An arch is supported on frame with different levels of supports as shown above. The frame is supported on a hinge at support A and a roller at support C. A tie beam is used to connect node B and support C. Plot the bending moment, shear force, and axial force diagram die to the externally applied load.

**Solution**

**Support Reactions**

**∑M**

_{C}= 030Ay – 10Ax – (10 × 30

^{2})/2 = 0

30Ay – 10Ax = 4500 ------------- (1)

**∑F**

_{X}= 0Ax + 25 = 0

**Ax = -25 kN**(pointing in the reverse of assumed direction)

Plugging the value of Ax into equation (1);

30Ay – 10(-25) = 4500

30Ay = 4250

**Ay = 141.667 kN**

**∑M**

_{A}= 030Cy – (25 × 10) – (10 × 30

^{2})/2 = 0

30Cy = 4750

**Cy = 158.333 kN**

**∑M**

_{D}^{R}= 0
15Cy – (H × 5) – (10 × 15

15(158.333) – 5H – (10 × 15

– 5H + 1250 = 0

###

The ordinate of the arch at any given horizontal length section is given by;

y = [4y

y = [(4 × 5) × (30

dy/dx = y’ = 2/3 - 2x/45

y’ = 2/3 - 0/8 = 0.667

sinÎ¸ = y'/√(1 + y'

cosÎ¸ = 1/√(1 + y'

y = (2/3)

y’ = 2/3 - 2(7.5)/45 = 0.333

sinÎ¸ = y'/√(1 + y'

cosÎ¸ = 1/√(1 + y'

y = (2/3)

y’ = 2/3 - 2(15)/45 = 0

sinÎ¸ = y'/√(1 + y'

cosÎ¸ = 1/√(1 + y'

y = (2/3)

y’ = 2/3 - 2(22.5)/45 = -0.333

sinÎ¸ = y'/√(1 + y'

cosÎ¸ = 1/√(1 + y'

y’ = 2/3 - 2(30)/45 = -0.667

sinÎ¸ = y'/√(1 + y'

cosÎ¸ = 1/√(1 + y'

M

M

M

M

M

Coming from the right

M

M

Q

Q

Q

Q

Note that the horizontal forces of 25 kN will eliminate each other;

Q

Q

Q

Q

Q

Member AB = 141.667 kN (compression)

Member BC = 250 kN (Tension)

For the arch section;

N

N

N

N

N

N

Thank you for visiting Structville.

Connect with our facebook page at

^{2})/2 = 015(158.333) – 5H – (10 × 15

^{2})/2 = 0– 5H + 1250 = 0

**H = 250 kN**###
**Geometric Properties of the Arch Section**

The ordinate of the arch at any given horizontal length section is given by;y = [4y

_{c}(L*x*-*x*^{2})] / L^{2}^{}Where y_{c}is the height of the crown of the archy = [(4 × 5) × (30

*x*-*x*^{2})]/30^{2}= (2/3)*x*- (*x*^{2}/45)dy/dx = y’ = 2/3 - 2x/45

**At x = 0; y = 0**y’ = 2/3 - 0/8 = 0.667

sinÎ¸ = y'/√(1 + y'

^{2}) = 1/√(1 + 0.667^{2}) = 0.5546cosÎ¸ = 1/√(1 + y'

^{2}) = 1/√(1 + 0.667^{2}) = 0.8319**At x = 7.5m;**y = (2/3)

*x*- (*x*^{2}/45) = (0.667 × 7.5) - (7.5^{2}/45) = 3.75my’ = 2/3 - 2(7.5)/45 = 0.333

sinÎ¸ = y'/√(1 + y'

^{2}) = 0.333/√(1 + 0.333^{2}) = 0.43159cosÎ¸ = 1/√(1 + y'

^{2}) = 1/√(1 + 0.333^{2}) = 0.9487**At x = 15 m;**y = (2/3)

