Theory of Structures(1)

Lecture No. 5

Statically Determinate Arches

An arch: is a curved beam supported at its two ends. Most known types are: fixed- two hinged-

three hinged.

It can be analyzed by either analytical or graphical methods

Analytical MethodThree Hinged Arch

1-The reactions of the arch can be calculated from the three equations of equilibrium and conditional equation at the intermediate hinge.

2- The bending moment at any section determined by calculating the moments of all the forces to the right

or to the left of the section.

3 -Determining normal and shearing force requires resolving resultant of all forces to the left or to the

right of section along the tangent and the normal to it.

3-The slope of the tangent can be defined if the equation of all the arch is defined first.

4-The slope of the tangent is equal to the first derivative of the equation.

5-Substitute by the coordinates of the specified point in the first derivative to calculate the slope.

Determine the N.F, S.F., and M at points d and e of the three-hinged parabolic arch shown

The arch equation is Y= x – 0.025 x2

Solution

Σ Ma= 0 = Yb × 40 – 6 (10 + 20 + 30) Yb = 9 t ↑Σ Y= 0 = Ya + 9 – 3 × 6 Ya = 9 t ↑

= 0 = 9 × 20 – 6 × 10 – 10 Xa Xa = 12 t →

ΣX = 0 = 12 – Xb Xb = 12 t ←

c

acM

A section through point d

At x = 5, y = Yd = 5 - 0.025 × 52 = 4.375 m.

5

1tan

xdx

dy = 1-0.05x=1 – 0.05 * 5 = 0.75

1sin = 0.6 and 1cos = 0.8

Σ X = 0 = 12 – X X= 12t ←

Σ Y = 0 = 9 – Y Y= 9t ↓

ΣMd = 0 = 12 × 4.375 – 9 × 5 + M

M = - 7.5 m.t., i.e M= 7.5m.t. (clockwise)

Ysin Ɵ1

Ycos Ɵ1Xcos Ɵ1

Xsin Ɵ1

Ɵ1

Ysin Ɵ1

Ycos Ɵ1Xcos Ɵ1

Xsin Ɵ1

Ɵ1

At x = 30 y = ye = 30 – 0.025 × 302= 7.5 m.

30

xdx

dy = 1 – 0.05 × 30 = 2

1

2tan = 0.50, 2sin = 0.446 2cos = 0.893

At Point e

180-Ɵ2=153.43 Ɵ2=26.57

ΣMe = 0 = M + 12 × 7.5 – 9 × 10

M is the bending moment at point e = 0 m.t Ne= - (X cos Ɵ2+ Y sin Ɵ2 )

Ne = - (12 × 0.893 + 9 × 0.446) = - 14.74 t

Qe=X sin Ɵ2 - Y cos Ɵ2

Qe = 12 × 0.446 – 9 × 0.893 = - 2.685 t

X cos Ɵ2

X sin Ɵ2

Y cos Ɵ2 Y sin Ɵ2

For a section just to the right of e

ΣX = 0 = 12 – X X = 12 t →

ΣY = 0 = 9 – Y Y = 9 t ↓

7.5m

Ɵ2

For a section just to the left of e

ΣX = 0 = 12 – X X = 12 t →

ΣY = 0 = 9 – 6 – Y Y= 3 t ↓

Ne= - (X cos Ɵ2+ Y sin Ɵ2 )

Ne = - (12 × 0.893 + 3 × 0.446) = - 12.05t

Qe=X sin Ɵ2 - Y cos Ɵ2

Qe = 12 × 0.446 – 3 × 0.893 = 2.673 t

Y cos Ɵ2

X sin Ɵ2

Y sin Ɵ2

X cos Ɵ2

Ɵ2

Example 4.2 : Determine the N.F, S.F., and M at point d of the three-hinged circular arch shown

The equation of the arch is Y=ax2+bx+c At A x=0 y=0 c=0 At x=8 y=4 4=a(8)2 + 8b 1=16a+2b (1) A t x=16 y=0 0=a(16)2 + 16b -b = 16a (2) Substitute from 2 into 1 1=-b+2b b=1 a=-1/16 Y=-x2 /16 +x

Example 4.2 : Determine the N.F, S.F., and M at point d of the three-hinged circular arch shown

The equation of the arch is Y=-x2/16+x

ΣMb= 0 = Ya × 16+ 4 ×2 – 6 × 10 - 8 × 4- 2 ×2 Ya = 5.5t ΣY= 0 Yb=10.5t ΣMc= 0 for left part 6 ×2+4 ×2 + 4 Xa - 5.5 ×8 = 0 Xa=6t ΣX= 0 Xb=10t

Y a = 5.5t Yb =10.5t

Xa = 6t Xb = 10t

Y=-X2/16+X At X=11 Y = 3.44mdy /dx = -X / 8+1 dy/dx= tan(180- Ɵ) = -0.375 Ɵ=20.56

SinƟ= 0.351 CosƟ = 0.936

A section through point d

Ɵ

10.5 t

10t

10t

0.5t

10 cos Ɵ

0.5 sin Ɵ

10 sin Ɵ

0.5 cos Ɵ

Ɵ

Nd= - (10 cos Ɵ+ 0.5 sin Ɵ )

Ne = - (10 ×0.936 + 0.5 × 0.351 ) = - 9.53 t

Qd=10 sin Ɵ – 0.5 cos Ɵ

Qe = 10 × 0.351 – 0.5 × 0. 936 = + 3.04 t

Md + 8 ×1+ 2 ×3+ 10 ×3.44 = 10.5 ×5

Md = 4.1 clockwise

Ɵ

10.5 t

10t

10t

0.5t

10 cos Ɵ

0.5 sin Ɵ

10 sin Ɵ

0.5 cos Ɵ

3.44 m

5 m

3 m

Problem 4.2 : Determine the N.F, S.F., and M at point S of the three-hinged circular arch shown

S

4.8m

Solution

Σ MC= 0 = Yb × 6 – 3.6 Xb-2*6*3=0 YB = 0.6 XB + 6

Σ MA= 0 2.8 XB+14YB = 2*14*7 XB= 10 t

YB=12 t

S4.8m

10t

8t

60

608sin60

8cos60

10cos 60

10sin60

8t

16t

10t

4.8m

Σ X= 0 XA=10t

Σ Y= 0 YA=16 t

Σ MS=MS-16*4+ 10*4.8+ 8*2=0

MS=0

Ns= - (10 sin 60+ 8 cos60 )=-12.66t

Qs= 8 sin 60 -10 cos 60 = - 1.93t 4m

60