This document summarizes the bending moment (BMD) and shear force (SFD) calculations for three beam problems. The first problem involves calculating the SFD and BMD for a beam with various point loads. The second problem does the same for a beam with distributed loads. The third problem again calculates SFD and BMD, determining values at specific points along the beam. All problems show the free body diagrams, mathematical equations used, and tabulated results.

1. 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.

2. Find out the SFD and BMD of the diagram?
Question 01:
10k
Internal
hing
20k/ft
104k/ft
50k
Internal
hing
2k/ft

3. Solution 01:
10k
Internal
hing
20k/ft
104k/ft
50k
2k/ft
Internal
hing
4.166k 527.834k 1194k 10k
a b c d
e f g h i
4.166
-5.834
522
122
-1118
76
26
10
10
29.162
-29.17
6693.2278
-749.458
25
0
6410.83
-141.458
Load
diagram
Share
diagram
(kips)
Moment
diagram
(k-ft)
X
X
X
X1

4. Detailed calculation
From figure 1
ΣMRA =0
→ 7 × 10 − × 12 = 0
→ = 5.834 .
ΣFy = 0
→ − 10 + = 0
→ = 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
:

5. from figure 3
ΣM = 0
→ −5.834 × 5 + 800 × 20 + 840 × 33.334 − f × 40 + 50 × 48 + 16 × 52 + 10 × 56 = 0
→ f = 11994 K.
Σ = 0
→ −5.834 + − 800 + 1194 − 50 − 16 − 10 = 0
→ = 527.834 .
Share force calculation:
→ 0 + 4.166 = 4.166
→ 4.166 − 10 = −5.834
→ −5.834 + 527.834 = 522
→ 522 − 20 × 20 = 122
f → 122 − 0.5(20 + 104)20 = −1118k
f → −1118 + 1194 = 76k
→ 76 − 50 = 26
ℎ → 76 − (2 × 8) = 10
→ 10 − (2 × 10) = −10
→ −10 + 10 = 0
Bending moment calculation:
→ 0
→ 4.166 × 7 = 29.162 −
→ 29.162 − (5.834 × 5) = 0 −
→ −(5.834 × 5) = −29.62 −
→ −29.62 + 0.5(522 + 122)20 = 6410.83 −
To get moment in X point, we have to know the distance X
20k/ft
104k/ft
x
Y
Figure 4

6. From figure 4
 =
 =
Now V x =0
→ 4.166 − 10 + 527.834 − 20 × (20 + ) − = 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 − .
Md→ 4.166 × 57 − 10 × 50 + 527.834 × 40 − 800 × 20 − 840 × 6.667 = −749.458 −
.
Me→ −749.458 + (76 × 8) = −141.458 − .
Mf→ −141.458 + 0.5 × (26 + 10) × 8 = 0 −
To know moment in X1
10
10X
→ =
→ = 5
So, Mx1→ 0.5 × 5 × 10 = 25 − .
Mi→ 25 − 0.5 × 5 × 10 = 0 − .
(Solved)

7. Question 02:
4 K/ft
7 K/ft
10 K 10 K 3 K 4 K 3 K
A
B
C
D E
F
H I J
LKG

8. Solution:
4K/ft
7K/ft
10K 10K 3K 4K 3K
O
A
B
C D
E
G H I
KJF
0
Ra=70.55K
Rc=
114.11K
Re=-28.45
K
RL=4.79K
ShearForce
Diagram(K)
0 Bending
Moment
Diagram
(K-ft)

9. Details Calculation :
For O to B part
∑ = 0
=> −40 × 18 + RA ×13 -0.5×7×13×13/3=70.55
=> = 70.55
∑ = 0
=> = × 13 × 7 + 40 − 70.55
=> = 14.95
For G to K part,
∑ = 0
=> 14 = 3 × 4 + 4 × 7 + 3 × 11
=> = 5.21
∴ = 5.21
∑ = 0
=> = 3 + 4 + 3 − 5.21
=> = 4.79
For B to G part,
∑ = 0
=> 10 × 5 − 10 × + 10 × 15 + 5.21 × 20 − 14.95 × 13 − × 7 × 13 × 2 × = 0
∴ = −28.45
, ∑ = 0
=> = 14.95 + × 7 × 13 + 10 + 10 + 5.21 + 28.45
=> = 114.11