*x*- (*x*^{2}/45) = (0.667 × 15) - (15^{2}/45) = 5.00my’ = 2/3 - 2(15)/45 = 0

sinÎ¸ = y'/√(1 + y'

^{2}) = 0/√(1 + 0^{2}) = 0cosÎ¸ = 1/√(1 + y'

^{2}) = 1/√(1 + 0^{2}) = 1.0**At x = 22.5m;**y = (2/3)

*x*- (*x*^{2}/45) = (0.667 × 22.5) - (22.5^{2}/45) = 3.75my’ = 2/3 - 2(22.5)/45 = -0.333

sinÎ¸ = y'/√(1 + y'

^{2}) = -0.333/√(1 + 0.333^{2}) = -0.43159cosÎ¸ = 1/√(1 + y'

^{2}) = 1/√(1 + 0.333^{2}) = 0.9487**At x = 30m; y = 0**y’ = 2/3 - 2(30)/45 = -0.667

sinÎ¸ = y'/√(1 + y'

^{2}) = -0.667/√(1 + 0.667^{2}) = -0.5546cosÎ¸ = 1/√(1 + y'

^{2}) = 1/√(1 + 0.667^{2}) = 0.8319**Internal Stresses in the Arch Structure****Bending Moment**M

_{A}= 0M

_{B}^{B }= (25 × 10) = 250 kN.mM

_{B}^{UP }= (25 × 10) = 250 kN.mM

_{1}= (141.667 × 7.5) + (25 × 13.75) – (25 × 3.75) – (10 × 7.5^{2})/2 – (250 × 3.75) = 93.7525 kN.mM

_{D}= (141.667 × 15) + (25 × 15) – (25 × 5) – (10 × 15^{2})/2 – (250 × 5) = 0Coming from the right

M

_{2}= (158.333 × 3.75) – (250 × 5) – (10 × 12^{2})/2 = -31.252 kN.mM

_{C}= 0

**Shear force**

**Section A - B (Column)**Q

_{A }– Ax = 0Q

_{A }– 25 = 0Q

_{A }– Q_{B}^{B}= 25 kNQ

_{i}= ∑V cosÎ¸ - ∑H sinÎ¸ (for arch section)Note that the horizontal forces of 25 kN will eliminate each other;

Q

_{B}^{UP}= (141.667 × 0.8319) – (250 × 0.5546) = -20.797 kNQ

_{1}^{R}= Q_{1}^{L}= [141.667 – (10 × 7.5)] × 0.9487 – (250 × 0.43159) = -44.651 kNQ

_{D}= [141.667 – (10 × 15)] × 1.0 – (250 × 0) = -8.333 kNQ

_{2}^{R}= Q_{2}^{L}= [141.667 – (10 × 22.5)] × 0.9487 – (250 × –0.43159) = 28.839 kNQ

_{C}= [141.667 – (10 × 30)] × 0.8319 – (250 × -0.5546) = 6.932 kN**Axial force**Member AB = 141.667 kN (compression)

Member BC = 250 kN (Tension)

For the arch section;

N

_{i}= -∑V sinÎ¸ - ∑H cosÎ¸N

_{B}^{UP}= -(141.667 × 0.5546) – (250 × 0.8319) = -286.543 KN (Compression)N

_{1}^{L}= -[141.667 - (10 × 7.5)] × 0.43159 – (250 × 0.9487) = -265.848 KNN

_{D}= -[141.667 – (10 × 15)] × 0 – (250 × 1) = -250 KN (Compression)N

_{2}^{L}= N_{3}^{R}= -[141.667 – (10 × 22.5)] × -0.43159 – (250 × 0.9487) = -273.141 KNN

_{C}= -[141.667 – (10 × 30)] × -0.5546 – (250 × 0.8319) = -295.786 KN**Internal Stresses Diagram**Thank you for visiting Structville.

Connect with our facebook page at

__www.facebook.com/structville__
Very impressive

ReplyDeleteThank you, and feel free to share..

Delete