Determine Bending Moment

And

Share Force Diagram of beam.

Course Code: CE 311 Course name: Structural Analysis I

ABSTRACT Some bending moment and share force of critically loaded beam. Analysis of this beam by Approximate method are solved in this document.

The problem of this Document is solved by 7th batch student of Civil Engineering Department. Leading University, Sylhet, Bangladesh.

Find out the SFD and BMD of the diagram?

Question 01:

10k

Internalhing

20k/ft

104k/ft

50k

Internalhing

2k/ft

Solution 01:

10k

Internalhing

20k/ft

104k/ft

50k2k/ft

Internalhing

4.166k 527.834k 1194k 10ka b c d e f g h i

4.166

-5.834

522

122

-1118

7626

10

10

29.162

-29.17

6693.2278

-749.458

25

0

6410.83

-141.458

Loaddiagram

Sharediagram(kips)

Momentdiagram(k-ft)

X

X

X

X1

Detailed calculation From figure 1 ΣMRA =0 → 7 × 10 − 푅푐 × 12 = 0 → 푅푐 = 5.834 퐾. ΣFy = 0 → 푅푎 − 10 + 푅푐 = 0

→ 푅푎 = 4.166 K. From the figure 2 Rh = Ri =10 K.

10k

Figure:1Ra

Rc

20k/ft

104k/ft

50k 2k/ft5.834K 10K

Figure:3

Rd Rf

Rh Ri

Figure:2

2k/ft

:

from figure 3 ΣM푅푑 = 0 → −5.834 × 5 + 800 × 20 + 840 × 33.334− 푅f × 40 + 50 × 48 + 16 × 52 + 10 × 56 = 0 → 푅f = 11994 K. Σ퐹푦 = 0 → −5.834 + 푅푑 − 800 + 1194− 50− 16− 10 = 0 → 푅푑 = 527.834 퐾. Share force calculation: 퐹푎 → 0 + 4.166 = 4.166 푘 퐹푏 → 4.166− 10 = −5.834푘 퐹푑 → −5.834 + 527.834 = 522푘 퐹푒 → 522− 20 × 20 = 122푘 퐹f → 122 − 0.5(20 + 104)20 = −1118k 퐹f → −1118 + 1194 = 76k 퐹푔 → 76− 50 = 26푘 퐹ℎ → 76− (2 × 8) = 10푘 퐹푖 → 10− (2 × 10) = −10푘 퐹푖 → −10 + 10 = 0푘 Bending moment calculation: 푀푎 → 0 푀푏 → 4.166 × 7 = 29.162 푘 − 푓푡 푀푐 → 29.162− (5.834 × 5) = 0푘 − 푓푡 푀푑 → −(5.834 × 5) = −29.62 푘 − 푓푡 푀푒 → −29.62 + 0.5(522 + 122)20 = 6410.83푘 − 푓푡 To get moment in X point, we have to know the distance X

20k/ft

104k/ft

xY

Figure 4

From figure 4 =

푌 = Now V x =0 → 4.166− 10 + 527.834− 20 × (20 + 푋) − 푥푦 = 0 → −2.1x2 - 20x + 122 = 0 Solving the second degree equation ,get

X = 4.2254 and Y=17.747 Mx → 4.116 × 41.2254− 10 × 34.2254 + 527.834 × 24.2254− (20 × 24.2254) ×12.1127− 0.5 × 4.254 × 17.747 × 1.4 = 6693.2278 푘 − 푓푡. Md→ 4.166 × 57 − 10 × 50 + 527.834 × 40− 800 × 20− 840 × 6.667 = −749.458 푘 −푓푡. Me→ −749.458 + (76 × 8) = −141.458 푘 − 푓푡. Mf→ −141.458 + 0.5 × (26 + 10) × 8 = 0 푘 − 푓푡

