−1.32 β 3 6.05−16.69 β−3 β 2=0

Page 12 of 25

16.69± √ (−16.69)2−4 (6.05)(−3) β= 2(−3) = -5.9 or 0.34, Value of m=

β

can’t be negative

q b2 [1.67+0.84 (0.34)−0.33(0.34)2 ] 2 [15.16+ 4(0.34)+2 ( 0.34 ) ]

=0.1145 q b

2

KN m/m

4. Consider an isotropic slab with opening as shown in figure 1. The slab is simply supported along edge CD, fixed support along edges AB and BC and supported by a column at F. Edges AF, EF and DE are free. The slab is subjected to uniformly distributed load q KN/m2 and line load along edge DE with intensity, P=1.5qb (KN/m). Assume that m'= 2m (KN m/m) where m and m' are positive and negative ultimate moments respectively. If the proposed yield line fracture pattern is as shown below, determine the ultimate moment resistance, m per unit length for the slab.

B

A

m' Part-3

UDL=q

2.5b

m m'

J

Part-2 E

b

m Part-1

m L m' Part-5

m Part-4 K

2βb

P

F

βb

m' =2 m P=1.5 qb q=10 KN /m2

D

C 3.5b

1.5b

b=2 m

Page 13 of 25

Solution:

Assume maximum displacement of 1 unit along line JK

Displacement at E (5-2β)b 3.5b

δE

From similar triangles approach

E 

1 E 3.5b  1 (5 - 2  )b

3.5 (5 - 2  )

1 

E 3.5  b (5 - 2  )b

Rotations,  JK 1 2   (5 - 2  )b (5 - 2  )b  JK 1 3   (2.5b - b) (2.5 -  )b 1 4  2b 1 5  b Internal work done,

∑I

=

I 1 + I 2+ I 3

+

I 4 + I5

I =( m+m' ) lθ I 1 =m x 3.5 b x

=

=

δE b

m x 3.5 b x

3.5 (5−2 β ) b

12.25 m 5−2 β

Page 14 of 25

m+ m 1 ( b 5−2 β ) I 2=¿

(¿¿ ') x 3.5 b x

=

10.5 m 5−2 β '

I 3 =( m+m ) x 5 bx

=

15 m 2.5−β

I 4=( m+m ' ) xβbx

=

1 ( 2.5−β ) b

1 2 βb

3m 2

I 5 =( m+m ' ) x 2 βbx

1 βb

= 6m Internal work done, =

=

∑I

=

I 1 + I 2+ I 3

+

I 4 + I5

12.25 m 10.5 m 30 m 3 m + + + +6 m 5−2 β 5−2 β 5−2 β 2 52.75 m 15 m + 5−2 β 2

105.5 m+15 m(5−2 β) = 2(5−2 β) m [90.25−15 β ] = (5−2 β ) External work done,

∑E

=

E1 + E2 + E3

+

E4 + E 5

Page 15 of 25

1 3.5 1 1 3.5 E1=q x x 3.5 b x b x x +1.5 qb x b x x 2 5−2 β 3 2 5−2 β 2

2

=

2.04 q b 2.625 q b + 5−2 β 5−2 β

=

4.67 q b 2 5−2 β

E2=

2.04 q b2 1 1 1 +q x ( 5−2 β ) b xβ b x + q x x (2.5− β) b x(5−2 β) b x 5−2 β 2 2 3

=

( 12.5−10 β +2 β2 ) q b 2 2.04 q b2 +2.5 βq b2−β 2 q b2 + 5−2 β 6

=

2.04 q b +2.08 q b2 +0.833 βq b 2−0.67 β 2 q b2 5−2 β

2

E 3=

( 12.5−10 β +2 β 2 ) q b2 6 2

+ q x 2 βbx ( 2.5−β ) b x

2

2

2

1 2 2

2

= 2.08 q b −1.67 βq b +0.33 β q b + 2.5 βq b −β q b =

2

2.08 q b2 +0.833 βq b 2−0.67 β 2 q b2

1 2 E4 =q x x 2 βb x β b x 2 3 ¿ 0.67 β 2 q b2 1 2 E5=q x x 2 βb x β b x 2 3 2

2

¿ 0.67 β q b

External work done,

∑E

=

E1 + E2 + E3

+

E4 + E 5

= 4.67 q b 2 2.04 q b2 + +2.08 q b2 +0.833 βq b 2−0.67 β 2 q b2 +2.08 q b2 +0.833 βq b 2−0.67 β 2 q b2 +0.67 β 2 q b2 +0. 5−2 β 5−2 β

Page 16 of 25

2 = qb

[

6.71 + 4.16+1.66 β ] 5−2 β

2

qb 2 = 5−2 β [6.71+20.8−8.32 β +8.35 β−3.32 β ] q b2 2 = 5−2 β [27.51+0.03 β−3.32 β ]

