Strength of materials

Tutorial 2: Solution Continuous Beams


Question 1.

Calculate the bending moments and reaction forces at points A,B,C and D. Also calculate point(s) of
contraflecture (PCF’s) if there are any.

16 kN
A1 10 kN/m
12 A2 16 kNm A3 A4 10 kN

A B C D
A1 A2
2m A3 1m
x1 A4
4m 5m 4m x4
X2
x3
24kNm
19.2 kNm
MB
MC 7.5 kNm
6.4 kNm
MA= 0 kNm MD =0 kNm
-9.6 kNm
Spans AB & BC
 A1 x1 A2 x2 
 M L  2M L  L    M L 6 
a 1 b 1 2 c 2  L1 L 
 2 


[ ( ) + 1 2×2×6 . 4×2 3×2−1 2×3×9 . 6×(2+1 3×3)
]
2 1 2×2 1 1
×24 ×4×2 ×2×19 . ×2+ ×3×19 . 2× 2+ ×3
3 2 3 2 3
−0−2 M B ( 4 +5 ) −5 M C =6 +
4 5 5

−18 M B−5 M C =6 32+ [
25.6+86.4 8. 53−43.2
5
+
5 ]
−18 M B−5 M C =284.8 ......................... ( 1 )
Alternatively
3
wl Wb ( 2 2 ) M ( 2
−18 M B−5 M C = + L −b − L −3b 2 )
4 L L
3
12∗4 16∗2 ( 2 2 ) 16 ( 2
−18 M B−5 M C = + 5 −2 − 5 −3∗22 )
4 5 5
−18 M B−5 M C =284.8 ............................ ( 1 )
Spans BC & CD
 A1 x1 A2 x2 
 M L  2M L  L    M L 6 
a 1 b 1 2 c 2  L1 L 
 2 




[ ( ) ( )
]
1 2×2 1 1 −1 6×2 1 1
×3×19. ×3+ ×3×19.2× 3+ ×2 ×3×9. ×3+ ×2×6.4× 3+ ×2
−5 M B−2 M C ( 5+4 ) −0=6¿ 2 3 2 3
+
2 3 2 3 ¿ ¿
¿
5 5
¿
−5 M B−18 M C =6 [
57.6+70.4 −28.8+23. 4672 2.5+22.5
5
+
5
+
4 ]
−5 M B−18 M C =184.7 .......................... ( 2 )
M 2
−5 M B−18 M C = ( l −3 a 2 ) + Wab ( l+a )+ Wab (l+b )
l l l



1

, Tutorial 2: Solution Continuous Beams


16 2
−5 M B−18 M C = ( 5 −3×32 ) + 16×3×2 ( 5+3 ) + 10×3×1 ( 4+1 )
5 5 4
−32 768 150
−5 M B−18 M C = + + =184.7 ...................... ( 2 )
5 5 4

[
−18 −5
]{ } { }
M B = 248. 8 ⇒ AX=B
−5 −18 M C 184 .7 { }
M
X = B = A−1 B=
MC {−14 .06
−6 .36 }
M B= −14 . 06 kNm and C M = −6 .36 kNm
Taking Moments about supports B & C, we get the reaction forces as follows:
M B = −14 . 06 kNm=4 R A −12×4×2 ∴ R A =20 . 485 kN
M C = −6 .36 kNm=9 R A +5 R B +16−12×4×7−16×2 ∴ R B=32. 255 kN
M C = −6 .36 kNm=4 R D−10×3 ∴ R D=5 . 91 kN
M B= −14 . 06 kNm=9 R D +5 RC −16−16×3−10×8 ∴ RC =15 . 35 kN
Points of contraflecture:
M x 1= 0=R A x 1 −6 x 2 =20. 485 x 1 −6 x 1=x 1 ( 20 . 485−6 x1 )
Cut A1-A1 : 1

20 . 485
∴ x 1=0 m ∴ x 1= =3 . 414 m
or 6 all measured from support A.
M x 2= 0=R A x 2 + R B ( x 2−4 ) −6 x 2 −6 ( x 2 −4 )2
Cut A2-A2 : 2

0=20 . 485 x 2 + 32. 255 ( x 2−4 ) −6 x 2 −6 ( x 2 −8 x 2 +16 )=4 . 47 x 2 −33. 02
2 2

33 . 02
0=4 . 47 x 2 −33 . 02 ∴ x 2= =6 .966 m ≈ 7 m
4 . 74 from support A or 6 m from B.
M x 3= 0=R D x 3 + RC ( x 2 −4 )−10 ( x 3 −1 )
Cut A3-A3 :
0=5 .91 x 3 +15 . 35 ( x2 −4 ) −10 x 3 +10 ∴ x 3=4 . 565 m ∴ x 2=8. 435 m from A
from B.
M x 4 = 0=R D x 4 −10 ( x 4 −1 ) =5. 91 x 4 −10 x 4 +10
Cut A1-A1 :
∴ x 4 =2 . 444 m from B or ∴ x 4 =10 .556 m from A.

Question 2

Calculate the bending moments and reactions at points A, B and C. Calculate the point of
contraflecture between points “B and C”


10 kN 16 kN/m 20 kN


A B C
1.5 m
3m 4m 1m
32 kNm x
18 kNm
7.5 kNm MB
MA
MC=-20 kNm
MD=0


2