Question

Part 1 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Let a=3.0 ft, b=7.0 ft, c=5.0 ft, d=4.0 ft, w = 13 kips/ft and P = 90 kips. Construct the shear-force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. w B с D E b с d For this loading, calculate the reaction forces Ay and Ey acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Ax = 0, it has been omitted from the free-body diagram.) W 1111111111 x Am B | С D E a b d

Answer 1

SOLUTION:

- free Body Diagram is. a okips 13 kipstst. c ¥ Sft Dugt 3ft Tft AY equilibrium, using equations of I Fx = 0 (-) tve) TAX=0 Efy=0 crtue Ay+ Ey-90+ (1347) = 0. Ay+ Ey - 90 +91=0. Ay+ Ey + 1 = 0 EME (stved C Eyx19) - (90x1S)+(91x6.5)=0 Ey= 39.92 kips

from ean co, Ay = = -1- Ey Ay= -1 - 39.92 Ay= -40.92 kips Nocu Shear force and Bending Moment Diagram, 4o.az qo a 3-5 F С. E 3 ft 1 et set uft. 39.92 D SF calculations, Going from right to left. Sign convention - म tve

SFER = 0 SFEIL - 39.92 SEOIRE -39.92 S FOILE 50-08. StolR= so.08 SFCIL = 50.08 SEEL=-alt so-08=-40.92 SFAIR= -40.92 SFAL= O. Now, Bra Calculations Br & = 0, tue BM D = 159.68 -> (39-92х4) Brc = =[6.30 39.92) xa] - [ qoxs Brica -90.72 BM B = 639.92х16)-(90x12) + Caix3.5) BT1B = -122.78 BHA=0.

4o.az al go 3.5 ६. a ust &lt Ift Sht 39.92 so-08 so.08 + xu 36F B ult. ft I Sht - 39.92 -4o.az 740.92 159.68 O A с 90.72 122.78 283.64

nlow, calculating distance of point of contra flexure. Cuzo) let Bo=X OC = BC-Bo=7-x. Applying concept of Similar triangles, so.08 40.92 x 286.44 . 4o. 92x= So. 08x. 286.44 = 0 al x= 3.1477 ft. FRAGE Total distance from A is, D = AB+ 3.1477 - 3+ 3.147) 10=6-1477 ft Therefore, Oc= 7-3.1677 106.3 3.85231 Br lo curo) = =(39.92x12.8523) (qox.8.8523) I Brlo 283.641

ANSWERS:

1. REACTIONS:

Ay = 40.92 Kips ( ACTING DOWNWARD)

Ey = 39.92 Kips (ACTING UPWARD)

2. SHEAR FORCES:

A) -40.92 Kips

B) -40.92 Kips

C) +50.08 Kips

D) +50.08 Kips

E) -39.92 Kips

F) -39.92 KIps

3. BENDING MOMENT AT POINT B: -122.78 Kip Ft

4. LOCATION (X) = 6.1477 Ft

5. MAX BENDING MOMENT AT V=0,

BM (AT V=0) = -283.64 Kip Ft

6. BENDING MOMENT

A) AT POINT C: -90.72 Kip Ft

B) AT POINT D: 159.68 Kip Ft