∑ I =∑ E m [90.25−15 β ]=¿ (5−2 β )

q b2 [27.51+0.03 β−3.32 β 2 ] 5−2 β

2

Therefore,

m=

2

q b [ 27.51+0.03 β−3.32 β ] [90.25−15 β ]

The maximum value of m is obtained when

dm =0, dβ

[90.25−15 β ] ( 0.03−6.64 β )=[27.51+ 0.03 β−3.32 β 2 ](−15) 2

1.805−599.26 β−0.3 β +99.6 β =−412.65−0.3 β+ 49.8 β

2

2

49.8 β −599.26 β+ 414.46=0 β 2−12.03 β+8.32=0 β=0.74

or 11.29,

Therefore

β=0.74

10(2)2 [27.51+0.03( 0.74)−3.32 ( 0.74 )2 ] m= [90.25−15( 0.74)] =13.0 KN m/m

5. Find the ultimate moment resistance, m per unit length for the slab shown in figure. Page 17 of 25

Find β if m’=2m, α=0.2, b=2.0 m and q=20 KN/m2.

m'

m

Given, 1.5b

Part-3

m m' Part-4

UDL=q

P=0.5 qb

(line load)

W

Part-2

m'

Part-1

b

(point load)

αb

m

2

W =1.5 q b

βb

P

m' =2 m

3b

1.5b

α=0.2 m b=2.0 m

q=20 KN /m2 Solution: 3

Assume maximum vertical displacement=1 along apex of part

Angle of rotation of each panel is

1 

1 b

2 

1 3b

3 

1 1.5b

4 

1 b

Internal work done,

∑I

=

I 1 + I 2+ I 3

+

I4

I =( m+m' ) lθ

Page 18 of 25

I 1 =(m+m' ) x 3 b x

=

1 αb

9m α

I 2 =m x (1.5+α )b x

=

1 3b

0.5 m+ 0.33 αm

I 3 =m ( 3+ β ) b x

1 1 + m' (4.5 b) 1.5 b 1.5 b

=

2 m+0.67 β m+ 6 m

=

8 m+ 0.67 β m

I 4=(m+ m') x 1.5 b x

=

1 βb

4.5 m β

Internal work done, =

∑I

=

I 1 + I 2+ I 3

+

I4

9m 4.5 m +0.5 m+ 0.33 αm+8 m+ 0.67 β m+ α β

9 4.5 = m[ α + β +0.33 α +0.67 β+ 8.5] External work done,

∑E

=

E1 + E2 + E3

+

E4

1 1 1 E1=q x x 3 b x α b x + 0.5 qbx α b x +0.5 qb x (1−α ) bx 1+ 1.5q b2 x 1 2 3 2 =

0.5 α q b2+ 0.25 α q b 2+ 0.5 q b2−0.5 α q b2 +1.5 q b 2

=

0.25 α q b 2+ 2.0 q b2

Page 19 of 25

1 1 1 E2=0.5 αq b2 +qx ( 1−α ) bx 3 b x + q x x 3 b x 1.5b x 2 2 3 2

2

2

2

0.5 αq b +1.5 q b −1.5 αq b +0.75 q b

=

2 2 =2 .25 q b −αq b

1 1 1 2 E3=0.75 q b +q x (1.5−β ) b x 1.5 bx + q x x 1.5 bx β bx 2 2 3 2 2 2 2 = 0.75 q b +1.125 q b −0.75 βq b +0.25 βq b 2

2

1.875 q b −0.5 βq b

=

E4 =0.25 βq b2 External work done,

∑E

=

E1 + E2 + E3

+

E4

=

0.25 α q b 2+ 2.0 q b2 +2.25 q b 2−αq b2 +1.875 q b2−0.5 βq b2 +0.25 βq b2

=

6.125 q b2 −0.75 αq b 2−0.25 βq b2

2 = q b [6.125−0.75 α −0.25 β ]

∑ I =∑ E 9 4.5 m[ + +0.33 α +0.67 β+ 8.5]=q b2 [6.125−0.75 α−0.25 β ] α β 2

Therefore,

m=

q b [ 6.125−0.75 α−0.25 β] 4.5 [ 53.57+ +0.67 β ] β 2

=

q b [5.975−0.25 β ] β 2 53.57 β+ 4.5+0.67 β

The maximum value of m is obtained when

dm =0, dβ

( 53.57 β+ 4.5+0.67 β 2 ) ( 5.975−0.5 β )=[5.975 β−0.25 β 2 ](53.57+1.34 β) Page 20 of 25

2

2

3

2

2

320.08 β−26.79 β +26.89−2.25 β + 4 β −.34 β =320.08 β +8.01 β −13.39 β −0.34 β

3

−17.41 β 2−2.25 β+26.89=0 β=

2.25± √(−2.25)2−4 (26.89)(−17.41) 2(−17.41) β

= 1.17 or -1.31, Value of

can’t be negative

2

m=

20(2) [ 5.975−0.25 ( 1.17 ) ] 1.17 2

53.57(1.17)+ 4.5+0.67(1.17)

