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
∑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
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 )
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