Cylindrical storage tank

Piping systems

Some examples of process equipment in Chemical engineering

Spherical storage tank Round with flat bottom storage tank

These structures need for sound engineering design to prevent failure - We need to study the materials of construction and on the forces acting this structure

- In the case of deformable structure, the application of forces causes the structure to be deformed

- Most steel structures are undeformable, and we do not want the structure to fail while in service. We have to examine and study all the forces acting internally and externally to the structure

Structure in equilibrium, all forces and moments, external and internal must be in equilibrium

Equation of equilibrium,

∑F = 0 ∑M = 0In 3 dimensional axis,

∑Fx = 0 , ∑Fy = 0 , ∑Fz = 0

ReactionsSurface forces developed at the supports/points of contact between bodies

∑Mx = 0 , ∑My = 0 and ∑Mz = 0

For individual for forces and moment in an axis,

∑Fx = 0 ∑Fy = 0

∑Mo = 0

The free body diagram (gambar rajah badan bebas)

Wg

Wg

W1 W1

+ ∑ Fy = 0 ( all vertical forces, No horizontal forces

2W1 – Wg = 0

W1 = Wg/2 KN

Example 1

Wg

Example 2

W1

W2

30º

40º

∑ Fy = 0, + : Wg = - W1 sin30 - W2 Sin40

∑Fx =0, + : W2 Cos 40 - W1 Cos 30 = 0

∑M = 0, + : W2sin40 (X1) – W1Sin30 (X2) = 0

X1 X2

Example

Determine the resultant internal loadings acting on the section at point C of the beam shown in the diagram

Distributed loading at C is found by proportion,

mN1809

270

6 w

w

Magnitude of the resultant of the distributed load,

FBD

N540618021 F

which acts from C m2631

= F

Applying the equations of equilibrium we have

(Ans) mN 0108

02540 ;0

(Ans) 540

0540 ;0

(Ans) 0

0 ;0

C

CC

C

Cy

C

Cx

M

MM

V

VF

N

NF

CB

2 m 4 mVC

NC

540 N

MC

EXAMPLE

Determine the resultant internal loadings acting on the cross section at C of the machine shaft as shown in the figure. The shaft is supported by journal bearings at A and B, which only exert vertical forces on the shaft.

FBD

Vc0.25 m

0.025 mAy

Nc

Mc

From overall FBD, take moment about B,

- Ay (0.4)+120 (0.125) -225(0.1) = 0

Ay = -18.75 N

Taking equilibrium conditions on the FBD,

Fx = 0 + Nc = 0

Fy + = 0

-18.85-40-Vc = 0

Vc = -58.8 N

Mc = 0 +

Mc +40(0.025) + 18.75b(0.25) = 0

Mc = -5.69 N.m

Stress

Normal stress is intensity of force or force per unit area, acting normal to the surface. If Fx is acting normal to the surface area, then we can write σz = lim Fx / A A 0

Differentiate between compressive and tensile stress

Shear stressThe intensity of force acting tangentially to A is called “Shear stress” or

Units: N/m² , Pascal (Pa), ( 1 Pa = 1 N/m² )