Theory of Structures(1)

Lecture No. 4

Statically Determinate Rigid Frames

A Frame : is a structure composed of a number of members .connected together by joints all or some are rigid

Hinged or Pin-connection allows relative rotation Between the ends of the connected members.

Rigid connection does not allow relative rotationbetween the ends of the connected members.

Internal Stability and Determinacy

If No. of members= m No. of External reaction components = r

No. of unknowns = 3m + r.No. of joints= j No. of equilibrium Eq.(s) = 3j; No. of condition Eq.(s) =S

Total No. of available Eq.= 3j+S

IF 3j + S > 3m + r, the frame is unstable.

3j + S = 3m + r, the frame is statically determinate.

3j + S < 3m + r, the frame is statically indeterminate.

Equations Unknowns

Equations Unknowns

th deg 18

Equations Unknowns

Draw N.F, S.F. and B.M.D (s) for the shown frame

1) ∑ X = 0 XA = 3t

2) ∑ MB = 0

10 YA + 3 × 2 – 8 × 7= 0

YA= 5t

3) ∑ Y = 0 YB = 3t

Solution of Example

3.1

A B

r = 3 J=4 S=0 m=3

3J+S=12 3m+r=12

Stable & Determinate

• ∑ X = 0 XA = 4t

∑ MB = 0

12 YA – 24 × 6 – 4 × 6= 0

YA= 14t

∑ Y = 0 YB = 10t

Solution of Example

3.2

10144

Solution of Example

3.3

1) ∑ X = 0 XB = 4t

2) ∑ MB = 0

16 YA + 4 × 6 – 32 × 8– 4 × 18 = 0

YA= 19t

3) ∑ Y = 0 YB = 17t

r = 3 J=4 S=0 m=4

3J+S=15 3m+r=15

Stable & Determinate

N.F.D

B.M.DS.F.D

1719

4

Solution of Example

3.4

1) ∑ X = 0 XB = 2.5t

2) ∑ MA = 0

8 YB + 2.5 × 2.5+ 3 × 2

–3×0.75–1×10×3 = 0 YB= 2.5t

3) ∑ Y = 0 YA +YB = 16t

YA= 13.5t

t 2.5

3m

10t

BA

t 2.5

m 3.125m 3.125

WL2/ 8 = 8

11.75 t.m

Solution of Example

3.51) ∑ X = 0 XA = 0

2) YA = YB=6t

A B

6 sin α =5.7

α

α

6 cos α =1. 9

sin α =0.948cos α =0.316

α

α6 cos α =1. 9

3m 3m

t/m 3

t 12

4m

t 12

t 12t 4

Solution of External

Example∑ X = 0

Xa=12 t

∑ Ma = 0

12 × 2+ 16 × 3 –6Yb=0

Yb=12 t ∑ Y = 0 Ya+Yb=16

Ya= 4t

36

4t 4t

12t

12t 12t

t 12t 43m 3m

t/m 3

t 12

4m

t 12

t 12t 4

Solution of External Example

∑ X = 0 Xa =5t

∑ Mb = 0

8Ya+ 15×2–4 × 8×4 –5 × 6 =0

Ya =16t

∑ Y = 0 YA+Yb=32+15= 47

Yb= 31t

4t/m5t

16t 31t

5t

Wwwwwww31t

15t16t

16t

16t

16t 16t

5t

32t5t

Solution of External Example

∑ Mb = 0 for part ab

2×1.5+4 ×4.5 –Ya × 6 =0 Ya =3.5t

∑ Me = 0 Ya 15 ×3+ 3.5 × 8 +3 × 4 = 2 ×3.5+4× 6.5 + 8Yd

Yd= 6.5t

∑ X = 0 Xe =3t∑ Y = 0

Ya+Ye + Yd=4+2+15

Ye= 11t

3t

4t

3t11t

6.5t3.5t

15t5m

412.531.5 4

4m

5.25 3.75

B.M.D

14.1

8

12

F g

F g

c

c

MF=3.5 ×1.5=5.25t.mMg=3.5 ×4.5-4 × 3=3.75t.mMc=3.5 ×8-4 × 6.5 -2 × 3.5-3 × 1=-8t.m