To know moment in X1

10

10X

→ = → 푥 = 5 So, Mx1→ 0.5 × 5 × 10 = 25 푘 − 푓푡. Mi→ 25 − 0.5 × 5 × 10 = 0 푘 − 푓푡. (Solved)

Question 02:

4 K/ft

7 K/ft

10 K 10 K 3 K 4 K 3 K

AB

C D EF

H I JLKG

Solution:

4 K/ft

7 K/ft

10 K 10 K 3 K 4 K 3 K

OA

B C DE

G H IKJF

0

Ra = 70.55K Rc =114.11 K

Re= -28.45K

RL = 4.79 K

Shear ForceDiagram (K)

0 BendingMomentDiagram(K-ft)

Details Calculation : For O to B part ∑푀 = 0

=> −40 × 18 + RA ×13 -0.5×7×13×13/3=70.55

=> 푅 = 70.55퐾

∑퐹 = 0

=> 푅 = × 13 × 7 + 40 − 70.55

=> 푅 = 14.95 퐾

For G to K part,

∑푀 = 0

=> 14푅 = 3 × 4 + 4 × 7 + 3 × 11

=> 푅 = 5.21 퐾

∴ 푅 = 5.21 퐾

∑퐹 = 0

=> 푅 = 3 + 4 + 3 − 5.21

=> 푅 = 4.79 퐾

For B to G part, ∑푀 = 0

=> 10 × 5 − 10 × 푅 + 10 × 15 + 5.21 × 20 − 14.95 × 13 − × 7 × 13 × 2 × = 0

∴ 푅 = −28.45 퐾

푁표푤,∑퐹푦 = 0

=> 푅 = 14.95 + × 7 × 13 + 10 + 10 + 5.21 + 28.45

=> 푅 = 114.11

4 K/ft

7 K/ft

10 K 10 K

3 K 4 K 3 K

oA

B

c DE

G H I KJ

F

7 K/ft

Rb

Ra

Rc Re

RG

Rg Rk

Rb

Shear Force:

퐹 = 0 퐾

퐹 = −4 × 10 = −40퐾

퐹 = −40 + 70.55 = 30.55 퐾

퐹 = 30.55 − × 7 × 13 = −14.95 퐾

퐹 = −14.95 − × 7 × 13 = −60.45 퐾

퐹 = −60.45 + 114.1 = 53.66

퐹 = 53.66 − 10 = 43.66 퐾

퐹 = 43.66 − 28.45 = 15.21 퐾

퐹 = 15.21 − 10 = 5.21 퐾

퐹 = 5.21 − 3 = +2.21 퐾

퐹 = 2.21 − 4 = −1.79 퐾

퐹 = 1.79 − 3 = −4.79 퐾

퐹 = −4.79 + 4.79 = 0 퐾

=

=> 푦 =

푉 = 0 => 30.55 − × 푋 × ( ) = 0 => 푥 = 10.61

Bending Moment:

푀 = 0 푀 = −4 × 10 × 5 = −200 퐾 − 푓푡 푀 = −4 × 10 × (5 + 10.61) + 70.55 × (10.61) − × 10.61 × × . × . = +15.387 = +16.94 퐾 − 푓푡 푀 = −4 × 10 × (18) + 70.55 × (13) − × 13 × 7 × = −0.016 퐾 − 푓푡

푀 = −4 × 10 × (5 + 26) + 70.55 × (26) − × 26 × 7 × = −588.7 퐾 − 푓푡 푀 = −588.7 + 53.66 × 5 = −370.4 퐾 − 푓푡 푀 = −370.4 + 43.66 × 5 = −102.1 퐾 − 푓푡 푀 = −102.1 + 15.21 × 5 = −26.05 퐾 − 푓푡 푀 = −26.05 + 5.21 × 8 = +15.63 퐾 − 푓푡

푀 = +15.63 + 2.21 × 4 = +24.47 퐾 − 푓푡

푀 = +24.47− 1.79 × 3 = +19.1 퐾 − 푓푡

푀 = +19.1− 4.79 × 4 = 0 퐾 − 푓푡

(solved)

