Question 9.1: Draw S.F.D. and B.M.D. for simple supported beam as shown in...

First, support reactions are to be determined. Consider F.B.D. of the simple supported beam AE.

ΣY = 0,

R_{A}+R_{E}=10+12+4

= 26 kN

R_{E} \times 4=(10 \times 1)+(12 \times 2)+(4 \times 3)

R_{E} = 11.5  KN

R_{A} = 14 5  KN

Note:
(i) For S.F.D., all downward forces acting on right side of a selected section are taken as positive while upward forces are taken as negative.
(ii) In B.M.D., force causing concavity of beam (Sagging) are considered as producing positive moment while force causing convex shape of beam are considered as producing negative moment.
(iii) Each S.F. or B.M. general equation for selected section gives the values of S.F. or B.M. respectively of that portion in which the section is considered.

Shear Force Diagram:

Consider sections x_{1} x_{1}^{1}, x_{2} x_{2}^{1}, x_{3} x_{3}^{1}, x_{4} x_{4}^{1} at distances x_{1}, x_{2}, x_{3} and x_{4}, respectively, from the end E as shown in the Fig. 9.9 (a)
To draw Shear Force diagram (S.F.D.),
Consider algebraic sum of all forces acting right side of section x_{1} x_{1}^{1}

(S . F)_{x_{1} x_{1}^{1}}=-R_{E}=-11.5  kN

This shows that value of shear force is not depending on distance from end E and remains constant between points D and E.

Thus  (S . F .)_{E}=(S . F .)_{D}=-11.5  kN

Consider algebraic sum of all forces acting right side of section x_{2} x_{2}^{1},

(S . F .)_{x_{2} x_{2}^{1}}=-R_{E}+4

= -11.5 + 4

= -7.5 kN

Thus, (S . F .)_{D}=(S . F .)_{C}=-7.5  kN

Consider section x_{3} x_{3}^{1},

(S . F .)_{x_{3} x_{3}^{1}}=-R_{E}+4+12

= -11.5 + 16

= +4.5 kN

(S . F)_{C}=(S . F)_{B}=+4.5  kN

Similarly, section x_{4} x_{4}^{1},

(S . F .)_{x_{4} x_{4}^{1}}=-11.5+4+12+10

= 14.5 kN

(S F)_{B}=(S . F)_{A}=+14.5  kN

Bending Moment Diagram:
To draw B.M.D., moments will be taken from axes X_{1} X_{1}^{1}, X_{2} X_{2}^{1}, X_{3} X_{3}^{1} and X_{4} X_{4}^{1} , respectively for all forces acting before corresponding section form the point E,

(B . M)_{X_{1} X_{1}^{1}}=+R_{E} \times x_{1}

= +11.5 x_{1}

We can observe that this equation is dependent of distance x_{1} from end E. B.M. is taken as positive as it is sagging moment.

Thus, ( BM )_{ E }=0,(\text { B.M. })_{ D }=+11.5  kNm

Similarly,

(B . M)_{X_{2} X_{2}^{1}}=+R_{E} \cdot x_{2}-4\left(x_{2}-1\right)

Here B.M. due to 4 kN will negative as it is hogging moment.
For B.M. at D put x_{2} = 1 m,

\text { (B.M.) }_{ D }=+11.5 \times 1-4(1-1)

= +11.5 kNm (Unchanged)

For B. M. at C put x_{2} = 2 m,

(\text { B. M. })_{C}=+11.5 \times 2-4(2-1)

= 19 kNm

(B . M)_{X_{3} x_{3}^{1}}=+R_{E} \times x_{3}-4\left(x_{3}-1\right)-12\left(x_{3}-2\right)

=11.5 x_{3}-4\left(x_{3}-1\right)-12\left(x_{3}-2\right)

For (B.M.) at C put x_{3} = 2 m,

\text { (B.M. })_{C}=11.5 \times 2-4(2-1)-12(2-2)

= 19 kNm (Unchanged)

at x_{3} = 3 m,

(\text { B.M. })_{ B }=+11.5 \times 3-4(3-1)-12(3-2)

= + 14.5 kNm

(\text { B. } M)_{X_{4} x_{4}^{1}}=11.5 x_{4}-4\left(x_{4}-1\right)-12\left(x_{4}-2\right)-10\left(x_{4}-3\right)

at x_{4} = 3 m,

(\text { B.M. })_{ B }=+11.5 \times 3-4(3-1)-12(3-2)-10(3-3)

= 14.5 kNm

at x_{4} = 4 m,

\text { (B.M. })_{ B }=11.5 \times 4-4(4-1)-12(4-2)-10(4-3)

= 0