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Tutorial 4 MECH 101 Liang Tengfei Office phone : 2358-8811 Mobile : 6497-0191 Office hour : 14:00-15:00 Fri 1.

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Presentation on theme: "Tutorial 4 MECH 101 Liang Tengfei Office phone : 2358-8811 Mobile : 6497-0191 Office hour : 14:00-15:00 Fri 1."— Presentation transcript:

1 Tutorial 4 MECH 101 Liang Tengfei tfliang@ust.hk Office phone : 2358-8811 Mobile : 6497-0191 Office hour : 14:00-15:00 Fri 1

2 Outline Example about normal stress and normal strain Example about shear stress, shear strain, and Hooke’s law in shear

3 Example 1 The two bar truss ABC has pin supports at points A and C, which are 2.0m apart. Members AB and BC are steel bars, pin connected at joint B. The length of BC is 3m. A sign weighing 5.4kn is suspended from bar BC at D and E, which are located 0.8m and 0.4m, respectively, from the ends of the bar. Determine the required cross-sectional area of AB and the required diameter of pin at C if the allowable stresses in tension and shear are 125 MPa and 45 Mpa.

4 Draw Free Body Diagram

5 Return to free body diagram of the entire truss, Reaction force at A and C, Supposing counterclockwise moments are positive,

6 Tensile force in bar AB Shear force acting on the pin at C Required area of the bar Required diameter of the pin

7 practice

8 Hook’s law in shear Shear strain : change in the shape of the element

9 Example 2 A flexible connection consisting of rubber pads( t=9mm) bonded to steel plates is shown in the figure. The pads are 160mm long and 80mm wide. (a)Find the average shear strain γaver in the rubber if the force P=16kN and the shear modulus for the rubber is G=1250kPa. (B) Find the relative horizontal displacement δ between the interior plate and the outer plates.

10 Shear strain : In most practical situations the shear strain γ is a small angle, and in such cases tan γ can be replaced by γ.

11 Shear force V=P/2, Shear Strain: Horizontal displacement δ,

12 Shear stress and Bearing stress Shear stress : tangential to the surface Average shear stress: Average bearing stress: Where n V m q V p a L n m p

13 One shear surface Two shear surfaces n V m q V p L q V p L F F

14

15 Example 3 A steel strut S serving as a brace for a boat hoist transmits a compressive forces P=54KN to the deck of a pier. The strut has a hollow square cross section with wall thickness t=12mm, and angle θ is 40 °. A pin through the strut transmits the compressive force from the strut to two gussets G that welded to the base plate B. Four anchor bolts fasten the base plate to deck. The diameter of the pin is d pin =18mm, the thickness of the gussets is t G =15mm, the thickness of the base plate is t B =8mm, the diameter of the anchor bolts is dbolt=12mm. Determine the following stress: (a)the bearing stress between the strut and the pin;(b) the shear stress in the pin;(c) the bearing stress between the pin and the gussets,(d) the bearing stress between the anchor bolts and the base plate,(e) the shear stress in the anchor bolts.

16 Solution (a)the bearing stress between the strut and the pin = The force in the strut / the total bearing area of the strut (b) the shear stress in the pin( the pin tends to shear the two planes between the gussets and the strut) Bearing surface Shearing surface

17 (c) the bearing stress between the pin and the gussets The pin bear against the gussets at two locations, so the bearing area is twice the thickness of the gussets times the pin diameter; (d) the bearing stress between the anchor bolts and the base plate (The vertical component of force is transmitted to the pier by direct bearing between the base plate and the pier; the horizontal component, is transmitted through the anchor bolts.) Bearing force Bearing surface

18 (e) the shear stress in the anchor bolts Shearing force Shearing surface

19 Practice

20


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