1. PROBLEM-1
Using graphical method, draw the shear and bending moment diagrams for the beam shown in the figure. Determine the absolute maximum bending stress.
P1 = 5 kN C A B
P2 = 8 kN D
2 m
6 m
5 cm
8 cm
FREE BODY DIAGRAM and
SUPPORT REACTIONS Dy Ay B
2 m
P1 = 5 kN
P2 = 8 kN C D A
MA = 0;
Dy(6) – P2(4) – P1(2) = 0
Dy = ( )/6 = 7 kN
Fy = 0;
Ay – P1 – P2 + Dy = 0
Ay = P1 + P2 – Dy = 6 kN

2. PROBLEM-1 SHEAR FORCE DIAGRAM SHEAR FORCE AREA: A1 = (6)(2) = 12 kN.mDy Ay B
2 m
P1 = 5 kN
P2 = 8 kN C D A
SHEAR FORCE DIAGRAM
SHEAR FORCE AREA:
A1 = (6)(2) = 12 kN.m
A2 = (1)(2) = 2 kN.m
A3 = (–7)(2) = –14 kN.m
V (kN) 6 A1 A2 A3 1 –7

3. PROBLEM-1 Absolute maximum bending stress: BENDING MOMENT DIAGRAMDy Ay B
2 m
P1 = 5 kN
P2 = 8 kN C D A
BENDING MOMENT DIAGRAM
5 cm
8 cm
M0 = 0
M2 = M0 + A1
= = 12 kN.m
M4 = M2 + A2
= = 14 kN.m
M (kN.m)
M6 = M4 + A3
= 14 – 14 = 0 14 12
Absolute maximum bending stress: D A B C

4. PROBLEM-2 If the rod bar has a diameter of 100 mm,
P1=10 kN
2 m
P2=15 kN B C A
If the rod bar has a diameter of 100 mm,
Determine the absolute maximum bending stress in the shaft.
FREE BODY DIAGRAM and
SUPPORT REACTIONS
P1=10 kN
P2=15 kN
2 m Ay Cy B C A
MA = 0;
P1(2) – P2(2) + Cy(4) = 0
Cy = (30 – 20)/4 = 2.5 kN
Fy = 0;
Ay – P1 – P2 + Cy = 0
Ay = 22.5 kN

5. PROBLEM-2 SHEAR FORCE DIAGRAM SHEAR FORCE AREA:
P1=10 kN
P2=15 kN
SHEAR FORCE DIAGRAM
2 m
2 m
2 m
Ay = 22.5 kN
Cy = 2.5 kN
SHEAR FORCE AREA:
V (kN)
12.5
A1 = (–10)(2) = –20 kN.m
A2 = (12.5)(2) = 25 kN.m
A3 = (–2.5)(2) = –5 kN.m
-2.5 A1 A2 A3
-10

6. PROBLEM-2 Absolute maximum bending stress: BENDING MOMENT DIAGRAM
P1=10 kN
P2=15 kN
BENDING MOMENT DIAGRAM
M0 = 0
2 m
2 m
2 m
M2 = M0 + A1
= 0 – 20 = –20 kN.m
Ay = 22.5 kN
Cy = 2.5 kN
M4 = M2 + A2
M (kN.m)
= – = 5 kN.m 5
M6 = M4 + A3
= 5 – 5 = 0
-20
Absolute maximum bending stress:

7. EXAMPLE-3 If the shaft has a diameter of 50 mm,
Determine the absolute maximum
Bending stress in the shaft.
Reaction forces:
MA = 0;
12 kN
8 kN A B Ay By
Fy = 0;
Ay = – By
Free-body diagram
Ay = kN

8. EXAMPLE-3 Bending moment diagram: Absolute maximum bending stress:
12 kN
8 kN A B
0.25 m
0.4 m
19.75 kN
0.25 kN
Bending moment diagram:
M0 = 0
M0.25 = –12x0.25 = –3 kN.m
M0.65= (–12x0.65)+(19.75x0.4)=0.1 kN.m M
–3 kN.m
0.1 kN.m x
M1.05 = 0
Absolute maximum bending stress:

9. PROBLEM-4
Shaft is supported by smooth journal bearings at A and B. Due to transmission of power to and from the shaft, the belts on the pulleys are subjected to the tensions shown. Determine the smallest diameter of the shaft using the maximum-shear-stress theory, with allow = 50 MPa.

10. PROBLEM-4
Support reactions are calculated and shown on the free-body diagram of the shaft.

11. EXAMPLE 11.6 (SOLN) MA2 = 0 MA1 = 0 MC2 = (150)(0.250) = 37.5 N.my z x
Bending-moment diagrams for Mx and Mz are shown above.
MA2 = 0
MA1 = 0
MC2 = (150)(0.250) = 37.5 N.m
MC1 = (475)(0.250) = N.m
MB2 = (150)(0.50) = 75 N.m

12. EXAMPLE 11.6 (SOLN)
Torque diagram is also shown.

13. EXAMPLE 11.6 (SOLN)
By inspection, critical pts for bending moment occur either at C or B. Also, just to the right of C and at B the torsional moment is 7.5 N·m. Moment at B is
At C, resultant moment is

14. EXAMPLE 11.6 (SOLN)
Since the design is based on the maximum-shear-stress theory, Eqn 11-2 applies.
The radical √(M2 + T2) will be the largest at section just to the right of C. We have
Thus the smallest allowable diameter is

15. PROBLEM-8 Determine the length of the length
of the center portion of the bar so
that the maximum bending stress
at section A, B, and C is the same.
The bar has a thickness of 10 mm. W
W/2 = 175L
Reaction forces:
Since the total weight is located
in the middle of beam, the support
reaction has the same value;
W/2 = (350L)/2 = 175L
Free-body diagram

16. PROBLEM-8 K = 1.45 Stress concentration factor:
Referring to the graph:
K = 1.45

17. PROBLEM-8 Reaction forces:W
W/2 = 175L VA MA
Reaction forces:
W/2
W/2 = 175L VC MC
Free-body diagram for analyzing the bending moment at A, B, and C:
Bending moment at section A:
Bending moment at section C:
MA = (175L)(0.3) = 52.5 L
MC = (175L)(0.3+L/2) – (175L)(L/4)
= 52.5 L L2
Bending moment at section B:
MB = MA = 52.5 L (symmetry)

18. PROBLEM-8 Maximum bending stress at sections A, B, and C
At either section A or B:
At section C:
Solving smax)A = smax)C will yield : L = m