10. 4 K/ft
7 K/ft
10 K 10 K
3 K 4 K 3 K
o
A
B
c D
E
G H I KJ
F
7 K/ft
Rb
Ra
Rc Re
RG
Rg Rk
Rb
Shear Force:
= 0
= −4 × 10 = −40
= −40 + 70.55 = 30.55
= 30.55 − × 7 × 13 = −14.95
= −14.95 − × 7 × 13 = −60.45
= −60.45 + 114.1 = 53.66
= 53.66 − 10 = 43.66
= 43.66 − 28.45 = 15.21

11. = 15.21 − 10 = 5.21
= 5.21 − 3 = +2.21
= 2.21 − 4 = −1.79
= 1.79 − 3 = −4.79
= −4.79 + 4.79 = 0
=
=> =
= 0
=> 30.55 − × × ( ) = 0
=> = 10.61
Bending Moment:
= 0
= −4 × 10 × 5 = −200 −
= −4 × 10 × (5 + 10.61) + 70.55 × (10.61) − × 10.61 ×
× .
×
.
=
+15.387
= +16.94 −
= −4 × 10 × (18) + 70.55 × (13) − × 13 × 7 × = −0.016 −
= −4 × 10 × (5 + 26) + 70.55 × (26) − × 26 × 7 × = −588.7 −
= −588.7 + 53.66 × 5 = −370.4 −
= −370.4 + 43.66 × 5 = −102.1 −
= −102.1 + 15.21 × 5 = −26.05 −
= −26.05 + 5.21 × 8 = +15.63 −
= +15.63 + 2.21 × 4 = +24.47 −
= +24.47 − 1.79 × 3 = +19.1 −
= +19.1 − 4.79 × 4 = 0 −
(solved)
13′
7

12. Question 03:
2 0 k
1 0 k / ft
Solution:
2 0 k
1 0 k / ft
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
a
b 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

13. details calculation:
From figure 1
Σ → 65 × 8.666 − 13 × = 0
→ = 43.33
Σ = 0
→ − 65 + 43.33 = 0
→ = 21.67
From figure 2
Σ → 65 × 8.666 − 13 × = 0
→ = 43.33
Σ = 0
→ − 65 + 43.33 = 0
→ = 21.67
10k-ft
Rb
figure:1
10k-ft
Re Rf
figure:2
43.33k43.33k
10k-ft
20k
Rc Rd
figure:3
Ra
Y
X
Y
X

14. From figure 3
Σ = 0
→ −43.33 × 8 + 280 × 5 − 10 + 43.33 × 18 = 0
→ = 183.3236
Σ = 0
→ 43.33 − + 280 − 183.3236 + 43.33 = 0
→ = 183.3364
Share force calculation
= 21.67
= 21.67 − 65 = −43.33
= −43.33 − (10 × 8) = −123.33
= −123.33 + 183.33 = 60
= 60 − (5 × 10) = 10
= 10 − 20 = −10
= −10 − (5 × 10) = −60
= −60 + 183.3236 = 123.3236
= 123.3236 − (10 × 8) = 43.33
ℎ = 43.33 − 65 = −21.67
ℎ = −21.67 + 21.67 = 0
Bending moment calculation:
Ma=0
From figure 1
=
10
13
=
10
13
Y=5.77
Vx=0
=> 21.67-0.5XY=0
= 7.5

15. = 21.6 × 7.5 − 0.5 × 7.5 × 5.77 × 1.92 = 108.525 −
= 21.6 × 13 − 0.5 × 13 × 10 × 4.334 = 0 −
= −0.5 × (43.33 + 123.33) × 8 = −666.64 −
= −666.64 + 0.5(60 + 10)5 = −491.64 −
= −491.64 − 0.5(60 + 10)5 = −666.64 −
= −666.64 + 0.5 × (43.33 + 123.33) × 8 = 0 −
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 = 108.525 −
= 21.6 × 13 − 0.5 × 13 × 10 × 4.334 = 0 − (Solved)