푥

13′

7 푦

Question 03:

2 0 k1 0 k / f t

Solution:

2 0 k1 0 k / f t

L o a d d ia g r a m

S F D K i p s

B M D k - f t

X X

Y Y

2 1 . 6 7

- 4 3 . 3 3

- 1 2 3 . 3 3

6 0

1 0

- 6 0

1 2 3 . 3 2 6 2

4 3 . 3 2 6 2

2 1 . 6 7

1 0 8 . 5 2 5

6 6 6 . 6 4

4 9 1 . 6 4

6 6 6 . 6 4

1 0 8 . 5 2 5

00

ab c c d e f 2 1 . 6 71 8 3 . 3 2 3 61 8 3 . 3 3 62 1 . 6 7

- 1 0

details calculation: From figure 1

Σ푀푅푎 → 65 × 8.666 − 13 × 푅푏 = 0

→ 푅푏 = 43.33푘

Σ퐹푦 = 0

→ 푅푎 − 65 + 43.33 = 0

→ 푅푎 = 21.67푘

From figure 2

Σ푀푅푒 → 65 × 8.666 − 13 × 푅푏 = 0

→ 푅푒 = 43.33푘

Σ퐹푦 = 0

→ 푅푓 − 65 + 43.33 = 0

→ 푅푓 = 21.67푘

10k-ft

Rbfigure:1

10k-ft

Re Rffigure:2

43.33k43.33k

10k-ft

20k

Rc Rdfigure:3

Ra

Y

X

Y

X

From figure 3

Σ푀푅푐 = 0

→ −43.33 × 8 + 280 × 5− 푅푑10 + 43.33 × 18 = 0

→ 푅푑 = 183.3236푘

Σ퐹푦 = 0

→ 43.33 − 푅푐 + 280 − 183.3236 + 43.33 = 0

→ 푅푐 = 183.3364푘

Share force calculation

퐹푎 = 21.67푘

퐹푏 = 21.67 − 65 = −43.33푘

퐹푐 = −43.33− (10 × 8) = −123.33푘

퐹푑 = −123.33 + 183.33 = 60푘

퐹푒 = 60 − (5 × 10) = 10푘

퐹푒 = 10 − 20 = −10푘

퐹푓 = −10 − (5 × 10) = −60푘

퐹푓 = −60 + 183.3236 = 123.3236푘

퐹푔 = 123.3236 − (10 × 8) = 43.33푘

퐹ℎ = 43.33 − 65 = −21.67푘

퐹ℎ = −21.67 + 21.67 = 0푘

Bending moment calculation:

Ma=0

From figure 1

푌푋

=1013

푦 =10푋13

Y=5.77

Vx=0

=> 21.67-0.5XY=0

푋 = 7.5

푀푥 = 21.6 × 7.5 − 0.5 × 7.5 × 5.77 × 1.92 = 108.525 푘 − 푓푡

푀푏 = 21.6 × 13− 0.5 × 13 × 10 × 4.334 = 0푘 − 푓푡

푀푐 = −0.5 × (43.33 + 123.33) × 8 = −666.64푘 − 푓푡

푀푑 = −666.64 + 0.5(60 + 10)5 = −491.64푘 − 푓푡

푀푒 = −491.64 − 0.5(60 + 10)5 = −666.64푘 − 푓푡

푀푓 = −666.64 + 0.5 × (43.33 + 123.33) × 8 = 0푘 − 푓푡

From figure 2

Similarly from previous calculation

We get X=7.5 And Y=5.77

Mx1=21.6 × 7.5− 0.5 × 5.5 × 5.77 = 108.525푘 − 푓푡

푀푓 = 21.6 × 13 − 0.5 × 13 × 10 × 4.334 = 0푘 − 푓푡 (Solved)