=7.81 KN m/m

6 Determine optimum value of moment resistance and β, if α=1, b=2.5m and q=15 KN/m2 C

B Part-3

1.8b

UDL=q P=0.8 qb

(line load)

Part-4 Part-2

W E

(point

αb

Part-1 F 2.5b

2

W =1.5 q b

D b

P

load) A

βb

1.5b

m' =2 m

Solution Assume point E has a virtual downward displacement of δ=1 along apex of part 3 Page 21 of 25

1 

1 b

2 

1 2.5b

3 

1 1.8b

4 

1 b

Internal work done,

∑I

=

I 1 + I 2+ I 3

+

I4

I =( m+m' ) lθ I 1 =m x 2.5b x

=

1 αb

2.5 m α

I 2 =m x (1.8+ α ) b x

1 1 +m' (2.8 b) x 2.5 b 2.5 b

=

0.72 m+

αm +2.24 m 2.5

=

2.96 m+

αm 2.5

I 3 =m ( 2.5+ β ) b x

1 1 + m' (4.0 b) 1.8 b 1.8 b

=

1.39 m+

βm + 4.44 m 1.8

=

5.83 m+

βm 1.8

Page 22 of 25

I 4=m x 1.8 b x

=

1 βb

1.8 m β

∑I

Internal work done,

I 1 + I 2+ I 3

=

+

I4

2.5 m αm βm 1.8 m + 2.96 m+ +5.83 m+ + = α 2.5 1.8 β 2.5 α β 1.8 m[ + + + +8.79] = α 2.5 1.8 β

∑E

External work done,

=

E1 + E2 + E3

+

E4

1 1 1 2 E1=q x x 2.5 b x α b x + 0.8 qbx α b x +0.8 qb x (1−α ) bx 1+ 1.5 q b x 1 2 3 2 =

0.42 α q b 2+ 0.4 α q b2+ 0.8 q b2 −0.8 α q b 2+1.5 q b2

=

0.02 α q b +2.3 q b

2

2

1 1 1 E2=0.42 αq b2 +qx ( 1−α ) bx 2.5 b x + q x x 2.5 b x 1.8 b x 2 2 3 =

0.42 αq b 2+1.25 q b2−1.25 αq b2 +0.75 q b 2

2 2 =2 .0 q b −0.83 αq b

1 1 1 2 E3=0.75 q b +q x (1.5−β ) b x 1.8 bx + q x x 1.8 bx β bx 2 2 3 2

2

2

= 0.75 q b +1.35 q b −0.9 βq b +0.3 βq b =

2

2.1 q b2−0.6 βq b2

E3=0.3 βq b2 External work done,

∑E

=

E1 + E2 + E3

+

E4

Page 23 of 25

2

2

2

=

0.02 α q b +2.3 q b +¿ 2 .0 q b −0.83 αqb

=

6.4 q b 2−0.81 αq b2−0.3 βq b 2

2

2

2

+ 2.1 q b −0.6 βq b +0.3 βq b

2

2 = q b [6.4−0.81 α −0.3 β ]

∑ I =∑ E 2.5 α β 1.8 2 + + + +8.79]=q b [ 6.4−0.81 α −0.3 β ] α 2.5 1.8 β

m[

Therefore ultimate moment resistance, 2

m=

q b [6.4−0.81 α −0.3 β ] 2.5 α β 1.8 [ + + + +8.79] α 2.5 1.8 β

For, α=1, b=2.5m and q= 15 KN/m2 m=

q b2 [6.4−0.81 α −0.3 β ] 2.5 α β 1.8 [ + + + +8.79] α 2.5 1.8 β 2

=

q b [5.59−0.3 β ] β 1.8 +11.69+ 1.8 β 2

=

q b [10.06−0.54 β ]β β 2 +21.04 β +3.24

The maximum value of m is obtained when

dm =0, dβ

( 21.04 β +3.24+ β 2 ) ( 10.06−1.08 β )=[10.06 β−0.54 β 2] (21.04+2 β) 211.66 β−22.72 β 2+32.59−3.50 β+ 10.06 β 2−1.08 β 3=211.66 β+20.12 β 2−11.36 β 2−1.08 β 3 −21.42 β 2−3.5 β+ 32.59=0

β=

3.5± √ (−3.5)2−4 (32.59)(−21.42) 2(−21.42)

Page 24 of 25

β

= 1.15 or -1.32, Value of

can’t be negative

2

m=

15(2.5) [ 10.06−0.54 ( 1.15 ) ] 1.15 2

21.04 (1.15)+ 3.24+( 1.15)

=35.39 KN m/m

Page 25 of 25